All About technology

morgan stanly placement papers

2013-10-12T02:19:00.002-07:00

Hellooo I m a student from Sardar patel institute of technology[SPIT]

I appeared for Following company .

Morgan Stanley :

There was three option c/c++/java to give the apti and two more section i.e computer sciene and quantative.

The apti question were totally from OCJP exam.so if u thoroughly prepared the katy sierra book then you can easily cleared it.

While the cs and quant where basic question.

AMDOCS:

The aptitude consist of quant,reading comprehension ,c/c++/java,SQL,and unix.

The c question example where simple for eg:

How many file pointers are created in the following code:

FILE *fp0,*fp1,*fp2;

Fp0 = fopen(“f1”);

Fp1 = fopen(“f1”);

Fp2 = fopen(“f1”);

Something like this was asked

The SQL and unix question were tough though

After clearing the apti the next was the technical interview round:

· The first question he asked me was about near, far and huge pointers

· Then he asked me to write a code about how to find the depth of the tree

· What is synonyms in SQL

· What is inode

· He then asked me about my project

· Need of recursion and its type with example

· etc

HR round which was a simple one

Aptitude format( had 6 sections):

1st section: 10 objective ques.

Analytical, logical reasoning and quant.

2nd section: 10 objective ques.

ques on C language

3rd section: 10 objective ques.

Java (5ques) + OOPS concept(5ques)

4th section: 5 subjective ques.

5 DSF ques.

5th section: 5 objective ques.

On SQL

6th section: write 2 codes out of 4 .

10 cleared aptitude. One was me....

Tech Interview was all logical.

-Started with How r u Apurva??

-Tell abt urself.

-Wat r ur negative points??

- Y Quinstreet??

-They asked how ll u design a search engine.

-Questions on AJAX , page ranking.

-equals and hashcodes in JAVA.

-How will u detect a new web page has come and add it into database...???

He gave a hint:-What ll u do if ur friend's number is unreachable...how ll u contact him??

answer:- Trigger wen an event occurs , use callback() etc.............

-Which data structure ll u use for storing keys and links.

-How ll u store it in the memory.

And more ques n more................

-Last question: Do u hav any ques to ask??

HR Interview:

It was nice..very nice.

-What does degree/education mean to u?

-Are u placed?? Reply- no.

-Not any non dream company also?? Reply- yes.

-Then y r u here???

Sir coz blah blah blah.... He was not fully satisfied.

- Abt family??

-If u r an interviewer who do u think is the most deserving/potential candidate among other candidates.

-Few questions from resume.

-Any question to ask????????

Microsoft Placement paper 2012

2013-09-01T02:57:00.003-07:00

Microsoft Placement paper 2012 IIT Guwahati

Written Test : It consists of two exams conducted by some organization(not Microsoft).

Exam 1 : It consists of 20 questions mostly are from data structure, graph theory, operating system, computer architecture, output related questions. The time given was 45 minute.

Note : The exam was simple, you can easily able to pass it out. This is just an eliminatory for second exam.

Exam 2 : It consists of two programming question. Total time was 60 minutes.

Q 1 : A large passage was given to you, Write a program to print last 10 lines ?. If the number of lines is less than 10, print all.

Q 2 : Print the binary tree level wise ?. For example if input is :

B C

/ \

C D

E A

Output is :

B C

C D

E A

Interview

Round 1 :

Q 0 : Tell me about your self ?, Tell me your interest ?

Q 1 : When you visit on your friend’s Facebook profile, there is a mutual friend section where common friends are listed, now let’s assume that your friend do the same thing, he/she visit his/her friend other then you, now the people other than common are connected to you by distance of two. Similarly think you are given two people on Facebook, how do you find this connectivity?. (Please give appropriate solution),

Now let’s think that some important people are given some weight(any), now do the same thing ?

Now calculate the most influential person? (Not an easy question, because of weights) ?

Q 2 : longest palindrome in a string ? (Need to tell in O(n) time complexity + O(1) space complexity)

Q 3 : What do you think, how the posts in Facebook are shown, to your page, as there are thousands of posts, likes, videos, images, links etc. shared by your friends, but not all are shown to you ? (Data mining question, have to tell appropriate solution which can work ?)

Q 4 : You have given a linked list in which each node have three items, 1) data, 2) a next pointer and 3) a random pointer which is pointing to its successor in sorted order. Replicate it ? (Need to generate a new linked list in O(n) + O(n) complexity)

After some discussion I ask a question to the interviewers

Q : Sir, Why can’t window also have similar system interaction as it is in Linux, for example they have given terminal, we also have command prompt, but both have a lot of difference in terms of system interaction ?

Round 2 :

Q 0 : What types of questions were asked in round 1 ?

Q 1 : You are the supervisor of an airport. What happens is that visitors are not visit your airport, instead they go to another one, which means your airport become unpopular nowadays, Now as a supervisor you need to find out what has happens ?, What went wrong ?,How do you find out ?, What is correct ?, How do you find correct one and at what cost ?

Note : All type of questions were discussed here, Hints for this questions, you need to tell with the concepts of data mining, you need to tell how we get information’s and at what cost, then you need to analysis information’s.

Q 2 : You are given finitely many intervals in 1D, you have to design a data structure an efficient data structure which can answer queries of the form “In how many intervals the point P belong ?”, P is an input point, and all intervals are closed. I answer B tree(think why) which is most efficient.

Then he asked me how to decide how many children should be there at a node ?, Design most efficient B-tree based upon information you have ?(need to think more about how to get information’s ) ?

Q 3 : you are given some nodes, and for each node a probability is given which will tell its importance, you need to design an efficient data structure such that the expected search time as minimum as possible. (Hint : Use dynamic programming + binary tree).

My question to interviewer :

Q : Sir, In searching, pages have some ranking number, then are assigned some probability of their access, now as there are billions of pages and every day around millions of queries comes, how would the system do the job, what kind of data structure they use to handle the efficiency ?

Round 3 :

Q 1 : How are you ?

Q 2 : Do you want to go for MS or PHD ?

Q 3 : What type of branch is yours ?(actually my branch name is mathematics and computing, and after 3 year Microsoft allowed our branch to appear for placement, so they ask this question)

Q 4 : One question is from my project “Garbage cleaner Robot” ?

Q 5 : You are given a binary search tree, and a value(data item), you need to tell the left most right cousin in as minimum time and as minimum space ?(need to minimize actual time complexity, he need minimum order of complexity as well as number of node access should be minimum)

Q 6 : You are given a ternary tree (a tree with 3 children at max with left, middle, right pointer at each node), create a singly linked list from it without using extra space ?

walmart lab recent papers 2012

2012-11-02T00:01:00.001-07:00

1. int fun(char *array ,int n)
{
return (array[n]);
}

a.it wont work
b.segmentation fault
c.it will not work without size of array
d.it does not work but we can fix

2.given array some three call functions to that functions after completing all what are values in array
easy only but time taking

3.given a tree we can find whether given tree is binary search tree are not in

a.O(n)
b.O(logn)
c.O(n^2)
d.O(log(logn))

4.merge sort algoritham we can apply for

i)singly linked list
ii)doubly linked list
iii)array

a.only iii
b.only i and ii
c.only ii,iii
d.i,ii,and iii

5.in chess tournament there are 100 players then what number of matches to find the winner

a.98
b.99
c.97
d.95

6.f(x+f(x))=4f(x) then find f(21) ,f(1)=4

a.64
b.25
c.88
d.12

7.find x
x=(x&0a0a0a0a0a)>>1 | (x&05050505)<<1 ;
x=(x&0c0c0c0c0c)>>2 | (x&03030303)<<2 ;
x=(x&0f0f0f0f)>>4 | (x&f0f0f0f0)<<4 ;
x=x&ffff0000)>>16| (x&0000ffff)<<16 ;

above statements will

a.reverse bits of x
b.interchange bits of x
c.flip bits if x
d.none

8.n and p are greater than 1 , 5n is square of number and 75np is cube of a number then what is smallest value of n + p

a.14
b.20
c.30
d.18

9. consider two series
17,21,25,.................................417
16,21,26,.................................666

find the number of terms common in two series

a.19
b.79
c.78
d.20

10. a painter needs to make solution of 2 parts of paint and 1.5 parts of water . by mistake he made solution
of 6 liters contain half paint and half water so how much amount he should add to make in correct
proportion

a.one liter paint
b.one liter water
c.half liter paint and one liter water
d.half liter water and one liter paint

11.there are two stacks A and B ,from stack A you can push element to B or print to screen . from b only u can print if A has four element and B is empty then how many number of sequence we can find in print screen

a.14
a.12
c.13
d.24
e.none

12. for A(m,n)

if m=0 and n!=0 then A(m,n) =n+1;
if m!=0 and n=0 then A(m,n)=A(m-1,1)
if m!=0 and n!=0 then A(m,n)=A(m-1,A(m,n-1))

then the value of A(1,1) is

a.3
b.2
c.1
d.4

13.given a singly linked list missed some code and asked to find missed code if code is for reversing linked
list

14. given code for binary search tree and array ={1,2,3,10,4,5) asked find the output if
bsearch(array,0,5,10)

a.-1
b.compile error
c.4
d.none

15. in LRU algorithm for cache replacement best data structure to read the data

a.linked list
b.array
c.hash table
d.none

coding round:

1.write a code to find diameter of given binary tree

2.write a code to
given array a={1,2,3,4,5,6}
and n value we have to rotate the array by n times
ex if: n=2
output={5,61,2,3,4}

write code for this function
int [ ] rotate_array ( int [ ] input ,int n )

thanks for cecil and shudhan shu for giving this information

zynga recent paper 2012

2012-09-12T23:49:00.000-07:00

Zynga test pattern :2012
posted by :24x7techbytes

test contains 15 questions

14 multiple choice questions and one programming question

Q1.simple puzzle

Q2.some zings are zangs ,some zangs are zongs which of following conclusion is correct

a.some zings are zongs
b.some zongs are zings
c.some zongs are zings
d.none of the above

ans :c

Q3.find next letter in sequence J,H,G,F,D,S

a.I
b.C
b.T
d,A

ans :d

Q4.what is ouput for following program
int main()
{
enum xxx{a,b,c=217483640,d}
printf("%d %d",b,d);
return 0;
}

a.2 214748364
b.1 217483641
c.compile time error
d.1 -214748364

5.what is output for following program
void f(int a)
{
static int x=a;
if(x==a)
printf("eqaul");
else if(x>a)
printf("greaterthan");
else
printf("lessthan");
}
int main()
{
f(2);
f(1);
return 0;
}

a.equal equal
b.equal greaterthan
c.equal lessthan
d.greaterthan lessthan

ans :b

Q6.which of following is equal to !(!A&&!B)

a.!(B&&A)
b.A&&B
c.A||B
d.!(A||B)

ans :c

Q7.what is output of following code
#define f1(a,b) a+b
#define f2(c,d) c*d
int main()
{
int x=f1(12,18)/f2(6,4);
printf("%d",x);
}
a.9
b.24
c.15
d.1

ans :b

Q8.you have a piece of tape n cms long fold should result in a tape whose length is an integer number of cms for eg:
a 7cm tape can be folded at the 3 cm mark to prodece 4cm tape what is min no of folds required to reach 1 cm tape
a.log2(n)
b.log2(n)-1
c.log2(n-1)+1
d.log2(n+1)-1

Q9.what is output
conside char takes 2 bytes and short takes 2 bytes

struct str
{
int a;
int b;
};
void main()
{
str s;
s.a=10;
s.b=10;
short *sh=(short *)((void *)&s);
cout<<*sh<<endl;
}

a.20
b.10
c.1290
d.2570

Q10.starting from a white corner cell on a 8*8 chessboard if king makes 100 moves what is number of final position he can achieve (king can move one position in west ,south and nort and east)

a.1
b.4
c.32
d.64

Q11.a perfectly balanced binary tree contains 1 to 1023 and root contains value 1 and value increasing one towards leaf then what level does 768 present
a.1
b.2
c.10
d.7

ans :c

Q12.find output for following code

#define add(a,b) a+b
#define mul(c,d) c*d

int main()
{
int a=10,b=20,c=30;
d=mul(c,add(a,b));
printf("%d",d);
}
a.320
b.900
c.610
d.none

ans :c

13.what is output of following program

int main()
{
int *a=0;
a++;
return 0;
}
a.compile time error
b.segmentation fault
c.no error
d.none of the above

14.what is output of program
#define f(a,b) a##b
int main()
{
int a1=1,a2=2,a3=3,a4=4.a5=5;
for(i=1;i<=5;i++)
printf("%d ",f(a,i);
}
a. 1 1 1 1 1
b. 2 2 2 2 2
c. 1 2 3 4 5
d.compile time error

15.write a program to given a linked list and input k

10->20->30->40->50->60->70->80->90->100

if k=2
output: 20->10->40->30->60->50->80->70->100->90

if k=3
output:30->20->10->60->50->40->90->80->70->100

if k=4
output:40->30->20->10->80->70->60->50->100->90

microsoft interview pattern and some questions

2012-09-11T01:35:00.000-07:00

microsoft interview pattern and some questions from nit trichy 2012

Two fields IT and IDC

technical interview:

1.implementation of malloc
2.given a string replace spaces with %20
3.atoi implementaion
4.given a list,a part of list is reversed write a optimized program to sort
1 2 3 6 5 4 7 8 9
so 6 5 4 is reversed
5.dubly linked list to binary search tree

6.given filte is going to land but conditions are not good then what u will u do (this is not technical questions)

Direct i paper pattern in 2010 and 2011

2012-09-11T01:31:00.000-07:00

Latest Directi Placement Papers 2010
Package: 10Lakh
Paper Type: Mostly Technical but they allow to appear all branches Students(CSE,ECE,CE,Civil..)
Experience: Question is base on C/C++/Algo ,OS(little) ,Puzzles mixed Apti.

Questions: 30 questions are there and you have given 45min to solve.
All Questions are multiple type. I just mention the questions options not remember.
1. What is Zombie process(From OS)

2. In decimal 184 which is equal to 1234 in x-ary format. Value of x is:
i. 3
ii. 4
iii. 5
iv. None of the above
Ans : iv

3. Find the complexity of the Equation: F(n)=F(n/2)+logn
i. logn
ii. loglogn
iii. log(n)^2
iv. (logn)^2

Ans i

4. You have 25 horses and you have to select fastest 3. You can make a race of 3 horses in one time. So how many time you have to make a race?
i. 6,
ii. 7
iii. 8
iv. 9
Ans ii

5. 4. You have 9 horses and you have to select fastest 3. You can make a race of 3 horses in one time. So how many time you have to make a race?
i. 3,
ii. 5
iii. 6
iv. 7
Ans ii

6. An AVL tree with height d. How many children it have?

7. 2^x=x^2 How many real roots are there?

8. In a ternary Tree No of leaves 28. How many nodes it have?

9. In tank with two pipes, filled within 24hrs and 16hrs and another pipe which empties it within 12hrs. If 3 pipes are opened together and what the time it requires to fill an empty thrice volumes Tank.
Ans: 144hours

10. In n-ary tree, how many leaves are there?

11. one man first go 1mile in east,1/2 mile in north, 1/4 mile in west, 1/8 mile in south, 1/16 mile in east and so on.. Now distance between destination and origin(0,0)?

12. How many 3-digit numbers are there whose sum is 13?

13. C program:
int x=150
for(y=x;y>0;y=(x&(y-1))
printf(Hi, There");
How many times how will be printed?
i. 0
ii. 1
iii. 75
iv. 150
ans: 150(not sure)

14. Get the output of the C program
int i=5;
printf("%u",&i);
What it will print:
i. 5
ii. Base address of the memory
iii. Physical address
iv. Logical address

15. A B C D ....Z
AA AB AC ....AZ
BA BB BC ....BZ
..
..
AAA AAB ... AAZ
Find the 1000th term of the sequence?

Directi Coding Question In IIT Guwahati 2010::::

In an n*n chess board you are given a position of a Knight(K) and the target position it has to reach(T). There are certain cells marked as X which the knight should not touched.
Find the least number of steps required for Knight to reach T from K.

Input:

First line n.
Second line K in x space y format
Third line T in x space y format
Until you encounter -1 space -1, each line will have x space y format numbers indicating the locations of X.
(X and Y will be 1 based)

Output:
Integer value
(-1 if the knight cant reach T)

Example 1:

Input:
3
11
22
33
32
-1-1
output:
-1

Example 2:

Input:
4
11
23
-1-1
Output:
1

Directi Placement Paper 2011 NSIT
1. What is the big-O time complexity of func(p)?
C++ code follows.
int get_power(int a, int b)
{
if(!b) return 1;
if(b%2) return a * get_power(a, b/2);
return get_power(a, b/2);
}
int func(int p)
{
int sum = 0;
for(int i = 1; i <= p; ++i) {
sum += get_power(i, 5);
}
return sum;
}
a. O(p log5)
b. O(plogp)
c. O(logp)
d. O(p)
e. O(p + logp)

2.Given a number N, generate a sequence S such that
S[ 0 ] = N
S[ i+1 ] = (3 * S[ i ] + 1) / 2 if S[ i ] is odd,
= S[ i ] / 2, if S[ i ] is even,
till you reach 1.
For example for N = 5, the sequence is 5, 8, 4, 2, 1
Given two numbers 20 and 35, generate the sequence S for A
and B, and find the number where the two sequences meet.
a.20
b.30
c.35
d.40

3.Question 6: Multiple Answer
If at least two of the statements below
are false, what is the probability that
statement 2 is true?
1. A is true.
2. B is true.
3. A is false and B is true.
a.0
b.0.25
c.0.5
d.0.75
e.1

4.What is the value of p, in main()? C code follows.
char* rev(char s[ ])
{
for(int i = 0, n = strlen(s); s[ i ]; ++i)
{
char c = s[ i ];
s[ i ] = s[ n-1-i ], s[ i ] = c;
}
return s;
}
int main()
{
char s[ ] = "uncommon ideas!";
char *p = rev(s);
}
a. !saedi nommocnu
b. ideas! uncommon
c. uncommon ideas!
d. nommocnu !saedi

5.Given 4 processes with their entering
time and execution time as follows :
Process
Name
Entering
Time
Execution
Time
P1 2 10
P2 3 8
P3 6 4
P4 10 6
A CPU following the shortest job first
algorithm(preemptive) is going to execute
them.
A shortest job first algorithm will execute
a process with the smallest CPU burst
first, because it is preemptive, If a
process is in excution and another
process comes in with CPU burst being
smaller than the remaining execution
time of the currently executing process,
then the new process will be alloted the
CPU and the process currently executing
will be sent to the waiting queue.
For the given processes above Calculate
the average waiting time.
a. 7.25
b. 6.75
c. 5.25
d. 6.25
e. None of the above

6. Given below is the post-order traversal of
a binary search tree. Which of the
following is its pre-order traversal?
(6, 5, 12, 13, 11, 15, 14, 20, 16, 10)
a. 10, 5, 6, 16, 14, 11, 13, 12, 15, 20
b. 10, 5, 6, 15, 14, 13, 11, 12, 16, 20
c. 10, 5, 6, 14, 13, 11, 12, 16, 15, 20
d. 10, 5, 6, 16, 13, 11, 12, 14, 15, 20

7. Insert the numbers in the sequence "8, 5, 1, 4, 2, 10, 12, 7, 3" into a min-heap. The order of insertion is as given. The insertion happens as described here.
If the root is said to be at level 1, the root's children at level 2, and so on, what is the level in which 8 occurs in the heap?
· a. 1
· b. 2
· c. 3
· d. 4
· e. 5
8.
You are given a wall that is 50m wide and a rope that is 20m in length. Place the two ends of the rope on the wall and form a figure such that the enclosed area between the wall and the rope is maximized. What is this area?
· a. 63.63
· b. 127
· c. 50
· d. 127.26
·
9. Two players are playing a game on a checker board (see image below). There is a queen at (i,j) and it can only move such that either "i" or "j" or both decrease (by any amount), that is, it can move only left, or down, or diagonally left-down. The players take turns moving the queen, and the player who first reaches origin wins. Which (i,j) values will result in a win for the first player, among the following? Assume that both players play optimally.
You cannot leave the board.
Optimal Play: A player is said to make an "optimal move" if he has considered all his moves, and all the opponent's and his subsequent moves till one of them wins. An optimal player chooses his current move assuming that the opponent will force him into a losing position, if he can.
Example: If the queen is initially at (2,1), your possible moves are to (1,1), (0,1), (2,0) or (1,0). Every one of these moves will make the opponent make his next move to (0,0) and hence (2,1) is a losing position for the first player.
The following image shows a checker board with the cell (2,3) marked in red.

a. 4,5
· b. 5,3
· c. 2,3
· d. 3,4
· e. 3,5

goldman sachs paper pattern and questions 2012

2012-09-11T01:27:00.006-07:00

Rounds:

1. Aptitude 25 questions : really Tough one Time: 25 min

i. You have 4218 no of balls. put 1 balls at top layer 3 at 2nd top layer then 6 and then 10 ... Continue in this way. How many layers are possible.?

ii. A B and C have chance of failure of 20%, 30% and 40%. To activate the machine at least two should be active. What is the probability that machine will be active?

iii. One circle two radius are perpendicular. Their connecting end of length 7root2. What is the area of that circle?

iv. Two mixture A and B mix in ratio 5:1. the price of each is 2:5. The labor takes .25 for producing per Kg. The rate per kg is 16.50. What is the price of B?

v. Two pipes P and Q fills a tank in 30 and 60 min. a person starts that at 5am to fill. he comes when it may fill but he saw that pipe R was also opened which empties that tank. Then he closes R. Now 20 min takes to fill the tank. What time R takes to empty the tank?

vi. A B and C are going for a party. A have 5 breads and B have 3 breads. But they ate same amount of bread. Then what amount C will pay to B?

vii. In a clock between 11 and 12, how many time two sticks will meet in integer no places?

2. C and C++ : 10 question Time: 10 min only

i. One b+ tree of order 3. insert 10 8 6 4 3 2 1 will require how many time to break the leaves maximum?

ii. Show the output:
mul(int b)
{
return(b*2);
}
main()
{
a=4;
a=mul(a=mul(a=mul(4)));
printf("\n %d",a)'
}

iii. one question from Normalization from database..

iv. one question from inheritance in C++. to show output. Parent child relations like that.

v. One task to complete take 50ms by single processor. But in pipeline processor will take 10 ms. What is the speed up for the processor when doing 240 ms sized tasks?

vi. Show the output:
main()
{
*str="C:\tc\bin\random"
prtintf("\n %s",str);
}

3. Algorithm writing Time: 25min

i. WAP to find out the day where you have given a date day/month/year format. You consider the date 1/1/2001 as Saturday as base.

ii. WAP to reverse words of a sentence. Like "August first is the Sunday" Will be shown as "Sunday the is first August"

4. Writing Essay Time: 5min

Topic: If you meet with CEO of goldman Sachs in an Elevator what what conversation will make good impression on You. 1 min conversation only

One thing must to say... Time is very less and don't waste time on only one problem others may be easy....

cisco written questions 2012

2012-09-09T23:26:00.001-07:00

1.extern keyword
2.working of extern keyword
3.linked list reverse and string reverse
4.some questions on subnetting
5.definition of cidr(class less interdomain routing)
6.what will happen when start bootup of computer in both diskless and diskfull computer
7.what will internally happen when type www.google.com
8.explain CIDR and ARP
9.in operating system zombie process definition and how to find zombie process
10.what File pointer what is declaration
11.difference between unix file system and windows file system
12.belady anamoly and thrashing and segmentaion and paging
13.process states(like ready and suspended and wait .....)
14.algorithms for routing(bellmanford and ford fulkerson)
15.multi tasking and multithreading ang multi process definitions and differences
16.can we use extern static at same time for a variable
17.diff between and static and extern

adobe recent paper nit warangal and nit trichy 2012

2012-09-09T23:21:00.000-07:00

1.number of multiplication neded for 5^12

int computeXn(int x int n)
{
    if(n%2=0)
   {
   return x*x;
   }
   else if(n%2=0)
   {
   int y computeXn(x n/2);
   return y*y;
   }
   else if(n%2=1)
   {
    int y computeXn(x n/2);
    return y*y*x;
    }
   }

2.finding heigth of tree

3.fork 2 times ask to find number child proecss

4.given a number we need to find what is the number formed by last k digits

5.some questions on subnetting

6.Diff between typedef and #define?

7.Polynomial A+Bx+Cx^2+....+Nx^(n-1) this representation is more suitable for which data structure. Then P and Q are two such polynomial and how to add that two using that data structure. WAP for that.

8. Specification of variables in one language: letter follow by letter or digit.
options:
1. (LUD)*
2. L.(LUD)* => this one right.
3. L.(L.D)+
4. L.(L.D)*

9. How Generic Swap of two elements can be implemented? that supports all type line int char float double etc..

10. Two rectangle is given with following data structure
rectangle {
int left_X;
int Left_X;
int right_X;
int Right_Y;
}
Two are in X-axis wise. How to find that they are intersected or not? WAP for that...

11. Free( P) How it decide how much space should be freed in c?

12. Head(aann)=a Tail(aann)=ann cont(aa,bb)=aabb;
For which of the following : cont(head(s),head(tail(tail(s)))='ac'

i. aaac
ii. abbc
iii. accc
iv. abdc

13.getbis function gives n bits from the position p of an binary no A.

14.You have to sort large data. But your memory does not have so much space. how you can sort that.

C Round:(10 question)

1. Diff between typedef and #define?

2. getbis function gives n bits from the position p of an binary no A.

3. You have to sort large data. But your memory does not have so much space. how you can sort that.

4. a[2][3][4] pointer representation

5. You have two threads T1 and T2 they are reader and writer respectively.
With some specification:
ADDNEW.Process
PROCESS.SET
PROCESS.RESET
ENTER CS
EXIT CS
LOOP
EXIT LOOP
WAIT# PROCESS

6. sprintf() function used how and what means?

7.
An array given Arr[] which is in decreasing order. How many swapping required in
for(int index=0;indexa[j+1])
{
swap(a[j],a[j+1]);
}
}
}

8. Finding Output:
int arr[]={10,20,30,40}
int varible_ptr=arr[0];
for(int index=0;index<4;index++) { printf(" arr[%d] = %d", index, *(varible_ptr+index)); varible_ptr+=sizeof(int); } Ans: output: arr[0]=10 arr[1]=30...

1. Given the binary representation of an integer in the form of a string of
zeros and ones, write a function to return the string representation of its
2’s complement. Use only one pass of the string.

2. An output based question which was very error prone and time consuming.

3. Given that file1.c contains a string definition as:
Char st[100];
Whereas file2.c contains following piece of code:
extern char * st;
func foo ()
{
st[0]++;
}
If file1 is now included in file2.c then what problems above function may
have and in which cases?

4. Define a doubly linked list. Write a function to reverse a doubly linked
list.

5. Write a function to return the maximum of three numbers using only
ternary operator (?:)

6. Given a 3D array as shown below:
int arr[3][2][4] = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21};
fine the value of the following:

**(*arr+1)

7. Give a one line description of the following explaining the difference
between the two:
Void (*ptr)(int * arr[], int n)
Void * (*ptr)(int * arr, int n)

8. Give Output:
int * x, y;
x = (int *)60;
y = (int *)20;
printf(“%d\n”,x-y);

9. #define ADD(X,Y) (X*Y)
#define MULTIPLICATION(X,Y) (X+Y)
Some expression was given, you had to give the output.

10. Write a recursive function to find the GCD of two numbers.
(Hint: use repeated division )

Engineering Paper

1. Given BAC7 in base 14. Convert it to base 7.
2. A tree traversal in preorder and inorder was given. You had to construct
the tree.
3. A prefix expression was given, you have to evaluate it.
4. A long Assembly code was given, you have to tell what it does.
5. A hash function is given:
H(x) = ((H(x)mod1000)div100 * 10 + H(x)mod10)

Some hash values were given you had to tell which one of them map to the
same hash keys.
6. A Boolean expression was given, you had to solve it.
7. Given to singly linked lists A and B, find whether linked list B points
to the middle of A or not? Use only one pass of linked list A and B.
(If linked list A contains 20 nodes then middle node = 11 and for 21 nodes
middle = 11)

8. Two unsorted arrays are given A[n] and B[n+1]. Array A contains n
integers and B contains n+1 integers of which n are same as in array B but
in different order and one extra element x. Write an optimized algorithm to
find the value of element x. Use only one pass of both arrays A and B.

9. Match the following that can be used to implement the values given on the
left column:

a) Dynamic Memory management i) Tree
b) Printer Spooling ii) Linked list
c) File system iii) stack
d) Function call iv) queue

implementation of hash table in c

2012-09-08T11:21:00.001-07:00

#include<stdio.h>
#include<malloc.h>
#include<conio.h>
#define INTCAPACITY 10

struct hashtable
{
    int size;
    int nelements;
    struct listnode **array;
};
struct listnode
{
    int data;
    struct node *next;
};
//creating hashtable

struct hashtable * createhashtbl()
{
    int i,position;
    struct hashtable *htbl;
    struct listnode *tmp;
    htbl=malloc(sizeof(struct hashtable));
    htbl->nelements=0;
    htbl->size=INTCAPACITY;
    htbl->array=malloc(INTCAPACITY*sizeof(struct listnode*));
    for(i=0;i<INTCAPACITY;++i)
    {
        tmp=malloc(sizeof(struct listnode));
        tmp->next=NULL;
        htbl->array[i]=tmp;
        }
        printf("\n hash table created successfully");
        return htbl;
}
//hash function u can use other hashfuntions
int hash(int key,int k)
{
    return key%k;
}
//for each index in hash table im maintaing linked list with sorted order
int listinsert(struct listnode *head,int i)
{
    struct listnode *temp,*q,*t;
    if(head->next==NULL)
    {
    temp=(struct listnode *)malloc(sizeof(struct listnode));
    temp->data=i;
    temp->next=NULL;
    head->next=temp;

    }

    else
    {
        temp=(struct listnode *)malloc(sizeof(struct listnode));
        temp->data=i;
        temp->next=NULL;
        t=head;
        q=head->next;
        while(i>q->data&&q->next!=NULL)
        {
        t=q;
        q=q->next;
         }
         if(i<q->data)
         {
         t->next=temp;
         temp->next=q;
        }
        else if(q->next==NULL)
        {
            q->next=temp;
            }
    }
    return 1;
}

//insert in to hashtable

int add(struct hashtable *htbl,int key)
{
    int lf=1,position;

    if(htbl->nelements/htbl->size>lf)
    {
        rehash(htbl);
        }


    position=hash(key,htbl->size);

    if(listinsert(htbl->array[position],key))
    {
        htbl->nelements++;
        return 1;
        }

        else
        return 0;
}

//rehash when number of elements pointed to index crosses loadfactor we will do rehash
int rehash(struct hashtable *htbl)
{
    int newsize,i,position;
    struct listnode **newarray;
    struct listnode *tmp,*current;
    newsize=2*htbl->size;
    newarray=malloc(newsize*sizeof(struct listnode *));
    printf("entered to rehash");
    for(i=0;i<newsize;i++)
    {
    tmp=malloc(sizeof(struct listnode));
            tmp->next=NULL;
            newarray[i]=tmp;
            }
    for(i=0;i<htbl->size;i++)
    {
        for(current=htbl->array[i];current->next!=NULL;)
        {
            tmp=current->next;
            position=hash(tmp->data,newsize);
            listinsert(newarray[position],tmp);
            current->next=tmp->next;
            }
        }
        htbl->size=newsize;
        htbl->array=newarray;
return 1;
     }
//find in likned list this will be used when searching a key in table
int findlink(struct listnode *root,int j)
{
    struct listnode *temp=root->next;
    if(temp==NULL) return 0;
    while(temp->data<j&&temp->next!=NULL)
    {
        temp=temp->next;
        }
    if(temp->data==j)
    return 1;
    else return 0;
}

int find(struct hashtable *htbl,int j)
{
    int k;
    k=hash(j,htbl->size);
    return findlink(htbl->array[k],j);
}
//remove node
int removenode(struct listnode *root,int j)
{
    struct listnode *temp1,*temp2;
    if(root->next==NULL) return 0;
    temp1=root->next;
    temp2=root;
    while(temp1->next!=NULL&&temp1->data!=j)
    {
        temp2=temp1;
        temp1=temp1->next;
        }
    if(temp1->data==j)
    {
    temp2->next=temp1->next;
    free(temp1);
    return 1;
     }
     else if(temp1->next==NULL)
     return 0;
}
//remove1 function
void remove1(struct hashtable *htbl,int j)
{
    int k,l;
    k=hash(j,htbl->size);
    l=removenode(htbl->array[k],j);
    if(l==1)
    {
        htbl->nelements--;
        printf("\n deleted successfully");
    }
    else printf("\n deletion not successfull given data not present in table");
}

void main()
{
    struct hashtable *htbl;
    int i,j;
    while(1)
    {
        printf("\n ");
    printf("\n enter 1 to initialize hashtable");
    printf("\n enter 2 to add elements to hashtable");
    printf("\n enter 3 to find whether elemnt is exist or not");
    printf("\n enter 4 to delete elment");
    scanf("%d",&i);
    switch(i)
    {

    case 1:
            htbl=createhashtbl();
            break;

    case 2:
        printf("\n enter value");
        scanf("%d",&j);
        i=add(htbl,j);
        if(i==1)
        printf("\n entered successfully");
        else printf("\n not success");
        break;

    case 3:
        printf("\n enter element to find");
        scanf("%d",&j);
        j=find(htbl,j);
        if(j) printf("\n element present in hashtable");
        else printf("\n element not present in hashtable");
        break;

    case 4:
        printf("\n enter elemnt to delete");
        scanf("%d",&j);
        remove1(htbl,j);
        break;
        }
    }
}

comment here if this is usefull for anyone

oracle database questions

2012-09-08T09:54:00.001-07:00

Oracle DBA Questions

SNAPSHOT is used for [DBA]
a. Synonym, b. Table space, c System server, d Dynamic data replication Ans : D
We can create SNAPSHOTLOG for[DBA]
a. Simple snapshots, b. Complex snapshots, c. Both A & B, d Neither A nor B Ans : A
Transactions per rollback segment is derived from[DBA]
a. Db_Block_Buffers, b. Processes, c. Shared_Pool_Size, d. None of the above
Ans : B
ENQUEUE resources parameter information is derived from[DBA]
a. Processes or DDL_LOCKS and DML_LOCKS, b. LOG_BUFFER, c. DB__BLOCK_SIZE..
Ans : A
LGWR process writes information into
a Database files, b Control files, c Redolog files, d All the above. Ans : C
SET TRANSACTION USE ROLLBACK SEGMENT is used to create user objects in a particular Tablespace
a True, b False Ans : False
Databases overall structure is maintained in a file called
a Redolog file, b Data file, c Control file, d All of the above.
Ans : C
These following parameters are optional in init.ora parameter file DB_BLOCK_SIZE, PROCESSES
a True, b False Ans : False
Constraints cannot be exported through EXPORT command
a True, b False Ans : False
It is very difficult to grant and manage common privileges needed by different groups of database users using the roles
a True, b False Ans : False
What is difference between a DIALOG WINDOW and a DOCUMENT WINDOW regarding moving the window with respect to the application window

a Both windows behave the same way as far as moving the window is concerned.
b A document window can be moved outside the application window while a dialog window cannot be moved
c A dialog window can be moved outside the application window while a document window cannot be moved
Ans : C
What is the difference between a MESSAGEBOX and an ALERT

A messagebox can be used only by the system and cannot be used in user application while an alert can be used in user application also.

A alert can be used only by the system and cannot be use din user application while an messagebox can be used in user application also.

An alert requires an response from the userwhile a messagebox just flashes a message
and only requires an acknowledment from the user

An message box requires an response from the userwhile a alert just flashes a message an only
requires an acknowledment from the user Ans : C

Which of the following is not an reason for the fact that most of the processing is done at the server ?
a To reduce network traffic. b For application sharing, c To implement business rules centrally,
d None of the above
Ans : D
Can a DIALOG WINDOW have scroll bar attached to it ?
a Yes, b No Ans : B
Which of the following is not an advantage of GUI systems ?
a. Intuitive and easy to use., b. GUI's can display multiple applications in multiple windows
c. GUI's provide more user interface objects for a developer d. None of the above
Ans :D
What is the difference between a LIST BOX and a COMBO BOX ?

a In the list box, the user is restricted to selecting a value from a list but in a combo box the user can type in value which is not in the list

b A list box is a data entry area while a combo box can be used only for control purposes

c In a combo box, the user is restricted to selecting a value from a list but in a list box the
user can type in a value which is not in the list

d None of the above
Ans : A
In a CLIENT/SERVER environment , which of the following would not be done at the client ?
a User interface part, b Data validation at entry line, c Responding to user events,
d None of the above Ans : D
Why is it better to use an INTEGRITY CONSTRAINT to validate data in a table than to use a STORED
PROCEDURE ?
a Because an integrity constraint is automatically checked while data is inserted into or updated in a table while a stored procedure has to be specifically invoked
b Because the stored procedure occupies more space in the database than a integrity constraint definition
c Because a stored procedure creates more network traffic than a integrity constraint definition
Ans : A
Which of the following is not an advantage of a client/server model ?
a. A client/server model allows centralised control of data and centralised implementation of business rules.
b A client/server model increases developer;s productivity
c A client/server model is suitable for all applications
d None of the above.
Ans : C
What does DLL stands for ?
a Dynamic Language Library
b Dynamic Link Library
c Dynamic Load Library
d None of the above
Ans : B
POST-BLOCK trigger is a
a Navigational trigger
b Key trigger
c Transactional trigger
d None of the above
Ans : A
The system variable that records the select statement that SQL * FORMS most recently used
to populate a block is
a SYSTEM.LAST_RECORD
b SYSTEM.CURSOR_RECORD
c SYSTEM.CURSOR_FIELD
d SYSTEM.LAST_QUERY
Ans: D
Which of the following is TRUE for the ENFORCE KEY field
a ENFORCE KEY field characterstic indicates the source of the value that SQL*FORMS uses to populate the field
b A field with the ENFORCE KEY characterstic should have the INPUT ALLOWED charaterstic turned off
a Only 1 is TRUE
b Only 2 is TRUE
c Both 1 and 2 are TRUE
d Both 1 and 2 are FALSE
Ans : A
What is the maximum size of the page ?
a Characters wide & 265 characters length
b Characters wide & 265 characters length
c Characters wide & 80 characters length
d None of the above
Ans : B
A FORM is madeup of which of the following objects
a block, fields only,
b blocks, fields, pages only,
c blocks, fields, pages, triggers and form level procedures,
d Only blocks.
Ans : C
For the following statements which is true
   1 Page is an object owned by a form
   2 Pages are a collection of display information such as constant text  and graphics.
     a Only 1 is TRUE
     b Only 2 is TRUE
     c Both 1 & 2 are TRUE
     d Both are FALSE
      Ans : B
The packaged procedure that makes data in form permanent in the Database is
a Post
b Post form
c Commit form
d None of the above
Ans : C
Which of the following is TRUE for the SYSTEM VARIABLE $$date$$
a Can be assigned to a global variable
b Can be assigned to any field only during design time
c Can be assigned to any variable or field during run time
d None of the above
Ans : B
Which of the following packaged procedure is UNRESTRICTED ?
a CALL_INPUT, b CLEAR_BLOCK, c EXECUTE_QUERY, d USER_EXIT
Ans : D
Identify the RESTRICTED packaged procedure from the following
a USER_EXIT, b MESSAGE, c BREAK, d EXIT_FORM
Ans : D
What is SQL*FORMS
a SQL*FORMS is a 4GL tool for developing & executing Oracle based interactive applications.
b SQL*FORMS is a 3GL tool for connecting to the Database.
c SQL*FORMS is a reporting tool
d None of the above.
Ans : A
Name the two files that are created when you generate a form using Forms 3.0
a FMB & FMX, b FMR & FDX, c INP & FRM, d None of the above
Ans : C
What is a trigger
a A piece of logic written in PL/SQL
b Executed at the arrival of a SQL*FORMS event
c Both A & B
d None of the above
Ans : C
Which of the folowing is TRUE for a ERASE packaged procedure
   1 ERASE removes an indicated Global variable & releases the memory associated with it
   2 ERASE is used to remove a field from a page
   1 Only 1 is TRUE
   2 Only 2 is TRUE
   3 Both 1 & 2 are TRUE
   4 Both 1 & 2 are FALSE
   Ans : 1
All datafiles related to a Tablespace are removed when the Tablespace is dropped
a TRUE
b FALSE
Ans : B
Size of Tablespace can be increased by
    a Increasing the size of one of the Datafiles
    b Adding one or more Datafiles
    c Cannot be increased
    d None of the above
    Ans : B
Multiple Tablespaces can share a single datafile
   a TRUE
   b FALSE
   Ans : B
A set of Dictionary tables are created
a Once for the Entire Database
b Every time a user is created
c Every time a Tablespace is created
d None of the above
Ans : A
Datadictionary can span across multiple Tablespaces
a TRUE
b FALSE
Ans : B
What is a DATABLOCK
a Set of Extents
b Set of Segments
c Smallest Database storage unit
d None of the above
Ans : C
Can an Integrity Constraint be enforced on a table if some existing table data does not satisfy the constraint
a Yes
b No
Ans : B
A column defined as PRIMARY KEY can have NULL's
a TRUE
b FALSE
Ans : B
A Transaction ends
a Only when it is Committed
b Only when it is Rolledback
c When it is Committed or Rolledback
d None of the above
Ans : C
A Database Procedure is stored in the Database
a In compiled form
b As source code
c Both A & B
d Not stored
Ans : C
A database trigger doesnot apply to data loaded before the definition of the trigger
a TRUE
b FALSE
Ans : A
Dedicated server configuration is
a One server process - Many user processes
b Many server processes - One user process
c One server process - One user process
d Many server processes - Many user processes
Ans : C
Which of the following does not affect the size of the SGA
a Database buffer
b Redolog buffer
c Stored procedure
d Shared pool
Ans : C
What does a COMMIT statement do to a CURSOR
a Open the Cursor
b Fetch the Cursor
c Close the Cursor
d None of the above
Ans : D
Which of the following is TRUE
1 Host variables are declared anywhere in the program
2 Host variables are declared in the DECLARE section
   a Only 1 is TRUE
   b Only 2 is TRUE
   c Both 1 & 2are TRUE
   d Both are FALSE
   Ans : B
Which of the following is NOT VALID is PL/SQL
   a Bool boolean;
   b NUM1, NUM2 number;
   c deptname dept.dname%type;
   d date1 date := sysdate
   Ans : B
Declare
fvar number := null; svar number := 5
Begin
goto << fproc>>
if fvar is null then
   << fproc>>
   svar := svar + 5
   end if;
   End;
What will be the value of svar after the execution ?
a Error
b 10
c 5
d None of the above
Ans : A
Which of the following is not correct about an Exception ?
a Raised automatically / Explicitly in response to an ORACLE_ERROR
b An exception will be raised when an error occurs in that block
c Process terminates after completion of error sequence.
d A Procedure or Sequence of statements may be processed.
Ans : C
Which of the following is not correct about User_Defined Exceptions ?
a Must be declared
b Must be raised explicitly
c   Raised automatically in response to an Oracle error
d   None of the above
Ans : C
A Stored Procedure is a
a Sequence of SQL or PL/SQL statements to perform specific function
b Stored in compiled form in the database
c Can be called from all client environmets
d All of the above
Ans : D
Which of the following statement is false
a Any procedure can raise an error and return an user message and error number
b Error number ranging from 20000 to 20999 are reserved for user defined messages
c Oracle checks Uniqueness of User defined errors
d Raise_Application_error is used for raising an user defined error.
Ans : C
Is it possible to open a cursor which is in a Package in another procedure ?
a Yes
b No
Ans : A
Is it possible to use Transactional control statements in Database Triggers?
a Yes
b No
Ans : B
Is it possible to Enable or Disable a Database trigger ?
a Yes
b No
Ans : A
PL/SQL supports datatype(s)
a Scalar datatype
b Composite datatype
c All of the above
d None of the above
Ans C
Find the ODD datatype out
a VARCHAR2
b RECORD
c BOOLEAN
d RAW
Ans : B
Which of the following is not correct about the "TABLE" datatype ?
a Can contain any no of columns
b Simulates a One-dimensional array of unlimited size
c Column datatype of any Scalar type
d None of the above
Ans : A
Find the ODD one out of the following
a OPEN
b CLOSE
c INSERT
d FETCH
Ans C
Which of the following is not correct about Cursor ?
a Cursor is a named Private SQL area
b Cursor holds temporary results
c Cursor is used for retrieving multiple rows
d SQL uses implicit Cursors to retrieve rows
Ans : B
Which of the following is NOT VALID in PL/SQL ?
a Select ... into
b Update
c Create
d Delete
Ans : C
What is the Result of the following 'VIK'||NULL||'RAM' ?
a Error
b VIK RAM
c VIKRAM
d NULL
Ans : C
Declare
a number := 5; b number := null; c number := 10;
Begin
if a > b AND a < c then
a := c * a;
end if;
End;
What will be the value of 'a' after execution ?
a 50
b NULL
c 5
d None of the above
Ans : C
Does the Database trigger will fire when the table is TRUNCATED ?
a Yes
b No
Ans : B
SUBSTR(SQUARE ANS ALWAYS WORK HARD,14,6) will return
a ALWAY
b S ALWA
c ALWAYS
Ans : C
REPLACE('JACK AND JUE','J','BL') will return
a JACK AND BLUE
b BLACK AND JACK
c BLACK AND BLUE
d None of the above
Ans : C
TRANSLATE('333SQD234','0123456789ABCDPQRST','0123456789') will return
a 333234
b 333333
c 234333
d None of the above
Ans : A
EMPNO ENAME SAL
A822 RAMASWAMY 3500
A812 NARAYAN 5000
A973 UMESH 2850
A500 BALAJI 5750
Use these data for the following Questions
Select SAL from EMP E1 where 3 > ( Select count(*) from Emp E2 where E1.SAL > E2.SAL ) will retrieve
a 3500,5000,2500
b 5000,2850
c 2850,5750
d 5000,5750
Ans : A
Is it possible to modify a Datatype of a column when column contains data ?
a Yes
b No
Ans B
Which of the following is not correct about a View ?
a To protect some of the columns of a table from other users
b Ocuupies data storage space
c To hide complexity of a query
d To hide complexity of a calculations
Ans : B
Which is not part of the Data Definiton Language ?
a CREATE
b ALTER
c ALTER SESSION
Ans : C
The Data Manipulation Language statements are
a  INSERT
b  UPDATE
c  SELECT
d  All of the above
Ans : D
EMPNO ENAME SAL
A822 RAMASWAMY 3500
A812 NARAYAN 5000
A973 UMESH
A500 BALAJI 5750
Using the above data
Select count(sal) from Emp will retrieve
a 1
b 0
c 3
d None of the above
Ans : C
If an UNIQUE KEY constraint on DATE column is created, will it accept the rows that are inserted with
SYSDATE ?
a Will
b Won't
Ans : B
What are the different events in Triggers ?
a Define, Create
b Drop, Comment
c Insert, Update, Delete
d All of the above
Ans : C
What built-in subprogram is used to manipulate images in image items ?
a Zoom_out
b Zoom_in'
c Image_zoom
d Zoom_image
Ans : C
Can we pass RECORD GROUP between FORMS ?
a Yes
b No
Ans : A
SHOW_ALERT function returns
a Boolean
b Number
c Character
d None of the above
Ans : B
What SYSTEM VARIABLE is used to refer DATABASE TIME ?
a $$dbtime$$
b $$time$$
c $$datetime$$
d None of the above
Ans : A
SYSTEM.EFFECTIVE.DATE varaible is
a Read only
b Read & Write
c Write only
d None of the above
Ans : C
How can you CALL Reports from Forms4.0 ?
a Run_Report built_in
b Call_Report built_in
c Run_Product built_in
d Call_Product built_in
Ans : C
When do you get a .PLL extension ?
a Save Library file
b Generate Library file
c Run Library file
d None of the above
Ans : A
What is built_in Subprogram ?
a Stored procedure & Function
b Collection of Subprogram
c Collection of Packages
d None of the above
Ans : D
GET_BLOCK property is a
a Restricted procedure
b Unrestricted procedure
c Library function
d None of the above
Ans : D
A CONTROL BLOCK can sometimes refer to a BASETABLE ?
a TRUE
b FALSE
Ans : B
What do you mean by CHECK BOX ?
a Two state control
b One state control
c Three state control
d none of the above
Ans : C - Please check the Correcness of this Answer ( The correct answeris 2 )
List of Values (LOV) supports
a Single column
b Multi column
c Single or Multi column
d None of the above
Ans : C
What is Library in Forms 4.0 ?
a Collection of External field
b Collection of built_in packages
c Collection of PL/SQl functions, procedures and packages
d Collection of PL/SQL procedures & triggers
Ans : C
Can we use a RESTRICTED packaged procedure in WHEN_TEXT_ITEM trigger ?
a Yes
b No
Ans : B
Can we use GO_BLOCK package in a PRE_TEXT_ITEM trigger ?
a Yes
b No
Ans : B
What type of file is used for porting Forms 4.5 applications to various platforms ?
a . FMB file
b . FMX file
c . FMT file
d . EXE file
Ans : C
What built_in procedure is used to get IMAGES in Forms 4.5 ?
a READ_IMAGE_FILE
b GET_IMAGE_FILE
c READ_FILE
d GET_FILE
Ans A
When a form is invoked with CALL_FORM does Oracle forms issues SAVEPOINT ?
a Yes
b No
Ans : A
Can we attach the same LOV to different fields in Design time ?
a Yes
b No
Ans : A
How do you pass values from one form to another form ?
a LOV
b Parameters
c Local variables
d None of the above
Ans : B
Can you copy the PROGRAM UNIT into an Object group ?
a Yes
b No
Ans : B
100. Can MULTIPLE DOCUMENT INTERFACE (MDI) be used in Forms 4.5 ?
a Yes
b No
Ans : A
Can MULTIPLE DOCUMENT INTERFACE (MDI) be used in Forms 4.5 ?
a Yes
b No
Ans : A
When is a .FMB file extension is created in Forms 4.5 ?
a Generating form
b Executing form
c Save form
d Run form
Ans : C
What is a Built_in subprogram ?
a Library
b Stored procedure & Function
c Collection of Subprograms
d None of the above
Ans : D
What is a RADIO GROUP ?
a Mutually exclusive
b Select more than one column
c Above all TRUE
d Above all FALSE
Ans : A
Identify the Odd one of the following statements ?
a Poplist
b Tlist
c List of values
d Combo box
Ans : C
What is an ALERT ?
a Modeless window
b Modal window
c Both are TRUE
d None of the above
Ans : B
Can an Alert message be changed at runtime ?
a Yes
b No
Ans : A
Can we create an LOV without an RECORD GROUP ?
a Yes
b No
Ans : B
How many no of columns can a RECORD GROUP have ?
a 10
b 20
c 50
d None of the above
Ans D
Oracle precompiler translates the EMBEDDED SQL statemens into
a Oracle FORMS
b Oracle REPORTS
c Oracle LIBRARY
d None of the above
Ans : D
Kind of COMMENT statements placed within SQL statements ?
a Asterisk(*) in column ?
b ANSI SQL style statements(...)
c C-Style comments (/*......*/)
d All the above
Ans : D
What is TERM ?
a TERM is the terminal definition file that describes the terminal from which you are using R20RUN
( Reports run time )
b TERM is the terminal definition file that describes the terminal from which you are using R20DES
( Reports designer )
c There is no Parameter called TERM in Reports 2.0
d None of the above
Ans : A
If the maximum records retrieved property of a query is set to 10, then a summary value will
be calculated
a Only for 10 records
b For all the records retrieved
c For all therecords in the referenced table
d None of the above
Ans : A
With which function of a summary item in the COMPUTE AT optio required ?
a   Sum
b   Standard deviation
c   Variance
d   % of Total function
Ans : D
For a field in a repeating frame, can the source come from a column which does not exist in
the datagroup which forms the base of the frame ?
a Yes
b No
Ans : A
What are the different file extensions that are created by Oracle Reports ?
a . RDF file & .RPX file
b . RDX file & .RDF file
c . REP file & .RDF file
d None of the above
Ans : C
Is it possible to Disable the Parameter form while running the report?
a Yes
b No
Ans : A
What are the SQL clauses supported in the link property sheet ?
a WHERE & START WITH
b WHERE & HAVING
c START WITH & HAVING
d WHERE, START WITH & HAVING
Ans : D
What are the types of Calculated columns available ?
a Summary, Place holder & Procedure column
b Summary, Procedure & Formula columns
c Procedure, Formula & Place holder columns
d Summary, Formula & Place holder columns
Ans.: D
If two groups are not linked in the data model editor, what is the hierarchy between them?
  a There is no hierarchy between unlinked groups
  b The group that is right ranks higher than the group that is to theleft
  c The group that is above or leftmost ranks higher than the group that is to right or below it
  d None of the above
Ans : C
Sequence of events takes place while starting a Database is
a Database opened, File mounted, Instance started
b Instance started, Database mounted & Database opened
c Database opened, Instance started & file mounted
d Files mounted, Instance started & Database opened
Ans : B
SYSTEM TABLESPACE can be made off-line
a Yes
b No
Ans : B
ENQUEUE_RESOURCES parameter information is derived from
a PROCESS or DDL_LOCKS & DML_LOCKS
b LOG BUFFER
c DB_BLOCK_SIZE
d DB_BLOCK_BUFFERS
Ans : A
SMON process is used to write into LOG files
a TRUE
b FALSE
Ans : B
EXP command is used
a To take Backup of the Oracle Database
b To import data from the exported dump file
c To create Rollback segments
d None of the above
Ans : A
SNAPSHOTS cannot be refreshed automatically
a TRUE
b FALSE
Ans : B
The User can set Archive file name formats
a TRUE
b FALSE
Ans : A
The following parameters are optional in init.ora parameter file DB_BLOCK_SIZE, PROCESS
a TRUE
b FALSE
Ans : B
NOARCHIEVELOG parameter is used to enable the database in Archieve mode
a TRUE
b FALSE
Ans : B
Constraints cannot be exported through Export command?
a TRUE
b FALSE
Ans : B
It is very difficult to grant and manage common priveleges needed by
different groups of database users using roles
a TRUE
b FALSE
Ans : B
The status of the Rollback segment can be viewed through
a DBA_SEGMENTS
b DBA_ROLES
c DBA_FREE_SPACES
d DBA_ROLLBACK_SEG
Ans : D
Explicitly we can assign transaction to a rollback segment
a TRUE
b FALSE
Ans : A
What file is read by ODBC to load drivers ?
a ODBC.INI
b ODBC.DLL
c ODBCDRV.INI
d None of the above
Ans : A

zynga previous papers 2011

2012-09-07T10:16:00.000-07:00

zynga previous papers 2011

1.Find duplicates in Array. I presented solution in O(n) time and space using hash.
we can do it in O(n) time and constant space. Assumption: range of numbers is known.

2.Reverse a Linked List. Iterative and then recursive

3.Reverse words : "I love to play" becomes "play to love I".

sol:
reverse each word in string then reverse total string

4.n=4 and k=3 . team name can be formed by four digits 1,2,3,4 . so n=4 now make team number using any three digits so that b1<b2<b3.
Basically print numbers which are in form b1<b2<b3..e.g. 123,124,234 etc

5. write a function that takes two numbers a and b and returns a.b eg: a=10 and b=12 output will be 10.12

NEVIS NETWORKs Interview Questions

2012-09-07T09:39:00.000-07:00

NEVIS NETWORKs Interview Questions 2010 IIT Guwahati
Test+ interview + Telephonic Interview

Selection criteria :
They prefer candidates with network project either in BTP or in MTP.
Test contains two parts, part A have C programming questions and part b has C questions and VLSI . U can choose either one.
Intreview (only Project)
Telephonic Interview(only abt Resume)
All the c questions are from Kernighan and Ritchie book .

Q. WAP for conversion string to float.
Q.Preorder Traversal for a given tree.
Q. WAP for DFS(depth first search).
Q. Question about storage classes.
Q. What is linkage error ?
Q. Program for Number conversion (octal to binary)

Q. Int i=0;
   While(++i<5)
    {
       For(j=0;j<10;j++)
           If(i==j)
             break;
What is the value of i at end of the program ?

Q. #define min(a,b) (a)<(b) ? a: b;
     Main()
      {
           Int a=5,b=4,k;
           K=min(a++,b--);
           Printf(“%d %d %d”,a,b,k);
        }
     What is the o/p;
Q.
     struct
      {
       Int a;
       Char ch;
      }x;
    Printf(“%d”,sizeof(x));

What is the o/p;

NetApps Written to Interview : IIT Guwahati

2012-09-07T09:29:00.002-07:00

Institute: IIT Guwahati

Date of test: 3/12/2010

Package: 9Lakh

Paper Type: CSE B Tech/M Tech

Time: 1 hour only

Hei,

Written Test:

They have 4 sections,
1. Apti and analytical
2. Coding knowledge : just showing the output
3. Computer c++/Os questions..
4. Computer Systems

My suggestion: Do not start from the beginning means section 1. Start from the technical round. Read the questions and no need to waste more time for long programs see others.

Some concepts from :

1. C++ virtual functions

2. Some from on network.. to transfer 1000 meter propagation delay and transmission delay was given. and asked for 1 bit delay equals to what meter?

3. One 1000ms time to short by one PC of 1000 values by quick sort. What time require to solve 100 values in that PC

Ans: 6.67 ms

4. One system have 8 processor and 8bit require to access one block. and 12 bit require for one instruction accessing within of a block. Amount of memory have?

5. Some Programming questions on C and algo. Just not remember those.

6. Like quick sort complexity

7. U have given the code of insertion sort. But one line missing only. Fill this line...
like A[i+1=key;

8. Max heap is given. u have to insert two more keys. then what will be the inorder sequence.

Interview:

Round 1:

4 question:

1. U have given two array A[n] and B[n+1]. A and B have same elements but B have only one elements more than A. Now U have to find out that element in O(n).

I did that

2. U have to check a number is 2^n or not in constant time.

I did that

3. U have to check a binary tree is BST or not?
Write the algorithm for that.

I did that

4. a,b,c,d are the four values. U have to find out the minimum with ternary operator.

I cant do that

Just within 5min they called for the 2nd round interview..

Round 2:

He asked many question:

1. Why u r doing M tech?

2. What are inside the process?

3. What use of stack in process?

4. What is thread? Why used in OS?

5. What difference with thread and process?

6. Why demand paging is required?

7. Write the code for writing the number of lines reading from a text file. And each line u have to print the number of words that have to print also.

8. What interests u to come in NetApps?

Round 3:HR round

1. Tell me about urself?

2. Scholling btech and mtech details?

3. Why u did BE in IT? Not in CSE?

4. What subjects u prefered?

5. What post u like? Development or Quality testing?

6. What u expects NetApps?

7. Will you have questions on NetApas?

Bit Magic

2012-09-04T02:10:00.000-07:00

1.Given a integer x, write a function that multiplies x with 3.5 and returns the integer result. You are not allowed to use %, /, *.

Examples:
Input: 2
Output: 7

Input: 5
Output: 17 (Ignore the digits after decimal point)

#include<stdio.h>
int main()
{
    int i,j,k;
    printf("enter number");
    scanf("%d",&i);
    j=i<<1;
    k=i>>1;
    i=i+j+k;
    printf("\n number after multiplying with 3.5 is %d ",i);
    return 0;
}

2.Count total set bits in all numbers from 1 to n

#include<stdio.h>
int main()
{
    int i=0,n;
    scanf("%d",&n);
    while(n>0)
    {
    i++;
    n=n&(n-1);
    }
    printf("%d",i);
}

3.Smallest of three integers without comparison operators

#include<stdio.h>
int main()
{
    int i,j,k;
    scanf("%d%d%d",&i,&j,&k);
    while(i&&j&&k)
    {
    !(--i)&&printf("first number is small")||!(--j)&&printf("second number is small")||!(--k)&&printf("3rd number is small");
    }
}

4.Program to count number of set bits in an (big) array

#include<stdio.h>
int fun(int n)
{
    int i=0;
    while(n>0)
    {
    i++;
    n=n&(n-1);
    }
    return i;
}
int main()
{
    int i,k,j,sum=0;
    printf("enter no of elements in array\n");
    scanf("%d",&i);
    for(j=0;j<i;j++)
    {
    scanf("%d",&k);
    sum=sum+fun(k);
    }
    printf("\ntotal number of set bits is %d",sum);
}

5.Count number of bits to be flipped to convert A to B
int fun(int n)
{
    int i=0;
    while(n>0)
    {
    i++;
    n=n&(n-1);
    }
    return i;
}
#include<stdio.h>
int main()
{
    int a,b,c;
    printf("enter a and b values");
    scanf("%d %d",&a,&b);
    c=a^b;
    printf("number of bits to filp to convert a to b is %d",fun(c));
}

6.Swap bits in a given number

Given a number x and two positions (from right side) in binary representation of x, write a function that swaps n bits at given two positions and returns the result. It is also given that the two sets of bits do not overlap.

Examples:

Let p1 and p2 be the two given positions.

Input:
x = 47 (00101111)
p1 = 1 (Start from second bit from right side)
p2 = 5 (Start from 6th bit from right side)
n = 3 (No of bits to be swapped)
Output:
227 (11100011)
The 3 bits starting from the second bit (from right side) are
swapped with 3 bits starting from 6th position (from right side)

#include<stdio.h>
int main()
{
    int n,i,j,k,m;
    printf("enter number");
    scanf("%d",&n);
    printf("enter positions");
    scanf("%d %d",&i,&j);
    m=j;
    for(k=i;k<m&&j<32;)
    {
    if(n&(1<<k)&&n&(1<<j));//when both digits 1 no need to swap
    else if(!(n&(1<<k))&&!(n&(1<<j)));//when both digits no need to swap
    else //when both digits are not same do swap
    {
        n=n^(1<<k);
        n=n^(1<<j);
    }
    k++;
    j++;
    }
    printf("%d",n);
    return 0;
}

7.Add 1 to a given number

Write a program to add one to a given number. You are not allowed to use operators like ‘+’, ‘-’, ‘*’, ‘/’, ‘++’, ‘–’ …etc.

Examples:
Input: 12
Output: 13

Input: 6
Output: 7

#include<stdio.h>
int main()
{
    int i,k=0;
    printf("\n enter numeber ");
    scanf("%d",&i);
    while(i&(1<<k))//go untill u find 0 digit in number then change it as 1
    {
    i=i^1<<k;
    k++;
    }
    i=i^1<<k;
    printf("%d",i);
}

   If im wrong please correct me or give comment and you can post your solutions and questions here

Some tricky c questions

2012-08-29T23:24:00.000-07:00

Some tricky and important c questions for placements

Index:

1. Data types
2. Expressions
3. Storage Classes
4. Arrays
5. Pointers
6. Strings
7. Input and Output
8. Structures and unions
9. Bitwise tricks

each topic consists level-1 and level -2

Assume integer takes 4 bytes ,character takes 1 byte, pointer take 4 bytes , float takes 4 bytes and double takes 8bytes
----------------------------------------------------------------------------------------------------------------------------------

DATA TYPES (Level-I)

1)
#include<stdio.h>
int main()
{
   int a=(1<<32)-1;
   a++;
   printf("%d",a);
}

Sol.: 0

2)
#include<stdio.h>
int main()
{
   char c=255;
   int i=c;
   printf("%d",i);
}

Sol.: -1

3)
#include<stdio.h>
int main()
{
   int a;
   char b='A';
   a=b;
   b=a;
   printf("%d",b);
}

Sol.: 65

4)
#include<stdio.h>
int main()
{
   int a;
   char b;
   float c;
   unsigned int d;
   printf("%d,%d,%d,%d",sizeof(a),sizeof(b),sizeof(c),sizeof(d));
   printf("\n%d,%d,%d",sizeof(1),sizeof('A'),sizeof(0.7));
}

Sol: 4,1,4,4
     4,1,8

5)
#include<stdio.h>
int main()
{

printf("\n%d,%d,%d",sizeof(1UL),sizeof(12U),sizeof(0.7f));
}

Sol: 8,4,4 (UL->unsigned long,U->unsigned,f->float)

6)
#include<stdio.h>
int main()
{
   float a,b,c;
   a=4/5;
   b=(float)(4/5);
   c=4.0/5;
   printf("%f,%f,%f",a,b,c);
}

Sol.: 0.0000,0.0000,0.8000

7)
#include<stdio.h>
enum{i=1,j=2};
int main()
{
   i++;
   printf("%d",i);
}

Sol.: error: lvalue required as increment operand

8)
#include<stdio.h>
enum days{SUN=3,MON,TUE=-1,WED};
int main()
{
   printf("%d,%d",MON,WED);
}

Sol: 4,0

9)
#include<stdio.h>
int main()
{
   signed char c=-64;
   unsigned char p=c;
   printf("%d",p);
}

Sol: 192 (i.e.256-64)

10)
#include<stdio.h>
int main()
{
   unsigned char c=-64;
   int b=-32;
   if(c>b)
    printf("pass");
   else
printf("fail");
}

Sol.: pass (Here value of c is 192)

11)
#include<stdio.h>
int main()
{
    int a=5,b=12;
float c=a;
double d=b;
printf("%f,%lf",c,d);
}

Sol: 5.0000,12.0000

DATA TYPES(LEVEL-II)

1)
#include<stdio.h>
int main()
{
   unsigned short a=10;
   signed short b=10;
   if(a>-1 && b>-1)
    printf("no surprise");
   else
   printf("surprise");
}

Sol: no surprise

2)
#include<stdio.h>
int main()
{
   unsigned int a=10;
   signed int b=10;
   if(a>-1 && b>-1)
printf("no surprise");
   else
printf("surprise");
}

Sol: suprise(Here a>-1 returns 0)

3)
#include<stdio.h>
int main()
{
   signed char c=-64;
   unsigned char d=64;
   short int p=c,q=d;
   printf("%d,%d",p,q);
}

Sol: -64,64

5)
#include<stdio.h>
int main()
{
   float a=0.7;
   double b=0.7;
   if(a==b)
printf("Equal");
   else
    printf("Unequal");
}

Sol: Unequal

6)
#include<stdio.h>
int main()
{
   signed char c;
   for(c=0;c<=127;c++)
   printf("%c",c);
}

Sol: Infinitly printing of characters.

7)
#include<stdio.h>
int main()
{
float f=0.0f;
   int i;
   for(i=0;i<10;i++)
f=f+0.1f;
   if(f==1.0f)
    printf("Equal");
   else
   printf("Unequal");
}

Sol: Unequal(Value of Float is not exactly stored in variable so f will give round off error)

8)
#include<stdio.h>
int main()
{
float f=0.0f;
int i;
for(i=0;i<5;i++)
   f=f+0.2f;
if(f==1.0f)
printf("Equal");
else
printf("Unequal");
}

Sol: Equal

9)
#include<stdio.h>
int main()
{
unsigned i;
for(i=5;i>=0;i--)
printf("%d",i);
}

Sol: Infinite printing of no.'s

11)
#include<stdio.h>
int main()
{
float a=5.6;
double b=12.345;
int c=a,d=b;
printf("%d,%d",c,d);
}

Sol: 5,12

12)
#include<stdio.h>
int main()
{
unsigned int a=12345;
unsigned char b=511,c=a;
printf("%d,%d",b,c);
}

Sol.: 255,57

13)
#include<stdio.h>
int main()
{
unsigned int a=12345;
signed char b=511,c=a;
printf("%d,%d",b,c);
}

Sol: -1,57

14)
#include<stdio.h>
int main()
{
signed char c=-64;
   int b=-32;
   unsigned int a=-16;
if(c>b)
{
    printf("pass1");
   if(c<a)
    printf("pass2");
else
   printf("fail2");
}
else
printf("fail1");
if(b<a)
   printf("pass2");
   else
printf("fail2");
}

Sol.: fail1pass2

------------------------------------------------------------------------------------------------------------------------------

Expression(LEVEL-I)

1)
#include<stdio.h>
int main()
{
   int a=1,b=2,c=3,d=-2,r;
   r=a+b*c/d;
   printf("%d\n",r);
   return 0;
}

Sol.: -2

2)
#include<stdio.h>
int main()
{
   int i,j=3,k=4;
   printf("%d\n",i=++j==k);
   return 0;
}

Sol.: 1

3)
#include<stdio.h>
int f(void)
{
   printf("f");
   return 1;
}
int g(void)
{
   printf("g");
   return 2;
}
int h(void)
{
   printf("h");
   return 3;
}
int main()
{
   int r;
   r=f()+g()*h();
   printf("%d\n",r);
   return 0;
}

Sol:. fgh7(NOTE:fgh will get printed in any order depending upon compiler)

4)
#include<stdio.h>
int main()
{
   int a=-2,b;
   b=-a--;
printf("%d,%d\n",a,b);
   return 0;
}

Sol.: -3,2

5)
#include<stdio.h>
int main()
{
   int i=+5,j=+4,k;
   k=i++*j++;
   printf("%d,%d,%d\n",i,j,k);
   return 0;
}

Sol.: 6,5,20(NOTE: Ans. will be compiler dependent)

6)
#include<stdio.h>
int main()
{
   int i=0;
   if(i++)
printf("%d\n",i);
printf("%d\n",i);
return 0;
}

Sol.: 1

7)
#include<stdio.h>
int main()
{
   int i=2,j=1,k=4,r1,r2;
   r1=(i,j,k,--k);
   r2=i,j,k;
   printf("%d,%d\n",r1,r2);
return 0;
}

Sol.: 3,2(NOTE:r1 will get the value from --k and r2 will get the value from i)

8)
#include<stdio.h>
int x(void)
{
printf("x");
return 2;
}
int y(void)
{
printf("y");
return 1;
}
int z(void)
{
printf("z");
return 0;
}
int main()
{
int r;
r=z() && y();
printf("%d\n",r);
return 0;
}

Sol.:z0

9)
#include<stdio.h>
int x(void)
{
printf("x");
return 2;
}
int y(void)
{
printf("y");
return 1;
}
int z(void)
{
printf("z");
return 0;
}
int main()
{
int r;
r=z() || x();
printf("%d\n",r);
return 0;
}

Sol.: zx1

10)
#include<stdio.h>
int main()
{
   int i=1,2;
   printf("%d\n",i);
   return 0;
}

Sol.: Compile time error

11)
#include<stdio.h>
int main()
{
   int i,j=1,k=0;
   i=(j++,++k ||--k);
   printf("%d,%d,%d\n",i,j,k);
   return 0;
}

Sol.: 1,2,1(NOTE:--k will not get evaluated)

12)
#include<stdio.h>
int main()
{
   int i=- -3,j=-(-4);
   printf("%d,%d\n",i,j);
   return 0;
}

Sol.: 3,4

13)
#include<stdio.h>
int main()
{
printf("abc");
printf("\b\b\bsi");
printf("\rha");
return 0;
}

Sol.: ha

14)
#include<stdio.h>
int main()
{
   int n=0;
   switch(n)
   {
   case 0: n+=1;
   case 1: n+=2;
     default: n+=3;
   }
   printf("%d",n);
   return 0;
}

Sol.: 6

15)
#include<stdio.h>
int main()
{
   int i,j;
   scanf("%d%d",&i,&j);
i+=2;
j-=2;
if((i=5) && (j=-3))
   printf("yes\n");
else
   printf("no\n");
   return 0;
}
Under what circumstances would the above program prints yes?

Sol.: Under every circumstances.

Expression(LEVEL-II)

1)
#include<stdio.h>
int main()
{
int i=1,j;
j=i--&&++i/2||++i%7;
printf("%d,%d\n",i,j);
return 0;
}

Sol.: 2,1

2)
#include<stdio.h>
int main()
{
int a=2,b=3;
a=a+b-(b=a);
printf("%d,%d\n",a,b);
return 0;
}

Sol.: This is compiler dependent.

3)
#include<stdio.h>
int main()
{
int i=0,j=2,k=0;
if(i && (j=k--))
k=1;
printf("%d,%d,%d",i,j,k);
return 0;
}

Sol.: 0,2,0(NOTE:j=k-- will not get evaluated)

4)#include<stdio.h>
int main()
{
   int i=3,j=-2,k;
   k=sizeof(1,--i + ++j);
   printf("%d,%d,%d\n",i,j,k);
   return 0;
}

Sol.: 3,-2,4(NOTE:Expressions inside sizeof() will not be evaluated because sizeof() is an operator)

5)
#include<stdio.h>
int main()
{
int a=1;
switch(a)
{
    int b=2;
    case 1:
printf("b=%d\n",b);
break;
    default:
printf("b=%d\n",b);
   break;
}
return 0;
}

Sol.: Garbage value(Inside switch the control will immideately go for case and will not read b)

6)
#include<stdio.h>
int main()
{
   int i=-2,r;
   r=++i++;
   printf("%d\n",r);
   return 0;
}

Sol.: error: lvalue required as increment operand

7)
#include<stdio.h>
int main()
{
   int cnt=5,a=1000;
    do
   {
     a/=cnt;
}while(cnt--);
   printf("%d\n",a);
   return 0;
}

Sol.: Floating point exception(Note:a is divided by 0 in the last iteration)

8)
#include<stdio.h>
int main()
{
   int i=-5,j=2,k=-3,r;
   r=i++&&(--j||k++);
   printf("%d,%d,%d,%d\n",i,j,k,r);
   return 0;
}

Sol.: -4,1,-3,1

9)Which one of the following expressions should be written in the blank so that the following program prints Yes as output?
#include<stdio.h>
int main()
{
   int i=1;
if(____,i--)
printf("Yes\n");
   else
   printf("No\n");
return 0;
}

Sol.: i++,++i

10)
int add(int i,int j)
{
return i+j;
}
#include<stdio.h>
int main()
{
int i=5;
printf("%d,%d\n",++i,add(i,2));
return 0;
}

Sol.: 6,7 or 6,8 (compiler dependent)

11)
#include<stdio.h>
int p(void)
{
printf("p");
return 20;
}
int q(void)
{
printf("q");
return -10;
}
int r(void)
{
printf("r");
return 0;
}
What logical operators should be placed in the blanks of the following program such that it produces the output pqr?
int main()
{
!p()__q()__r();
return 0;
}

Sol.: !p()||q()&&r()

12)
#include<stdio.h>
int main()
{
   int i=0,j=199;
   while(i<j)
   {
     --j;
     i+=2;
     }
printf("%d\n",i-j);
return 0;
}

Sol.: 2

13)
#include<stdio.h>
int main()
{
int i=-1,j=-2,k;
k=-i++ * ++j;
printf("%d,%d,%d\n",i,j,k);
return 0;
}

Sol.: 0,-1,-1(Compiler dependent code)

14)
int x=-3;
int f(void)
{
x=x+2;
return x;
}
int g(void)
{
x=x*3;
return x;
}
#include<stdio.h>
int main()
{
printf("%d\n",f()+g());
return 0;
}

Sol.: -4 or -16(compiler dependent)

15)
#include<stdio.h>
int main()
{
int i=12,j=3,k;
k=i+++j;
printf("%d,%d,%d\n",i,j,k);
return 0;
}

Sol.: 13,3,15

---------------------------------------------------------------------------------------------------------------------------------

STORAGE CLASSES(LEVEL-I)

1)
#include<stdio.h>
int f(void)
{
int a=5;
return a;
}
main()
{
int b=2;
b=f();
printf("%d",b);
}

Sol.: 5

2)
#include<stdio.h>
main()
{
static int a=2;
if(a)
{
printf("%d\n",a);
a--;
main();
}
}

Sol.: 2,1

3)
#include<stdio.h>
static int a;
int b;
main()
{
a++,b++;
printf("%d,%d",a,b);
}

Sol.: 1,1

4)
#include<stdio.h>
void f()
{
a=10;
printf("%d",a);
}
int a=5;
main()
{
a++,b++;
printf("%d,%d",a,b);
f();
}

Sol.: Compiler Error

5)
#include<stdio.h>
char f()
{
static char c='A';
c+=2;
return c;
}
main()
{
f();
printf("%c",f());
}

Sol.: E

6)
#include<stdio.h>
main()
{
int i=2;
{
int i=3;
i++;
printf("%d\n",i);
}
printf("%d",i);
}

Sol.: 4,2

7)
#include<stdio.h>
int i=0;
void g()
{
i=100;
i++;
}
main()
{
i++;
g();
i+=2;
g();
printf("%d",i);
}

Sol.: 101

8)
#include<stdio.h>
main()
{
register int a=2;
register float c=3.4;
printf("%d,%f",a,c);
}

Sol.: 2,3.4000

9)
#include<stdio.h>
main()
{
int a=2;
if(a)
{
printf("%d\n",a);
a--;
main();
}
}

Sol.: infinite printing of 2.

10)
#include<stdio.h>
int i=-3;
void f(int i)
{
i++;
printf("%d\n",i);
}
main()
{
int i=2;
f(i);
printf("%d",i);
}

Sol.: 3,2

11)
#include<stdio.h>
int a=1,b=3;
void f()
{
printf("%d,%d",a,b);
}
main()
{
int a=-1,b=2;
printf("%d,%d\n",a,b);
f();
}

Sol.: -1,2
      1,3

12)
main()
{
printf("A");
main();
}

Sol.: Infinite printing of A

13)
Consider the following 'C' code in two files.
        f1.c
#include<stdio.h>
int i=2;
extern void f();
main()
{
f();
printf("%d",i);
}
        f2.c
extern int i;
void f()
{
i++;
}
What would be the output after linking and execution of above two files?

Sol.: 3

14)
Consider the following 'C' code in two files.
        f1.c
#include<stdio.h>
int i=-2;
extern int f();
main()
{
printf("%d",f());
}
        f2.c
extern int i;
int f()
{
static int i;
i++;
return i;
}
What would be the output after linking and execution of above two files?

Sol: 1

15)
Consider the following 'C' code in two files.
        f1.c
#include<stdio.h>
extern int i;
main()
{
i++;
printf("%d\n",f());
printf("%d",i);
}
        f2.c
static int i;
int f()
{
i++;
return i;
}
What would be the output after linking and execution of above two files?

Sol.: Link time error

16)
#include<stdio.h>
main()
{
extern int p;
printf("%d\n",p);
}

Sol.: Link time error

17)
#include<stdio.h>
extern int p;
void f(void)
{
printf("%d",++p);
}
static int p=4;
main()
{
f();
}

Sol.: Compile time error

18)
#include<stdio.h>
int f(int p)
{
return (p%2)?2:f(p/2);
}
static int p=4;
main()
{
printf("%d",f(36));
}

Sol.: 2

19)
int q=2;
void fun()
{
int *p=(int*)malloc(q*sizeof(int));
return p;
}
Which part of the memory would be accessed on execution of above program?

Sol.: q->data segment, p->stack segment, malloc()->heap segment.

STORAGE CLASSES(LEVEL-II)

1)
#include<stdio.h>
main()
{
int i,r=0;
int f();
for(i=0;i<4;i++)
r+=f();
printf("%d",r);
}
int f(void)
{
static int x=-2;
return x++;
}

Sol.: -2

2)
Consider the 'C' code in the following two files:
        f1.c
#include<stdio.h>
extern int i;
extern int f();
main()
{
i++;
printf("%d\n",f());
printf("%d",i);
}
        f2.c
int i;
static int f()
{
i++;
return i;
}
What would be the output after linking and execution of above two files?

Sol: Link time error.

3)#include<stdio.h>
int p=65;
main()
{
char p;
printf("%d",p);
}

Sol.: garbage value

4)
#include<stdio.h>
int g(int x)
{
if(x%2)
return g(x-1);
return x/2;
}
main()
{
printf("%d",g(11));
}

Sol.: 5

5)
#include<stdio.h>
int recursionMethod(int x)
{
int call;
if(x!=-1)
{
call=recursionMethod(x-1)+1;
printf("%d",call);
}
return x;
}
main()
{
recursionMethod(5);
}

Sol.: 012345

6)
#include<stdio.h>
int f(int k)
{
if(k>=0)return 1;
return f(k+1)+f(k+2);
}
main()
{
f(-2);
}
How many calls to function 'f' are made while executing the above program?
Sol.: 5(NOTE: f(-2) will call f(-1) & f(0). f(-1) will call f(0) & f(1))

7)
#include<stdio.h>
main()
{
int a=-1,b=2,r;
int f(int);
r=f(a+b);
printf("%d,%d,%d",a,b,r);
}
int b=-4;
int f(int a)
{
a++;
b++;
return a+b;
}

Sol.: -1,2,-1

8)
#include<stdio.h>
int zap(int n)
{
int s=0;
if(!n) return 0;
s+=n%10;
zap(n/10);
return s;
}
main()
{
printf("%d",zap(12345));
}

Sol.: 5

9)
Compare the following 'c' code in two files:
        f1.c
#include<stdio.h>
main()
{
double x,y,z;
x=3.5;
y=4.2;
z=f(x,y);
}
        f2.c
double f(double a,double b)
{
return b-a;
}
Does the linking and execution of above two files result in any errors. If yes what changes do you suggest to eliminate them?

Sol.: No error while linking and execution

10)
#include<stdio.h>
int f(int n)
{
if(n<=1) return n;
return f(n-1)+f(n-2)+f(n-3);
}
main()
{
printf("%d",f(5));
}

Sol.: 3

11)
#include<stdio.h>
static int v=5;
void f(int v)
{
{
static int v=0;
for(;v<5;v++)
   printf("%d\n",v);
}
}
main()
{
while(v--) f(v);
printf("%d\n",v);
}

Sol.: 0
      1
      2
      3
      4
      -1

12)
Fill up the blank in the following program such that it prints 15 on the screen.
#include<stdio.h>
int sample(int i)
{
static int s=0;
if(i==2) return s;
s+=5;
i++;
return sample(i);

}
main()
{
printf("%d\n",sample(__));
}

Sol.: -1

----------------------------------------------------------------------------------------------------------------------------------

ARRAYS(LEVEL-I)

1)
main()
{
int a[]={1,2,3,4};
printf("%d,%d",*a+2,*(a+1));
}

Sol.: 3,2

2)
main()
{
int a[]={1,2,3};
int b[3][4]={1,5,3,6};
printf("%d,%d\n",sizeof(a),sizeof(b));
}

Sol.: 12,48

3)
main()
{
float a[2];
printf("%u,%u,%u,%u\n",a,a+1,&a,&a+1);
}

Sol.: 1000,1004,1000,1008

4)
main()
{
int a[2][3]={
        {3,2,1},
        {6,5,4}
         };
int (*ap)[3];
ap=a;
printf("%u",ap);
ap++;
printf("%u",ap);
printf("%d,%d",(*ap)[0],(*ap)[1]);
}

Sol.: 1000,1012,6,5

5)
main()
{
int i,r=0;
int a[10]={1,2,-3,-4};
for(i=0;i<10;++i)
if(a[i]%2) r+=a[i];
printf("%d",r);
}

Sol.: -2(NOTE: if an array is partially initialize then all its left element becomes 0)

6)
main()
{
unsigned int a[2][3]={ {3,2,1},
            {6,5,4}
              };
printf("%u,%u,%u,%u\n",a,a+1,&a,&a+1);
}

Sol.: 1000,1012,1000,1024

7)
main()
{
int a[]={1,2,3,4,5};
int *p=a;
printf("%d,%d",p[1],p[2]);
}

Sol.: 2,3

8)
main()
{
static int a[2][3][4];
int (*ap)[3][4];
ap=a;
printf("%u\n",ap);
ap++;
printf("%u",ap);
}

Sol.: 1000,1048

9)
main()
{
int a[]={1,-1,2,-2,3,-3,4,-4,5,-5};
int *p=a+1;
int *q=a+6;
printf("%d",q-p);
}

Sol.: 5

10)
main()
{
int a[]={1,2,3,4,5};
int p,*s;
s=a;
printf("%d\n",*(s++));
p=*++s;
printf("%d",p);
}

Sol.: 1,3

11)
int f(int*p)
{
int i,s=0;
for(i=0;i<sizeof(p);i++)
s+=(p+i);
return s;
}
main()
{
static int a[5]={1,-2,3,-4,5};
int r;
r=f(a);
printf("%d",r);
}

Sol.: -2

12)
int a[2][3]={{1}};
main()
{
int b;
b=a[0][0]+a[0][2]+a[1][3];
printf("%d",b);
}

Sol.: undefine behaviour

13)
main()
{
int i,r=0;
static int a[5]={0x1,0x2,0x10,0x20,0x30};
for(i=0;i<5;++i)
if(!(i%2)) r+=a[i];
printf("%d",r);
}

Sol.: 65

14)
main()
{
int i,r=0;
for(i=0;i<10;++i)
if(!a[i]) r+=(!a[i]);
printf("%d",r);
}

Sol.: 5

15)
#define rows 2
#define cols 3
void test(int y[][cols]);
main()
{
static int a[rows][cols]={-1,2,-3,4,-5,6};
test(a);
}
void test(int x[][cols])
{
int i,j,k;
for(j=0;j<cols;++j)
{
k=0;
for(i=0;i<rows;++i)
   if(x[i][j]>k) k=x[i][j];
printf("%d\n",k);
}
}

Sol.: 4,2,6

16)
main()
{
int b=0;
int a[2][3]={{1,2}};
b=a[0][1]+a[0][2]+a[1][2];
printf("%d",b);
}

Sol.: 2

17)
void foo(int n,int a[])
{
n=0;a[n]=3;
n=1;a[n]=9;
n=2;a[n]=27;
n=3;a[n]=81;
}
main()
{
int n=1,a[]={2,4,8,16};
foo(n,a);
printf("%d",a[n]);
}

Sol.: 9

18)
main()
{
float a[5];
printf("%u",&a[3]-3);
}

Sol.: 1000

19)
main()
{
int a[2][3]={-1,-2,0,1,2,3};
printf("%d",sizeof((int*)a));
}

Sol.: 4

20)
main()
{
int a[]={10,20,30,-30,-20,-10};
int *b=&a[4];
printf("%d",b[-3]);
}

Sol.: 20

ARRAYS(LEVEL-II)

1)
#include<stdio.h>
int a[2][3]={010,020,030,040,050};
main()
{
int i,j,min=100;
for(i=0;i<2;++i)
for(j=0;j<3;++j)
   if(a[i][j]<min) min=a[i][j];
printf("%d",min);
}

Sol.: 0(NOTE:last element will be zero)

2)
#include<stdio.h>
int array[]={23,34,12,17,204,99,16};
main()
{
unsigned d;
for(d=-1;d<=5;d++)
printf("%d\n",array[d+1]);
}

Sol.: No output

3)
#include<stdio.h>
main()
{
int a[]={1,2,3,4,5},b;
b=(char*)(a+5)-(char*)a;
printf("%d",b);
}

Sol.: 20(NOTE:this is the no. of bytes between 5th element of a and 0th element)

4)
Fill up the blank in the following program such that it prints the address of element 5 without subscription operator([])
#include<stdio.h>
main()
{
unsigned int a[3][3]={
            {1,2,3},
            {2,3,4},
            {3,4,5}
              };
printf("%d",_____);
}

Sol.: (*a)+4

5)
#include<stdio.h>
int a[]={1,2,3};
int* f(void)
{
int i;
for(i=0;i<3;i++)
return a+1;
}
main()
{
*f()=5;
printf("%d,%d,%d",a[0],a[1],a[2]);
}

Sol.: 1,5,3(Note: f will return address of 2nd array variable and that is made equal to 5 in main function.

6)
#include<stdio.h>
void f(float(*a)[3])
{
printf("%d",sizeof(a[0]));
}
main()
{
float a[2][3];
f(a);
}

Sol.: 12

7)
Fill in the blank in the following program s.t. the contents of cell a[1][1] will become 8
#include<stdio.h>
void f(char *a)
{
________________
}
main()
{
int a[2][3]={
        {1,2,3},
        {4,5,6}
         };
f((char*)a);
printf("%d",a[1][1]);
}

Sol.: *(a+4*(1*3+1))=8;

8)
#include<stdio.h>
int arr[2][3]={
        {1,2,3},
        {-1,-2,-3}
          };
main()
{
int i=2;
int *a=(int*)&arr;
printf("%d",(a++)[i++]++);
}

Sol.: 3

9)
#include<stdio.h>
main()
{
int a[][4]={
        {1,2,3},
        {4,5,6},
        {7,8,9}
        };
int b;
b=(char*)a+2-(char*)a;
printf("%d",b);
}

Sol.: 2

10)
main()
{
int a[2][2][3]={
        {
        {1,2,3},
        {4,5,6},
        },
        {
        {7,8,9},
        {6,5,4}
        }
        };
printf("%d,%d",*(*(*(a+1))),*(a[0][1]+2));
}

Sol.: 7,6

11)What changes do you suggest in the following program so that it displays 5,10,15 on the screen. Try to make required changes as minimal as possible.
#include<stdio.h>
int a[]={1,2,3};
int* f(void)
{
int i;
return a + i++;
}
main()
{
*f()=5;
*f()=10;
*f()=15;
printf("%d,%d,%d",a[0],a[1],a[2]);
}

Sol.: in function f we will declayer int i as static int i.

12)
#include<stdio.h>
main()
{
int a[2][3]={
        {2,3,4},
        {5,6,7}
         };
int *pa[2]={a[0],a[1]};
printf("%d,%d",pa[0],*pa[1]);
}

Sol.: 1000,5

13)
#include<stdio.h>
main()
{
int a[][3]={
        {-1,2,3},
        {-4,5,6}
         };
printf("%d",*(*(a+1)-1)+2);
}

Sol.: 5(3+2)

14)
#include<stdio.h>
main()
{
int a[2][3]={
        {2,3,4},
        {5,6,7}
         };
int *p=a[0];
printf("%d",p[2]);
}

Sol.: 4

15)
#include<stdio.h>
void f(int a[][4])
{
printf("%d",*a[0]);
}
main()
{
int a[2][4]={
        {1,2,3,4},
        {5,6,7,8}
         };
f(a);
}

Sol.: 1

-------------------------------------------------------------------------------------------------------------------------------

POINTERS(Level-I)

1)
#include<stdio.h>
main()
{
int a=-10,b=-2;
int c,*pa,*pb;
pa=&a;
*pa=2*a;
pb=&b;
c=3*(*pb-*pa);
printf("%d,%d",a,c);
}

Sol.: -20,54

2)
#include<stdio.h>
main()
{
char x='A',y;
char *px=&x,*py;
*px=x+1;
y=*px+1;
printf("%c",y);
}

Sol.: C

3)
#include<stdio.h>
main()
{
int r,*p=NULL;
r=*p++;
printf("%d",r);
}

Sol.: Segmentation fault(Run time error).

4)
#include<stdio.h>
char f(char s,char t)
{
s='B';
t='A';
return s<t?s:t;
}
main()
{
char a='A',b='B',r;
r=f(a,b);
printf("%d,%d,%c",a,b,r);
}

Sol.: 65,66,A

5)
#include<stdio.h>
char f(char *s,char *t)
{
*s='A';
*t='B';
return *s==*t?*s:*t;
}
main()
{
char a='x',b='y',r;
r=f(&a,&b);
printf("%c,%c,%d",a,b,r);
}

Sol.: A,B,66

6)
#include<stdio.h>
#include<malloc.h>
int f(int *p)
{
int i,s=0;
for(i=0;i<sizeof(p);i++)
s+=*(p+i);
return s;
}
main()
{
int i,r;
static int *a;
a=malloc(5*sizeof(int));
for(i=0;i<5;i++)
*(a+i)=i+1;
r=f(a);
free(a);
printf("%d",r);
}

Sol.: 10

7)
#include<stdio.h>
main()
{
int i=65;
int *p;
p=i;
p++;
printf("%d",p);
}

Sol.: 69(NOTE:at p=i warning: assignment makes pointer from integer without a cast [enabled by default]
              at printf("%d",p) warning: format ‘%d’ expects argument of type ‘int’, but argument 2 has type ‘int *’ [-Wformat] )

8)
#include<stdio.h>
main()
{
int *x;
*x=100;
printf("%d",*x);
}

Sol.: Segmentation fault(Run time error)

9)
#include<stdio.h>
void f1(int *k)
{
*k+=15;
}
void f2(int *x)
{
int m=*x,*n=&m;
*n*=10;
}
main()
{
int i=20,*j=&i;
f1(j);
*j+=10;
f2(j);
printf("%d,%d",i,*j);
}

Sol.: 45,45

10)
#include<stdio.h>
main()
{
short int q=351;
unsigned char *p=(unsigned char *)&q;
printf("%d,",*p);
*p+=5;
printf("%d",q);
}

Sol.: 95,356(NOTE: 95=351-256)

11)
#include<stdio.h>
main()
{
float x,*p,*q;
p=q=&x;
q++;
printf("%d",p-q);
}

Sol.: -1

12)
#include<stdio.h>
main()
{
int *p,q=5;
p=&q;
++*p=6;
printf("%d",*p);
}

Sol.: Compile time error(lvalue required as left operand of assignment at ++*p=6)

13)
#include<stdio.h>
main()
{
int a=4,b=5;
void *p=&a,*q=&b;
p++;
printf("%u,%u",p,*q);
}

Sol.: Compile time error

14)
#include<stdio.h>
#include<malloc.h>
main()
{
unsigned int *p,*q;
p=malloc(10);
q=calloc(10,1);
printf("%d,%d",*p,*q);
}

Sol.:0,0

15)
#include<stdio.h>
#include<malloc.h>
void alloc(int *a)
{
a=(int*)malloc(sizeof(int));
}
main()
{
int *a;
alloc(a);
*a=6;
printf("%d",*a);
}

Sol.: Segmentation Fault(Run Time Error)

16)
#include<stdio.h>
main()
{
int *p;
{
int q=2;
p=&q;
}
printf("%d",*p);
}

Sol.: 2

POINTERS(LEVEL-II)

1)What is the difference between following declerations?
int* f()
int (*g)()

Sol.: int* f() is the decleration of a function f that will return integer pointer.
      Whereas int (*g)() is the decleration of a function pointer that will return int.

2)
#include<stdio.h>
void test(void)
{
printf("Algorithmica\n");
}
main()
{
void (*f)(void);
f=test;
(*f)();
f();
}

Sol.: Algorithmica
      Algorithmica

3)
Explain the meaning of the following declerations:
a. double *(*a[10])(double)
b. char (*p)(int(*a)[3])
c. float (*x(int*a))[5]

Sol.: a) it is an array of 10 variable which will contain pointer to a function which will receive double and return double pointer
      b) it is a pointer to a function which will receive adress of an int array of 3 size and will return char
      c)

4)
#include<stdio.h>
void g()
{
printf("hi");
}
void h(void(*f)())
{
printf("%d\n",sizeof(f));
(*f)();
}
main()
{
h(g);
}

Sol.: 4
      hi

5)
#include<stdio.h>
#include<malloc.h>
main()
{
int *a,i;
a=(int*)malloc(5*sizeof(int));
for(i=0;i<5;i++)
*(a+i)=i*i;
for(i=0;i<5;i++)
printf("%d,",*a++);
free(a);
}

Sol: 0,1,4,9,16,

6)
#include<stdio.h>
int *f()
{
int a=6,*b=NULL;
b=&a;
return b;
}
main()
{
int *p=NULL;
p=f();
printf("%d",*p);
}

Sol.: 6

7)
#include<stdio.h>
void f()
{
printf("f");
}
void g()
{
printf("g");
}
void h()
{
printf("h");
}
main()
{
int i;
void f(),g(),h();
_________________
for(i=0;i<=2;i++)
(*t[i])();
}
What statements should be added to the above code so that the output will be fgh?

Sol.: void(*t[3])()={f,g,h};

8)
What does following function do?
int test(unsigned int n)
{
unsigned char* p=(unsigned char*)&n;
return *p&&*(p+1)&&*(p+2)&&*(p+3);
}

Sol.: this function will do && of all the 4 bytes of n and will return the value.

9)
#include<stdio.h>
int m=6;
void foo(int *p)
{
++*p++;
p=&m;
++*p++;
}
main()
{
int n=5;
int *ptr=&n;
foo(ptr);
printf("%d,%d",m,n);
}

Sol.: 7,6(NOTE:here binding will be ++(*(p++)) )

10)
#include<stdio.h>
void f(int *p,int *q)
{
int *temp;
temp=p;
p=q;
q=temp;
}
main()
{
int a=10,b=20;
int *p,*q;
p=&a;
q=&b;
f(p,q);
printf("a=%d,b=%d,*p=%d,*q=%d",a,b,*p,*q);
}

Sol: a=10,b=20,*p=10,*q=20(NOTE: Changes done by f will not get reflected on main)

11)
#include<stdio.h>
void test(unsigned char *p)
{
unsigned char tmp=*p;
*p=*(p+1);
*(p+1)=tmp;
}
main()
{
unsigned short int i=2;
test(&i);
printf("%u",i);
}

Sol.:512(NOTE: short int contain 2 bytes. Here test will swap the position of both bytes)

12)How do you declare an array of 10 pointers to functions returning pointers to functions returning pointers to characters?

Sol.:

13)
Fillup the blank in the following program so that contents of x & y are swapped without using temp variable.

Sol.:
#include<stdio.h>
void swap(int *a,int *b)
{
*a=(*a)^(*b);
*b=(*a)^(*b);
*a=(*a)^(*b);
}
main()
{
int x=4,y=3;
swap(&x,&y);
printf("%d,%d",x,y);
}

14)
#include<stdio.h>
#include<malloc.h>
main()
{
int i,*p;
for(i=0;i<10;i++)
p=malloc(1024);
}

Sol.: This program will execute sucessfully but in every itteration of for loop a new chunk of memory will be allocated from heap segment which will go unused.

-------------------------------------------------------------------------------------------------------------------------------

STRINGS(LEVEL-I)

1)
main()
{
char p[]="java";
char q[]={'j','a','v','a'};
char *r="java";
printf("%d,%d,%d",sizeof(p),sizeof(q),sizeof(r));
}

Sol.: 5,4,4

2)
main()
{
char s[]="c\0java";
char p[]="abc\0\0";
printf("%s,%s",s,p);
}

Sol.: c,abc

3)
main()
{
char s[]="language\0\0\0";
printf("%d,%d",sizeof(s),sterlen(s));
}

Sol.: 12,8

4)
main()
{
char s[]="c language";
printf("%s\n",s+2);
printf("programming"+3);
}

Sol.: language
      gramming

5)
main()
{
char *s="yahoo";
printf("%d,%d",sizeof("yahoo"),sizeof(*s));
}

Sol.: 6,1

6)
main()
{
char a[10];
a="down";
printf("%s",a);
}

Sol.: Error:Lvalue required

7)
main()
{
char str1[]="program";
if(str1=="program")
printf("Equal");
else
printf("Unequal");
}

Sol.: Unequal

8)
main()
{
char *str1="program";
if(str1=="program")
printf("Equal");
else
printf("Unequal");
}

Sol.: Equal

9)
main()
{
char *str1="program";
char *str2="program";
if(str1==str2)
printf("Equal);
else
printf("Unequal");
}

Sol.: Equal

10)
main()
{
char a[3]="xyz";
printf("%s",a);
}

Sol.: Undefined behaviour

11)
main()
{
char s[]="Algorithmica";
printf("%c\n",(*s)++);
printf("%c",++*s);
}

Sol.: AC

12)
main()
{
static char s[]="Algorithmica";
printf("%c",*(s+strlen(s)));
}

Sol.: Not display any thing

13)
main()
{
char str[10];
int i;
for(i=0;i<3;i++)
i[str]='Z'-i;
i[str]='\0';
printf("%s",str);
}

Sol.: ZYX

14)
main()
{
char str[10];
int i=0;
i++[str]='B'+1;
i++[str]='C';
i[str]='\0';
printf("%s",str);
}

Sol.: CC

15)
main()
{
char s[]="Algorithmica";
printf("%s",&s[10]-2);
}

Sol.: mica

16)
main()
{
char *months[4]={ "january",
           "february",
           "march",
           "april"
        };
printf("%c",months[1][2]);
}

Sol.: b

17)
main()
{
char *a="down";
a[0]='D';
printf("%s",a);
}

Sol.: run time error

18) What will be the o/p of following program, if string liteeral "abcdef" starts at address 1000.
main()
{
printf("%u",&"abcdef"+1);
}

Sol.: 1007

19)
main()
{
char a[] ="language";
char p,*s;
s=a;
printf("%c\n",*(s++));
p=*++s;
printf("%c",p);
}

Sol.: l
      n

20)
void f(char a[])
{
a[0]='A';
}
main()
{
f("ada");
}

Sol.: Run Time error

STRINGS(LEVEL-II)

1)
#include<stdio.h>
main()
{
char c;
c="abcdef"[3]-3["abcdef"];
printf("%c",c);
}

Sol.: No output(c will have 0 value)

2)
#include<stdio.h>
main()
{
char *days[4]={"sunday",
        "monday",
        "tuesday",
        "wednesday"
           };
printf("%s,%s",*(days+2),*days+1);
}

Sol.: tuesday,unday

3)a)
#include<stdio.h>
main()
{
char *days[4]={"sunday",
        "monday",
        "tuesday",
        "wednesday"
           };
printf("%s,%s",*(days+2),*days+1);
}

Sol.: Segmentation fault(Run time error)

3)b)
#include<stdio.h>
#include<string.h>
main()
{
char *s="NULL";
if(strcmp(s,NULL))printf("Yes");
else printf("No");
}

Sol.: Segmentation fault

4)
#include<stdio.h>
#include<string.h>
#include<malloc.h>
void alloc(char *s1)
{
s1=malloc(20);
memset(s1,'\0',20);
}
main()
{
char *s1,*s2="down";
alloc(s1);
strcat(s1,s2);
printf("%s",s1);
}

Sol.: Seg. fault

5)
#include<stdio.h>
main()
{
switch("c"=="c")
{
case 0:
   printf("alas!");
   break;
case 1:
   printf("hurray");
   break;
default:
   printf("superb");
}
}

Sol.: hurray

6)Fill up the blank in the following program such it outputs b onto screen.
#include<stdio.h>
void f(char*s)
{
printf("%c",____);
}
main()
{
char a[]="abcdefgh";
f(a+5);
}

Sol.: *(s-4)

7)
#include<stdio.h>
#include<string.h>
void f(char s[])
{
printf("%d,%d\n",sizeof(s),strlen(s));
}
main()
{
char s[]="abcde";
f(s);
f("algorithmica");
}

Sol.: 4,5
      4,12

8)
#include<stdio.h>
#include<string.h>
main()
{
char *str1="Lang";
char str2[20];
char str3[30]="Comp";
if(strcmp(strcat(str3,strcpy(str2,str1)),"Comp.Lang"))
printf("Yes");
else
printf("No");
}

Sol.: Yes

9)
#include<stdio.h>
#include<string.h>
void f(char *s)
{
s="New string";
}
main()
{
char *s="Old string";
f(s);
printf("%s",s);
}

Sol.: Old string

10)
#include<stdio.h>
#include<string.h>
main()
{
char s[][5]={
        {'a','b','c','d'},
        {'w','x','y','z'}
         };
printf("%d",sizeof(s[0]));
}

Sol.: 5

11)
#include<stdio.h>
#include<string.h>
main()
{
char s[][5]={
        {'a','b','c'},
        {'x','y','z'}
         };
printf("%c",**s[1]);
}

Sol.: Compilation error

12)
#include<stdio.h>
#include<string.h>
main()
{
char s1[]="xyz";
char *s2="xyz";
if(s1==s2)
printf("Equal");
else
printf("Unequal");
}

Sol.: Unequal.Change if statement to 'if(strcmp(s1,s2)==0)'

13)
#include<stdio.h>
#include<string.h>
void f(char **days)
{
printf("%c",*(days[2]+2));
}
main()
{
char *days[4]={"sunday",
        "monday",
        "tuesday",
        "wednesday"
           };
f(days);
}

Sol.: e

14)Fill up the blank in the following program such that it will print m as output
#include<stdio.h>
#include<string.h>
void f(char *p[])
{
printf("%c",____);
}
main()
{
char *months[4]={"january",
        "february",
        "march",
        "april"
           };
f(months);
}

Sol.: *(p[2])

15)
#include<stdio.h>
#include<string.h>
main()
{
char bool[][6]={"true","false"};
printf("%s",bool[(unsigned)-1 == ~0]);
}

Sol.: false

16)
#include<stdio.h>
#include<string.h>
int f(char *str)
{
int i,j,len;
len=strlen(str);
for(i=0;i<len;i++)
{
char tmp;
str[i]^=str[len-i-1];
str[len-i-1]^=str[i];
str[i]^=str[len-i-1];
}
printf("%s",str);
}
main()
{
char s[]="ADA";
f(s);
}

Sol.: A(Note: D& A(first) will xor with itself and create '\0'and A which is printed is the last A which after final itteration comes to 1st position)

17)
#include<stdio.h>
#include<string.h>
char* f(void)
{
char *s="c++";
return s;
}
main()
{
char *p="java";
p=f();
printf("%s",p);
}

Sol.: c++

18)
#include<stdio.h>
#include<string.h>
main()
{
char *days[]={"sunday","monday","tuesday"};
printf("%d,%d,%d",sizeof(days),sizeof(days[0]),sizeof(*days[0]));
}

Sol.: 24,4,1(Note:tuesday will determine the size of each array as it is the longest)

19)
#include<stdio.h>
#include<string.h>
char* f(void)
{
char s[]="c";
return s;
}
main()
{
char *p="java";
p=f();
printf("%s",p);
}

Sol.: Garbage output

----------------------------------------------------------------------------------------------------------------------------------

INPUT & OUTPUT(LEVEL-I)

1)
#include<stdio.h>
main()
{
int i=10,b=20
printf("%d");
}

Sol.: compiler error or unpredictable output(depend upon compiler)

2)
#include<stdio.h>
main()
{
int a;
char b;
scanf("%d%c",&a,&b);
printf("a=%d,b=%c",a,b);
}

Sol: a will take the first integer entered through keyboard and b will have white space character

3)
What is the o/p of the following program if the i/p provided is 25 a from keyboard
#include<stdio.h>
main()
{
int a;
char b;
scanf("%d\n%c",&a,&b);
printf("a=%d,b=%c",a,b);
}

Sol.: a=25,b=a

4)
#include<stdio.h>
main()
{
int i;
for(i=8;i<12;i++)
printf("%2d",i);
}

Sol.: 8 91011

5)How do you give input to the program sothat it reads and displays the date properly.
#include<stdio.h>
main()
{
int d,m,y;
scanf("%d-%d-%d",&d,&m,&y);
printf("%d-%d-%d",d,m,y);
}

Sol.: 6-6-1986

6)What does the scanf function in the following program trying to do?
#include<stdio.h>
main()
{
char buf[50];
scanf("%[^\n]",buf);
printf("%s",buf);
}

Sol.: It will scan with all the white character

7)
#include<stdio.h>
main()
{
printf("%d"+1,245);
}

Sol.: d

8)
#include<stdio.h>
main()
{
int r;
r=printf("stupid")+printf("program");
printf("%d",r);
}

Sol.: stupidprogram13

9)
#include<stdio.h>
main()
{
int i,j;
if(printf("%d",scanf("%d%d",&i,&j)))
printf("Yes");
else
printf("No");
}

Sol.: It will take input two int through keyboard and will display 2Yes.

10)What is the difference betwee the o/p of the following code snippets?
a)#include<stdio.h>
main()
{
printf("%10s","expert");
}

b)
#include<stdio.h>
main()
{
printf("%10.3s","expert");
}

Sol.: a)    expert, b)       exp

11)
The following program is trying to read an integer and a character from keyboard.Does it correctly reads intented input? If not,why so?
main()
{
int a;
char b;
scanf("%d%c",&a,&b);
}

Sol.: No,after entering the first value, we will press any whitespace character(,i.e.,space,enter,tab). b will contain the value of that character.

12)
How will you print % character using printf statement only?

Sol.: #include<stdio.h>
main()
{
printf("%%");
}

13)
What wiil be the o/p of the following program if the given i/p is: 12345 67
#include<stdio.h>
main()
{
int a,b;
scanf("%3d%4d",&a,&b);
printf("a=%d,b=%d",a,b);
}

Sol.:123,45

14)
#include<stdio.h>
main()
{
float a=135.23567,b=233.45;
printf("%3.2f,%.4f",a,b);
}

Sol.: 135.24,233.4500

15)The following is a sample C program to read and print an integer. But it is not working properly.What is(are) the mistake(s)?
#include<stdio.h>
main()
{
int n;
printf("Enter a number:\n");
scanf("%d\n",&n);
printf("n=%d",n);
}

Sol.: Program will run correctly only it will read the first whitespace character. In order to correct it remove \n from scanf.

16)
#include<stdio.h>
main()
{
char q='A';
printf("%d",printf("%c\n",q));
}

Sol. A
     2

17)
#include<stdio.h>
main()
{
char *p=NULL;
printf("%s",p);
}

Sol.:(null)

18)
#include<stdio.h>
#include<string.h>
main()
{
FILE *in,*out;
int x='a';
out=fopen("sample","w");
putc(toupper(x),out);
fclose(out);
in=fopen("sample","r");
printf("%c",getc(in));
fclose(in);
}

Sol.: A

19)#include<stdio.h>
main()
{
FILE *in,*out;
int count=0;
out=fopen("sample","w");
fprintf(out,"%s","one");
fclose(out);
in=fopen("sample","r");
while(!feof(in))
{
fgetc(in);
count++;
}
fclose(in);
printf("%d\n",count);
}

Sol.: 4

INPUT & OUTPUT(LEVEL-II)

1)#include<stdio.h>
main()
{
char c;
while((c=getchar())!=EOF)
putchar(c);
};

Sol.: This program will take acharecter and display it. This process will be continued until ctrl+c is pressed.

2)The following program generates 100 unique filenames i.e.,file0 - file99 and stores them in fname array. Fill up the missing logic.
#include<stdio.h>
#include<string.h>
main()
{
int i;
char *c;
char fname[100][10];
for(i=0;i<100;i++)
{
c[0]=i+0;
c[1]='\0';
strcpy(fname[i],strcat("file",c));
}
for(i=0;i<100;i++)
printf("\n%s",fname[i]);
}

3)
#include<stdio.h>
main()
{
int k,max=4;
char *s="Algorithmica";
printf("%.*s",max,s);
}

Sol.: Algo

4)
#include<stdio.h>
main()
{
int p;
if(printf("%d"+2,scanf("%d",&p)))
printf("Yes");
else
printf("No");
}

Sol.: No (it will first waight for an i/p through keyboard)

5)What is the o/p of following program for the input: 10
                              xy
#include<stdio.h>
main()
{
int a;
char b[15];
scanf("%d",&a);
gets(b);
printf("%d,%s",a,b);
}

Sol.: After entering 10 when we will press enter key gets(b) will feed enter value in b and will display value 10 insted if we will give i/p like 10xy then it will show 10,xy

6)
#include<stdio.h>
main()
{
int x=0;
for(;x<20 && printf("%d ",x);x++)
{
switch(x)
{
   case 0: x++,x*=2;
   case 20: x+=2;
   case 70: x+=6;
   default: x+=3;
            break;
}
}
}

Sol.: 0 14 18

7)
#include<stdio.h>
main()
{
char c;
while(c=getchar()!='a')
printf("%d ",c);
}

Sol.: Here if we give any input accept a and press enter it will print 1 1 (1 for the character and 1 for enter) when we will give a as input it will get outside the while loop

8)What is the output of the following, if i/p provided is: Life is beautiful
#include<stdio.h>
main()
{
char dummy[80];
printf("Enter a string:\n");
scanf("%[^a]",dummy);
printf("%s\n",dummy);
}

Sol.: Life is be

9)Predict the o/p of the code with the following inputs:
a)2c3
b)2 5
#include<stdio.h>
main()
{
int i=0,j=4;
scanf("%dc%d",&i,&j);
printf("%d %d",i,j);
}

Sol.: a)2 3
      b)2 4

10)What is the condition so that the following code snippet prints HelloWorld?
#include<stdio.h>
main()
{
if(_________)
printf("Hello");
else
printf("World");
}

Sol.: !(printf("Hello"))

11)
#include<stdio.h>
main()
{
printf("%s%s"+2,"Hello","World",2);
}

Sol.: Hello

12)
#include<stdio.h>
#include<malloc.h>
main()
{
int i;
int size=3;
int total=0;
int numbers[]={001,010,014};
for(i=0;i<size;i++)
total+=numbers[i];
toString(total);
}
int toString(int i)
{
char* temp=(char*)malloc(sizeof(int)*3);
sprintf(temp,"%d",i);
printf("%s",temp);
return i;
}

Sol.: 21(NOTE:When the no. starts from 0 then it is octal representation)

13)
Which of the following statements about the display function is/are true? Assume that message is'\0' terminated and memory allocated for it.
int display(const char *message)
{
int count=0;
count=printf(message);
return count;
}

Sol.: I)Returns the numbers of characters that were printed successfully.

14)
#include<stdio.h>
main()
{
FILE *out1,*out2,*in;
char tmp[20];
out1=fopen("sample","w");
out2=fopen("sample","w");
fprintf(out1,"%s","one");
fprintf(out2,"%s","two");
fclose(out1);
fclose(out2);
in=fopen("sample","r");
while(!feof(in))
{
fscanf(in,"%s",tmp);
printf(tmp);
}
}

Sol.: two(NOTE:The Sequence depend on the order of fclose)

15)
#include<stdio.h>
#include<string.h>
#define MAX 50
main()
{
FILE *fp1,*fp2;
int count=0;
char *this1,*this2;
fp1=fopen("ip1.txt","r");
fp2=fopen("ip2.txt","r");
while((getline1(this1,fp1)!=0) && (getline1(this2,fp2)!=0))
{
if(!strcmp(this1,this2))
{
   count++;
   continue;
}
else
   break;
}
printf("%d",count);
fclose(fp1);
fclose(fp2);
}
int getline1(char *line,FILE *fp)
{
if(fgets(line,MAX,fp)==NULL)
return 0;
else
return strlen(line);
}

Sol.: Segmentation Fault in getline1

16)
#include<stdio.h>
main()
{
FILE *out,*in1,*in2;
char a,b;
int count=0;
out=fopen("sample","w");
fprintf(out,"%s","one");
fclose(out);
in1=fopen("sample","r");
in2=fopen("sample","r");
while(!feof(in1) && !feof(in2))
{
a=fgetc(in1);
b=fgetc(in2);
count++;
}
fclose(in1);
fclose(in2);
printf("%d\n",count);
}

Sol.: 4

----------------------------------------------------------------------------------------------------------------------------------

Structures & Union(level-I)

1)
#include<stdio.h>
struct book_s
{
int bno;
char bname[20];
};
typedef struct book_s book;
main()
{
book b1={100,"java"},*pb1;
pb1=&b1;
printf("%c,%s",pb1->bname[1],pb1->bname+2);
}

Sol.: a,va

2)
#include<stdio.h>
typedef struct student_s
{
int sno;
char sname[20];
char *cname;
}student;

main()
{
student s={111,"xyz","stanford"},*ps;
ps=&s;
printf("%s\n",(*ps).sname+1);
ps->cname++;
printf("%s",ps->cname+3);
}

Sol.: yz
      ford

3)
#include<stdio.h>
#include<malloc.h>
struct node_s
{
int element;
struct node_s* next;
};

main()
{
struct node_s *p,*q;
p=malloc(sizeof(*p));
q=malloc(sizeof(*p));
p->element=10;
p->next=q;
q->element=-15;
q->next=NULL;
printf("%d\n",p->element);
p=p->next;
printf("%d",p->element);
}

Sol.: 10
      -15

4)
#include<stdio.h>
#include<malloc.h>
struct test
{
int a;int b;
};
void f(struct test t)
{
t.a=10;
t.b=15;
}
main()
{
struct test t={4,5};
f(t);
printf("%d,%d",t.a,t.b);
}

Sol.: 4
      5

5)
#include<stdio.h>
struct circle
{
float r;
};
void f(struct circle *t)
{
t[0].r=12.3;
t[1].r=13.5;
}
main()
{
struct circle t[]={4.5,5.0};
f(t);
printf("%f,%f",t[0].r,t[1].r);
}

Sol.: 12.30000,13.50000

6)
#include<stdio.h>
struct test
{
int a; int b;
};
void f(struct test *t)
{
t->a=10; t->b=15;
}
main()
{
struct test t={4,5};
f(&t);
printf("%d,%d",t.a,t.b);
}

Sol.: 10,15

7)
#include<stdio.h>
union test{short int a;char b;};
main()
{
union test u;
u.a=258;
u.b=10;
printf("%hd",u.a);
}

Sol.: 266(this is due to 256+10)

8)#include<stdio.h>
union register1 {short int a;char b[2];};
void f(union register1* pu)
{
printf("%d,%d",pu->b[1],pu->b[0]);
}
main()
{
union register1 u;
u.a=300;
f(&u);
}

Sol.: 1,44 (due to 300=256+44)

9)
#include<stdio.h>
struct example {int a;char b;};
main()
{
struct example *s;
s->a=10;
s->b='A';
printf("%d,%c",s->a,s->b);
}

Sol.: Segmentation Fault(Run time error)

10)Does the program generates any error? If yes, what changes do you suggest in the program to remove them? If no,what is the o/p?
#include<stdio.h>
struct student
{
int sno;
char sname[20];
struct date d;
};
struct date { int day;int month;int year;};
main()
{
struct student s={121,"xyz",15,8,1947};
printf("%d/%d/%d",s.d.day,s.month,s.year);
}

Sol.: yes,the suggest changes are:
#include<stdio.h>
struct date { int day;int month;int year;};
struct student
{
int sno;
char sname[20];
struct date d;
};
main()
{
struct student s={121,"xyz",15,8,1947};
printf("%d/%d/%d",s.d.day,s.d.month,s.d.year);
}

11)Which of the following decleration of structure is legal? why?
a)struct node1
{
int element;
struct node1 next;
};
b)struct node2
{
int element;
struct node2 *next;
};

Sol.: b) (Note:in a) field ‘next’ has incomplete type,hence it will give compile time error)

12)Is the following code decleration legal? If not, what changes do you suggest to make it legal?
typedef struct node
{
int element;
listnode *next;
}listnode;

Sol.: no it is not legal. It will give compile time error.The correct way is:
typedef struct node
{
int element;
struct node *next;
}listnode;

13)Does the following program generates any error? If Yes, what changes do you suggest to remove remove them?
#include<stdio.h>
main()
{
typedef struct point_s point;
point p={4,5};
struct point_s{int x;int y;};
printf("%d,%d",p.x,p.y);
}

Sol.: Yes it will generate error. Changes done are:
#include<stdio.h>
main()
{
struct point_s{int x;int y;};
typedef struct point_s point;
point p={4,5};
printf("%d,%d",p.x,p.y);
}

14)
#include<stdio.h>
main()
{
struct stud
{
char *name;
int rollno;
};
struct subject
{
struct stud s;
char *sub[3];
};
struct subject s1={"xyz",123,"C","DS","C++"};
printf("%s,%s",s1.s.name+1,s1.sub[1]+1);
}

Sol.: yz,S

15)
#include<stdio.h>
typedef union
{
struct
{
char c1,c2;
}s;
unsigned int t;
float f;
}u;
main()
{
u sample;
sample.s.c1='A';
sample.s.c2='B';
sample.t=5;
printf("%d,%d,%d",sample.s.c1,sample.s.c2,sample.t);
}

Sol.: 5,0,5

16)
#include<stdio.h>
struct student_s
{
char *name;
int rollno;
int age;
};
main()
{
struct student_s s={"xyz"};
printf("%d,%d",s.rollno,s.age);
}

Sol.: 0,0(NOTE: if a structure is partially initialised then all of its uninitialised value becomes 0)

17)
#include<stdio.h>
struct tag
{
int i;
union
{
int a;
int b;
}c;
}a;
main()
{
a.c.a=10;
printf("%d",a.c.b);
}

Sol.: 10

18)
#include<stdio.h>
typedef struct s{ int a[10];}s;
typedef struct t{ int *a;}t;
main()
{
s s1,s2;
t t1;
s1.a[0]=0;
t1.a=s1.a;
s1.a[0]++;
s2=s1;
s1.a[0]++;
printf("%d,%d,%d",s1.a[0],s2.a[0],t1.a[0]);
}

Sol.: 2,1,2

19)
#include<stdio.h>
struct sample
{
int x;
int y[3];
};
main()
{
struct sample s={10,12,-8,11};
printf("%d",s.x+s.y[1]);
}

Sol.: 2(i.e.,10-8)

20)
#include<stdio.h>
struct test
{
int a;
int b;
};
struct test f(void)
{
struct test t;
t.a=20;
t.b=25;
return t;
}
main()
{
struct test t={4,5};
t=f();
printf("%d,%d",t.a,t.b);
}

Sol.: 20,25

Structures & Union(level-II)

1)
#include<stdio.h>
struct test
{
int a;
int b[3];
int *c;
};
main()
{
struct test p={0,1,2,3},*q;
q=&p;
q->c=&q->b[2];
printf("%d",++*q->c);
}

Sol.: 4

2)
#include<stdio.h>
struct book
{
int n;
char* bname;
};
main()
{
struct book b={100,"java"};
struct book *p=&b;
printf("%c\n",*p->bname++);
printf("%c\n",(*p->bname)++);
}

Sol.: j
      Segmentation fault(Run Time error as ++ operator can't be applied to constants)

3)
#include<stdio.h>
struct s1
{
int a;
char b;
float c;
char d;
};
main()
{
printf("%d",sizeof(struct s1));
}

Sol.: 16(i.e.,4*4)

4) Suggest an ordering for members of structure s such that the wastage of storage space will be minimal.

Sol.: Ordering should be done in the increasing order of the storage space of the particular element

5)
#include<stdio.h>

main()
{
struct s2
{
int a;
char b;
double c;
};
struct s2 *ps1;
printf("%d",sizeof(*ps1));
}

Sol.: 16(4+4+8)

6)
#include<stdio.h>
void f(char **s)
{
*s="java";
}
main()
{
struct s
{
int s;
char *t;
}s1={200,"c++"},*ps1=&s1;
f(&s1.t);
printf("%c",ps1->t[1]);
}

Sol.: a

7)
#include<stdio.h>
struct node
{
int value;
struct node *next;
}
main()
{
struct node *k,*m;
k=malloc(sizeof(struct node));
m=malloc(sizeof(int));
printf("%d,%d",sizeof(k),sizeof(m));
}

Sol.: 8,8

8)
#include<stdio.h>
struct point
{
int x;
int *y;
};
main()
{
int a=-3,b=-4;
struct point p[2];
struct point *ps=p;
p[0].x=3;p[0].y=&a;
p[1].x=4,p[1].y=&b;
printf("%d\n",ps++->x);
printf("%d",++(*ps->y));
}

Sol.: 3 (Note:this 3 corresponds to p[0].x)
      -3 (Note:this -3 corresponds to b+1 )

9)
#include<stdio.h>
struct test
{
int x;
char *y;
}
s[2][2]={1,"sun",2,"jan",3,"sep",4,"moon"};
main()
{
char *t;
__________
printf("%s",t);
}
Fill up the blank in the above program so that it prints only oon as output.

Sol.: t=s[1][1].y+1;

10) Write a macro to compute the size of any type t without using sizeof operator

Sol.:
#include<stdio.h>
#define object_size(t) ((char *)(&(t) + 1) - (char *)&(t))
main()
{
int i;
struct s { int x; char y;}s1;
printf("%d,%d",object_size(i),object_size(s1));
}

12)Write a macro to compute the offset of any member m of structure type t.

Sol.: #define offsetof(m,t) ((size_t)((char *)&((t *)0)->m-(char *)0))

13)
#include<stdio.h>
void f(void) {printf("f");}
void g(void) {printf("g");}
struct test
{
int a;
int b;
void (*t[2])(void);
};
main()
{
int i;
struct test q,*p=&q;
___________
___________
for(i=2;i>=1;i--)
___________
}
Write C statements in the above program to initialize the array of function pointers in a structure and to call the functions through these function pointers such that the output will be fg.

Sol.:
#include<stdio.h>
void f(void) {printf("f");}
void g(void) {printf("g");}
struct test
{
int a;
int b;
void (*t[2])(void);
};
main()
{
int i;
struct test q,*p=&q;
q.t[1]=f;
q.t[0]=g;
for(i=2;i>=1;i--)
p->t[i-1]();
}

14)In the following program, union u consists of different types. It is trying an approach through which it knows the type of member stored in that union at any time. Do you think it is correct approach? If not, explain the flaw in that approach and suggest a correct way of doing it.
union u
{
int type;//0 if char is stored, 1 if integer, 2 if float
char c;
int a;
float b;
};

---------------------------------------------------------------------------------------------------------------------------------

Bitwise trick(LEVEL-I)

1) Write an equivalent expression for x%32 using only bitwise operators?

Sol.:
#include<stdio.h>
main()
{
int i;
scanf("%d",&i);
i=i&31;
printf("%d",i);
}

2)What does the following function do?
unsigned findwhat(unsigned int x,int p,int n)
{
return((x>>(p+1-n))&~(~0<<n);
}

Sol.: this function will divide x by 2^(p+1-n) and then do % of it with 2^(n+1).

3)
#include<stdio.h>
void fun(int a)
{
printf("A");
if(!(a&(a-1))) return;
fun(a & (a-1));
}
main()
{
fun(0xA35F);
}

Sol.:AAAAAAAAAA(At every call of fun the least most significant 1 from a will be deleted).

4)
Write a C function that rounds the given integer to next higher multiple of 8 and returns the value.

Sol.:
int HigherMult*(int n)
{
return (n+7)& -8;
}

5) Write a C function to check whether a given unsigned integer is having zero bytes(,i.e., the value of any byte is zero) or not? Your function should return 1 true, or 0 otherwise.

Sol.:
int HasZeroByte(int n)
{
while(n)
{
if((n&1)==0)
return(0);
n>>=1;
}
return(1);
}

BITWISE TRICKS(LEVEL-II)

1) Write a C function to set 5th and 7th bits of given integer n and returns the resultant n value? Your function should not assume specific size for integer.

Sol.:
int setBits(int n)
{
n|=(1<<5);
n|=(1<<7);
return n;
}

2)Which one of the following is true about the following function?
unsigned test2(int x)
{
x|=(x>>1);
x|=(x>>2);
x|=(x>>4);
x|=(x>>8);
x|=(x>>16);
return x+1;
}

Sol.: b)returns the lest power of 2 greater than x.

3)Write a C function that returns the position of the rightmost one in a given unsigned integer.

Sol.:
#include<stdio.h>
#include<math.h>
unsigned RightmostOne(unsigned n)
{
return log2(n&-n)+1;
}

2012-08-27T12:08:00.003-07:00

Write an efficient function that deletes all the characters from string src which are given in string remove. Your function should not change the relative order of characters in src. What are the time and space complexities of your solution?

consider function type

void RemoveChar(char [] src, char []remove)

Input: Gopipodila<-src

ila <-remove

Output: Goipodi <-src

Algorithm:

im using hashtable and for storing data from remove string then for every charcter in src string whether charecter is present in hash table or not if it is avilable skip that charcter

Code:

#include<stdio.h>
#include<malloc.h>
#include<string.h>
#define INIT_SIZE 26
#define LF 10
struct ListNode
{
int data;
struct ListNode *next;
};
struct HashTbl
{
int size;
int nelement;
struct ListNode **array;
};
struct HashTbl* createHashTable();
int add(struct HashTbl*,int);
int ListInsert(struct ListNode*,int);
void Rehash(struct HashTbl*);
void insert(struct ListNode*,struct ListNode*);
int find(struct HashTbl*,int);
int ListFind(struct ListNode*,int);
int removelst(struct HashTbl*,int);
int ListDelete(struct ListNode*,int);

struct HashTbl* createHashTable()
{
int i;
struct ListNode *temp;
struct HashTbl *htbl;
htbl=(struct HashTbl*) malloc(sizeof(struct HashTbl));
htbl->nelement=0;
htbl->size=INIT_SIZE;
htbl->array=(struct ListNode**) malloc(INIT_SIZE * sizeof(struct ListNode*));
for(i=0;i<INIT_SIZE;i++)
{
temp=(struct ListNode*) malloc(sizeof(struct ListNode));
temp->next=NULL;
htbl->array[i]=temp;
}
return(htbl);
}

int add(struct HashTbl *htbl,int key)
{
int pos;
if((htbl->nelement)/(htbl->size)>LF)
Rehash(htbl);
pos=hash(key,htbl->size);
if(ListInsert(htbl->array[pos],key))
{
(htbl->nelement)++;
return(1);
}
else
return(0);
}
int hash(int k,int s)
{
return(k%26);
}

int ListInsert(struct ListNode *p,int k)
{
struct ListNode *current,*temp;
current=p;
while(current->next!=NULL)
{
if(current->data==k)
   return(0);
current=current->next;
}
if(current->data==k)
return(0);
temp=(struct ListNode*) malloc(sizeof(struct ListNode));
temp->data=k;
temp->next=NULL;
current->next=temp;
return(1);
}

void Rehash(struct HashTbl *htbl)
{
int i,newsize,pos;
struct ListNode *temp,**newarray,*current;
printf("\nrehashing\n");
newsize=htbl->size*2;
newarray=(struct ListNode**) malloc(newsize * sizeof(struct ListNode*));
for(i=0;i<newsize;i++)
{
temp=(struct ListNode*) malloc(sizeof(struct ListNode));
temp->next=NULL;
newarray[i]=temp;
}
for(i=0;i<htbl->size;i++)
{
for(current=htbl->array[i];current->next!=NULL;)
{
   temp=current->next;
   current->next=temp->next;
   pos=hash(temp->data,newsize);
   insert(newarray[pos],temp);
}
}
htbl->size=newsize;
free(htbl->array);
htbl->array=newarray;
}

void insert(struct ListNode *a,struct ListNode *t)
{
t->next=a->next;
a->next=t;
}

int find(struct HashTbl *htbl,int key)
{
int pos;
pos=hash(key,htbl->size);
return(ListFind(htbl->array[pos],key));
}

int ListFind(struct ListNode *a,int k)
{
struct ListNode *current=a->next;
while(current!=NULL)
{
if(current->data==k)
   return(1);
current=current->next;
}
return(0);
}

int removelst(struct HashTbl *htbl,int key)
{
int pos;
pos=hash(key,htbl->size);
return ListDelete(htbl->array[pos],key);
}

int ListDelete(struct ListNode *a,int k)
{
struct ListNode *current,*prev;
prev=a;
current=a->next;
while(current!=NULL)
{
if(current->data==k)
{
   prev->next=current->next;
   free(current);
   return(1);
}
current=current->next;
prev=prev->next;
}
return(0);
}

void main()
{
int i,l=0;
char a[20],b[5],c[20];
struct HashTbl *head;
head=createHashTable();
printf("\nHash Table created\n");
printf("\nEnter string A:");
scanf("%s",a);
for(i=0;i<strlen(a);i++)
add(head,a[i]);
printf("\nEnter the string B:");
scanf("%s",b);
for(i=0;i<strlen(b);i++)
{
if(find(head,b[i])==1)
   removelst(head,b[i]);
}
for(i=0;i<strlen(a);i++)
{
if(find(head,a[i])==1)
{
   c[l]=a[i];
   l++;
}
}
c[l]='\0';
printf("\nString after modification: %s\n",c);
}

convert Bst to sorted circular linked list

2012-08-27T12:00:00.001-07:00

Write an efficient function to convert binary search tree into sorted circular doubly linked list. Your function should return the address of first node of resultant list. What are the time and space complexities of your solution?

note:
im building tree here statically

code:
#include<stdio.h>
#include<malloc.h>
struct node
{
    struct node *prev;
int data;
   struct node *next;
};
struct tree
{
    struct tree *left;
int data;
    struct tree *right;
};
void append(struct tree**,int);
void treelink(struct tree*,struct node*);
void insertlink(struct node*,int);
void display(struct node*);

void main()
{
    struct tree *head;
struct node *headlink;
    headlink=(struct node*) malloc(sizeof(struct node));
headlink->prev=headlink;
    headlink->next=headlink;
head=NULL;
    append(&head,18);
   append(&head,23);
   append(&head,12);
   append(&head,8);
   append(&head,15);
   append(&head,22);
   append(&head,5);
   append(&head,1);
   append(&head,45);
   append(&head,35);
   append(&head,6);
   append(&head,19);
   append(&head,33);
   append(&head,26);
   append(&head,14);

   treelink(head,headlink);
   display(headlink);

}
void display(struct node *q)
{
    struct node *current=q->next;
while(current!=q)
    {
   printf(" %d ",current->data);
        current=current->next;
   }
}

void append(struct tree **q,int n)
{
   struct tree *current,*temp,*parent;
   if(*q==NULL)
{
   temp=(struct tree*) malloc(sizeof(struct tree));
   temp->data=n;
   temp->left=NULL;
           temp->right=NULL;
          *q=temp;
       }
else
   {
   current=*q;
    temp=(struct tree*) malloc(sizeof(struct tree));
   temp->data=n;
      temp->left=NULL;
      temp->right=NULL;
   while(current!=NULL)
      {
            parent=current;
if(n<=current->data)
            current=current->left;
          else
      current=current->right;
}
     if(n<=parent->data)
   parent->left=temp;
     else
parent->right=temp;
      }
}

void treelink(struct tree *q,struct node *a)
{
if(q==NULL)
return;
treelink(q->left,a);
insertlink(a,q->data);
   treelink(q->right,a);
}

void insertlink(struct node *a,int i)
{
   struct node *current=a,*temp;
   while(current->next!=a)
   current=current->next;
   temp=(struct node*) malloc(sizeof(struct node));
   current->next=temp;
   temp->data=i;
   temp->prev=current;
   temp->next=a;
   a->prev=temp;
}

lca in binary tree

2012-08-27T11:53:00.000-07:00

Least Common Ancestor(LCA) in a tree is defined as the first node that comes common for both the given nodes while travelling towards the root. Write an efficient function “TreeNode FindLca(TreeNode t, TreeNode p, TreeNode q)” to find the least common ancestor of nodes p and q in a binary tree t. You are not allowed to modify the structure of tree node.

What are the time and space complexities of your solution? Assume that p and q points to valid nodes in a given binary tree.

Input:

3                                      p=ADDR(9) q=ADDR(8)
    / \
4    7
/ \    / \
5 1 6   8
       /
     9

Return value: ADDR(7)

Code:

#include<stdio.h>
#include<conio.h>
#include<malloc.h>

struct treenode
{
   int data;
   struct treenode *left;
   struct treenode *right;
} *root=NULL;
//building tree

struct treenode * buildtree()
{
   struct treenode *temp1,*temp2,*temp3,*temp4;
   root=(struct treenode *)malloc(sizeof(struct treenode));
    temp1=(struct treenode *)malloc(sizeof(struct treenode));
    temp2=(struct treenode *)malloc(sizeof(struct treenode));
   temp3=(struct treenode *)malloc(sizeof(struct treenode));
   temp4=(struct treenode *)malloc(sizeof(struct treenode));
     root->data=4;
     root->left=temp1;
     temp1->data=5;
     root->right=temp2;
     temp2->data=6;
     temp1->left=temp3;
     temp3->data=7;
     temp3->left=temp3->right=NULL;
     temp1->right=temp4;
     temp4->data=8;
     temp4->left=temp4->right=NULL;
      temp2->left=temp2->right=NULL;
printf("tree buildedsuccessfull");
       return root;
}

//this for given value find address which having those values
struct treenode * findaddress1(struct treenode *root,int i)
{
   if(root==NULL) return NULL;
   if(root->data==i) return root;
   root->left=findaddress1(root->left,i);
   root->right=findaddress1(root->right,i);
   if(root->left!=NULL)
   return root->left;
   else if(root->right!=NULL)
   return root->right;
   else return NULL;

}

//finding lca
struct treenode * lcabinary(struct treenode *root,int i,int j)
{
   struct treenode *rh,*lh;
   if(root==NULL) return NULL;

   if(root->data==i) return root;
   if(root->data==j) return root;

   lh=lcabinary(root->left,i,j);
   rh=lcabinary(root->right,i,j);


   if(lh!=NULL&&rh!=NULL) return root;
   else if(lh!=NULL) return lh;
   else if(rh!=NULL) return rh;
   else if(lh==NULL&&rh==NULL) return NULL;


}
void preorder(struct treenode *root)
{
   if(root==NULL)
   return;

   printf("%d|%u     ",root->data,root);//this for our reference
   preorder(root->left);
   preorder(root->right);
}

void main()
{
   int i,k=1,j;
   struct treenode *root,*temp1,*temp2,*temp;
   while(k)
   {
       printf("\n enter 1 to build tree");
       printf("\n enter 2 to print in preorder");
       printf("\n enter 3 to find lca");
       scanf("%d",&k);
       switch(k)
       {
           case 1:
               root=buildtree();
               break;

           case 2:
               preorder(root);
               break;

           case 3:

               printf("enter value of address 1");
               scanf("%d",&i);
                   printf("enter value of address 2");
               scanf("%d",&j);
                temp=lcabinary(root,i,j);
                printf("lca is %u and data %d ",temp,temp->data);
           }
       }
}

Finding max levelsum in binary tree

2012-08-27T11:42:00.003-07:00

Given a binary tree, write an efficient function that returns the maximum level sum. If tree is empty return 0. Assume that all values in binary tree are positive integers. What are the space and time complexities of your solution?

Input:
3

/ \

4 7

/ \ / \

5 1 1 2

Return value: 11

Code:

#include<stdio.h>
#include<conio.h>
#include<malloc.h>
#define QUEUESIZE 20

struct treenode
{
    int data;
    struct treenode *left;
    struct treenode *right;
} *root=NULL;

int *queue[QUEUESIZE];
int rear=-1,front=-1;
//queue finctions

void enque(struct treenode *root)
{
    if(rear<QUEUESIZE)
    {
        queue[++rear]=root;

        }
        else printf("queue overflow");
    }

int deque()
{
    int i;
    if(front!=rear)
    i=queue[++front];
    else printf("underfolw");
    return i;
}

int isqempty()
{
    if(front==rear)
    return 1;

    return 0;
}

//for building tree statically
struct treenode * buildtree()
{
    struct treenode *temp1,*temp2,*temp3,*temp4,*temp5,*temp6;
    root=(struct treenode *)malloc(sizeof(struct treenode));
    temp1=(struct treenode *)malloc(sizeof(struct treenode));
       temp2=(struct treenode *)malloc(sizeof(struct treenode));
        temp3=(struct treenode *)malloc(sizeof(struct treenode));
         temp4=(struct treenode *)malloc(sizeof(struct treenode));
          temp5=(struct treenode *)malloc(sizeof(struct treenode));
           temp6=(struct treenode *)malloc(sizeof(struct treenode));

    root->data=4;
     root->left=temp1;
     temp1->data=5;
     root->right=temp2;
     temp2->data=6;
     temp1->left=temp3;
     temp3->data=7;
     temp3->left=temp3->right=NULL;
     temp1->right=temp4;
     temp4->data=8;
     temp4->left=NULL;
     temp4->right=NULL;
      temp2->left=temp5;
      temp5->data=6;
      temp2->right=temp6;
      temp6->data=10;
      temp5->left=temp5->right=temp6->left=temp6->right=NULL;
printf("\n tree buildedsuccessfull");
       return root;
}
//levelsum

void printlevel(struct treenode *root)
{
    int sum=0,maxsum=root->data;
    struct treenode *temp;
    enque(root);
    enque(NULL);
    temp=deque();
    while(!isqempty())
    {
        sum=0;
        while(temp!=NULL)
        {
            sum=sum+temp->data;
        if(temp->left!=NULL)
        enque(temp->left);

        if(temp->right!=NULL)
        enque(temp->right);

    //    printf("%d   ",temp->data);
        temp=deque();

         }
         if(maxsum<sum)
         maxsum=sum;

         if(temp==NULL)
        {
        enque(NULL);
        temp=deque();
        }
        }
        printf("\n maximum sum is %d",maxsum);
}

//main

void main()
{
    struct treenode *root;
    int i=1;
    while(i)
    {
        printf("\n enter 1 to build elements");
        printf("\n enter 2 print elements in level wise");
        scanf("%d",&i);

    switch(i)
    {
    case 1:
        root=buildtree();
        break;

    case 2:
        printlevel(root);
        break;
    }
     }
}

2012-08-27T11:37:00.003-07:00

Write an efficient function that takes a binary tree as input and displays the elements of tree level by level, but from last level to first. What are the time and space complexities of your solution?

Input:

3

/ \

4 7

/ \ / \

5 1 6 8

Output: 5 1 6 8 4 7 3

code:

#include<stdio.h>
#include<conio.h>
#include<malloc.h>
#define QUEUESIZE 20

struct treenode
{
    int data;
    struct treenode *left;
    struct treenode *right;
} *root=NULL;

//queue functions
int *queue[QUEUESIZE];
int rear=-1,front=-1;

void enque(struct treenode *root)
{
    if(rear<QUEUESIZE)
    {
        queue[++rear]=root;

        }
        else printf("queue overflow");
    }

int deque()
{
    int i;
    if(front!=rear)
    i=queue[++front];
    else printf("underfolw");
    return i;
}

int isqempty()
{
    if(front==rear)
    return 1;

    return 0;
}

struct treenode * buildtree()
{
    struct treenode *temp1,*temp2,*temp3,*temp4,*temp5,*temp6;
    root=(struct treenode *)malloc(sizeof(struct treenode));
    temp1=(struct treenode *)malloc(sizeof(struct treenode));
       temp2=(struct treenode *)malloc(sizeof(struct treenode));
        temp3=(struct treenode *)malloc(sizeof(struct treenode));
         temp4=(struct treenode *)malloc(sizeof(struct treenode));
          temp5=(struct treenode *)malloc(sizeof(struct treenode));
           temp6=(struct treenode *)malloc(sizeof(struct treenode));

    root->data=4;
     root->left=temp1;
     temp1->data=5;
     root->right=temp2;
     temp2->data=6;
     temp1->left=temp3;
     temp3->data=7;
     temp3->left=temp3->right=NULL;
     temp1->right=temp4;
     temp4->data=8;
     temp4->left=NULL;
     temp4->right=NULL;
      temp2->left=temp5;
      temp5->data=6;
      temp2->right=temp6;
      temp6->data=10;
      temp5->left=temp5->right=temp6->left=temp6->right=NULL;
printf("\n tree buildedsuccessfull");
       return root;
}
//no nodes
int noofnodes(struct treenode *root)
{
    if(root==NULL) return 0;

    return noofnodes(root->left)+noofnodes(root->right)+1;
}

//
void printfrombottam(struct treenode *root)
{
int *array[QUEUESIZE],i=0,j=0;
struct treenode *k;
array[j++]=root;
//k=noofnodes(root);
while(i<j)
{

    if(root->right!=NULL)
    array[j++]=root->right;

    if(root->left!=NULL)
    array[j++]=root->left;

    root=array[++i];

}
for(i=j-1;i>=0;i--)
{
    k=array[i];
    printf("%d   ",k->data);
}
}//end

//main

void main()
{
    struct treenode *root;
    int i=1;
    while(i)
    {
        printf("\n enter 1 to build elements");
        printf("\n enter 2 print elements in level wise");
        scanf("%d",&i);

    switch(i)
    {
    case 1:
        root=buildtree();
        break;

    case 2:
        printfrombottam(root);
        break;
    }
     }
}

Zigzag traversal of tree or spiralorder traversal of binary tree

2012-08-27T11:24:00.004-07:00

Idea:
maintain array keep all theelemnts in arry then traverse array in way to get spiral order

example:

For the given tree,
(1)
(2) (3)
(4) (5) (6) (7)
(8) (9)
Spiral traversal is
1, 2, 3, 7, 6, 5, 4, 8, 9 ….

Array elements after inserting elements will be

1 null 2 3   null 7 6 5 4 null 8 9 null

In array after every level insert null then traverse in forward direction till null and go next null traverse backward direction to null continue same process

Code:
#include<stdio.h>
#include<conio.h>
#include<malloc.h>
#define QUEUESIZE 20

struct treenode
{
    int data;
    struct treenode *left;
    struct treenode *right;
} *root=NULL;

//building tree

struct treenode * buildtree()
{
    struct treenode *temp1,*temp2,*temp3,*temp4,*temp5,*temp6;
    root=(struct treenode *)malloc(sizeof(struct treenode));
    temp1=(struct treenode *)malloc(sizeof(struct treenode));
       temp2=(struct treenode *)malloc(sizeof(struct treenode));
        temp3=(struct treenode *)malloc(sizeof(struct treenode));
         temp4=(struct treenode *)malloc(sizeof(struct treenode));
          temp5=(struct treenode *)malloc(sizeof(struct treenode));
           temp6=(struct treenode *)malloc(sizeof(struct treenode));

    root->data=4;
     root->left=temp1;
     temp1->data=5;
     root->right=temp2;
     temp2->data=6;
     temp1->left=temp3;
     temp3->data=7;
     temp3->left=temp3->right=NULL;
     temp1->right=temp4;
     temp4->data=8;
     temp4->left=NULL;
     temp4->right=NULL;
      temp2->left=temp5;
      temp5->data=6;
      temp2->right=temp6;
      temp6->data=10;
      temp5->left=temp5->right=temp6->left=temp6->right=NULL;
printf("\n tree buildedsuccessfull");
       return root;
}

//print
void print(int *a[],int n)
{
    struct treenode *temp;
    int k=0,i,j,m;
    for(j=0;j<n;)
    {
        if(a[j]==NULL)
        {
            k++;
            j++;
            }

        else if(k%2==0)
        {
            temp=a[j];
            j++;
            printf("%d   ",temp->data);
          }
        else if(k%2==1)
        {
            i=j;
            while(a[i]!=NULL)
            i++;
            for(m=i-1;m>=j;m--)
            {
                temp=a[m];
            printf("%d    ",temp->data);
             }
            j=i;

            }

        }
}
//zizag order

void zigzag(struct treenode *root)
{
int *array[QUEUESIZE],i=0,j=-1;
struct treenode *k;
array[++j]=root;
array[++j]=NULL;
while(i<j)
{
    if(root!=NULL)
    {
    if(root->left!=NULL)
    array[++j]=root->left;

    if(root->right!=NULL)
    array[++j]=root->right;

    root=array[++i];
    }
    else
    {
    array[++j]=NULL;
    root=array[++i];
    }
   }

print(array,j);
}//end

//main

void main()
{
    struct treenode *root;
    int i=1;
    while(i)
    {
        printf("\n enter 1 to build elements");
        printf("\n enter 2 print elements in zigzag order");
        scanf("%d",&i);

    switch(i)
    {
    case 1:
        root=buildtree();
        break;

    case 2:
        zigzag(root);
        break;
    }
     }
}

2012-08-27T11:13:00.000-07:00

Finding number nodes in a tree and height of tree and find number of full nodes

fullnodes:

nodes which have both left and right childrens if a node contains left or right is null or both are
null those nodes are not full nodes

code:

#include<stdio.h>
#include<conio.h>
#include<malloc.h>

struct treenode
{
    int data;
    struct treenode *left;
    struct treenode *right;
} *root=NULL;
//static tree building function
struct treenode * buildtree()
{
    struct treenode *temp1,*temp2,*temp3,*temp4,*temp5;
    root=(struct treenode *)malloc(sizeof(struct treenode));
    temp1=(struct treenode *)malloc(sizeof(struct treenode));
       temp2=(struct treenode *)malloc(sizeof(struct treenode));
        temp3=(struct treenode *)malloc(sizeof(struct treenode));
         temp4=(struct treenode *)malloc(sizeof(struct treenode));

    root->data=4;
     root->left=temp1;
     temp1->data=5;
     root->right=temp2;
     temp2->data=6;
     temp1->left=temp3;
     temp3->data=7;
     temp3->left=temp3->right=NULL;
     temp1->right=temp4;
     temp4->data=8;
     temp4->left=NULL;
     temp4->right=NULL;
      temp2->left=temp2->right=NULL;
printf("\n tree buildedsuccessfull");
       return root;
}
//no nodes
int noofnodes(struct treenode *root)
{
    if(root==NULL) return 0;

    return noofnodes(root->left)+noofnodes(root->right)+1;
}
//height of tree
int height(struct treenode *root)
{
    if(root==NULL) return 0;
    return max(height(root->left),height(root->right))+1;
}
//
int max(int i, int k)
{
    if(i>k)
    return i;
    return k;
}

int fullnodes(struct treenode *root)
{
    int count=0;
    if(root==NULL ) return ;
    else if(root->left!=NULL&&root->right!=NULL)
    count=1;

    return fullnodes(root->left)+fullnodes(root->right)+count;
}
int d=0;
int diameter(struct treenode *root,int l)
{
    d=l;
    int lh,rh;
    if(root==NULL) return 0;
    lh=diameter(root->left,d);
    rh=diameter(root->right,d);
    if(d<(lh+rh))
    d=lh+rh;
    return max(lh,rh)+1;

}

void main()
{
    struct treenode *root;
    int i=1;
    while(i)
    {
    printf("\n enter 2 to build tree");
    printf("\n enter 1 to find no of nodes of given tree");
    printf("\n enter 3 to find height");
    printf("\n enter 4 find full nodes of given tree");
    printf("\n enter 5 find diameter");
    scanf("%d",&i);
    if(i==5)
        {
            diameter(root,d);
         printf("\n %d is diameter",d);
         }
    if(i==4)
    printf("\n full node is %d",fullnodes(root));
    if(i==2)
    root=buildtree();
    if(i==3)
    printf("\n height if tree is %d",height(root));
    if(i==1)
    printf("\n %d is no of nodes in given tree",noofnodes(root));
}
}

Traversing a tree inorder and post and preorder without using recursion

2012-08-27T11:00:00.000-07:00

Traversing a tree inorder and post and preorder using non recursive technique:

#include<stdio.h>
#include<conio.h>
#include<malloc.h>
#define STACKSIZE 20

struct treenode
{
    int data;
    struct treenode *left;
    struct treenode *right;
} *root=NULL;

// stack functions
int top=-1; struct treenode *stack[STACKSIZE];
void push(struct treenode *temp)
{
    if(top<STACKSIZE)
    {
        top++;
        stack[top]=temp;
        }
        else
        printf("stack overflow");
}

int pop()
{
    int i;
    if(top==-1)
    {
        printf("stackunderflow");
        }
    else
    i=stack[top--];
    return i;

}
int isstackempty()
{
    if(top==-1)
    return 1;
    else return 0;
}
//end of stackfunctions

struct treenode * buildtree()
{
    struct treenode *temp1,*temp2,*temp3,*temp4,*temp5;
    root=(struct treenode *)malloc(sizeof(struct treenode));
    temp1=(struct treenode *)malloc(sizeof(struct treenode));
       temp2=(struct treenode *)malloc(sizeof(struct treenode));
        temp3=(struct treenode *)malloc(sizeof(struct treenode));
         temp4=(struct treenode *)malloc(sizeof(struct treenode));

    root->data=4;
     root->left=temp1;
     temp1->data=5;
     root->right=temp2;
     temp2->data=6;
     temp1->left=temp3;
     temp3->data=7;
     temp3->left=temp3->right=NULL;
     temp1->right=temp4;
     temp4->data=8;
     temp4->left=NULL;
     temp4->right=NULL;
      temp2->left=temp2->right=NULL;
printf("\n tree buildedsuccessfull");
       return root;
}

void inorder(struct treenode *root)
{
    struct treenode *temp;
    temp=root;
    while(1)
    {
    while(temp!=NULL)
       {
            push(temp);
            temp=temp->left;
        }
    if(isstackempty()) return;
    temp=pop();
    printf("%d    ",temp->data);
    temp=temp->right;
     }
}
void preorder(struct treenode *root)
{
    struct treenode *temp;
    temp=root;
    while(1)
    {
    while(temp!=NULL)
       {
            printf("%d   ",temp->data);
            push(temp);
            temp=temp->left;
        }
    if(isstackempty()) return;
    temp=pop();
    temp=temp->right;
     }
}

void postorder(struct treenode *root)
{
    struct treenode *temp,*q;
    temp=root;
    while(1)
    {
    while(temp!=NULL)
       {
            push(temp);
            temp=temp->left;
        }
    if(isstackempty()) return;
    temp=pop();
    if(((int)temp)<0)
    {
        temp=((unsigned int)temp)*-1;
    printf("%d    ",temp->data);
    temp=NULL;
     }
     else
     {
            q=temp;
            temp=temp->right;
        q=((unsigned int)q)*-1;
        push(q);
            }
}
}


//main
void main()
{
    int k=1,j;
    struct treenode *root;
    while(k)
    {
        printf("\n enter 1 to build tree");
        printf("\n enter 2 to print preorder");
        printf("\n enter 3 to print inorder");
        printf("\n enter 4 print postoreder");
        scanf("%d",&k);

        switch(k)
        {
        case 1:

            root=buildtree();
            break;

        case 2:
            preorder(root);
            break;

           case 3:
                inorder(root);
                break;

        case 4:
            postorder(root);
            break;

            }


        }
}

Cisco written question paper 2011:

2012-08-27T10:35:00.000-07:00

Cisco written question paper 2011:

The written test had two parts.
Test Date : 21 August 2010
20 questions from aptitude and 30 technical, no negative marking.
Aptitude was quite a lot tricky and technical was Electronics + CS.

Three C programs
1) main()
{
int const a=100;
int *p;
p=&a;
(*p)++;
printf("%d%d",a,*p);
}
Ans: 101 101

Explanation: Pointer can modify an object it pointing to but the variable can't change it when the variable defined as const.

2) ARP is used for?

The Address Resolution Protocol (ARP) is used for determining a network host's Link Layer or hardware address (MAC Address) when only its Internet Layer (IP) or Network Layer address is known. You have written it reversely..

3. End to End delivery is responsibility of which layer?

Transport Layer

4. If virtual address is[0,514] then if page table contains an entry (032) then physical address is:
Ans append binary form of 32 with binary form of 514

5. Floating point representation: Simple one
If 23 bits are used for IEEE representation then whats the range of normalized nos. which can be represented?

Ans.

Smallest Number can be : +/- 2^(-126) to
Highest number +/- (2-2^23)*2^127.
This table will be helpful to identify the range of normalized numbers

Type                  Exponent     Fraction
Zeroes                     0              0
Denormalized numbers       0           non zero
Normalized numbers       1 to 2e − 2     any
Infinities                2e − 1          0
NaNs                      2e − 1       non zero

One semaphores.

Generally 2 technical interviews and one HR. But for some 3 technical and one HR.
Technical interview contained questions like:

1. Full form of SQL
Ans. Structured Query Language

2. Various functions in SQL
Ans. All SQL functions return a single value that is either a number or a string.
a. Standard: These return a value from an expression.
example:
ABS Returns the absolute value of the numeric expression.
num = ABS( expression )

ALL Tests against the values in the value_list. Returns 1 (true) if matches every item. Equivalent to individual tests against each item in the list joined by AND operators. ALL is used particularly to test against sets of values returned by subqueries.
value = ALL ( value_list | subquery )

ANY Tests against the values in the value_list. Returns 1 (true) if matches any item. Equivalent to individual tests against each item in the list joined by OR operators. ANY is used particularly to test against sets of values returned by subqueries.
value = ANY ( value_list | subquery )

INT Returns the truncated integer value for the numeric expression.
num = INT ( expression )

LEN Returns the number of characters (including trailing blanks) in the string expression.
num = LEN( string )

LOWER Returns the string with all uppercase letters converted to their lowercase equivalent.
str = LOWER( string )

b. Aggregation: These compute a value from a number of rows and thus alter the number of rows produced in the output table.
example:
AVG ([UNIQUE] numeric_col) Returns the average or mean value of the non-missing values for numeric columns. If UNIQUE is specified then only unique values are used to calculate the mean.

COUNT ( [ UNIQUE ] col | * ) Returns the number of non-missing values encountered. If UNIQUE is specified, then only the unique values add to the count. An asterisk as the argument returns the number of all rows selected regardless of whether the values are valid, missing or undefined.

MAX (col) Returns the maximum non-missing value encountered. The type of variable returned corresponds to the type of the variable being referenced.

MIN (col) Returns the minimum value of the non-missing values. The type of variable returned corresponds to the type of the variable being referenced.

Decode, ltrim, rtrim, lpad, rpad, truncate etc..

3. Java...
Object oriented analysis and design: usecases, collaboration and sequential diagrams, difference purpose class diagram.

5. Data warehousing?

6. Networking client server basic concepts.

7. C program to count the no. of elements which are repeating more than size/2 times in an array.

8. Derive a mathematical equation to calculate the angle between two hand of clock.

Ans
Angel= 1/2(6H+M) - 6M

H is an integer in the range 0–11 and M is the minute

9. What are different segments in a program.
Ans:

The computer program memory is organized into the following:

  1. Data Segment (Data + BSS + Heap)
             The data area contains global and static variables
 used by the program that are initialized. 
             BSS(Block Started by Symbol): uninitialized data 
starts at the end of the data segment and contains all uninitialized 
global variables and static variables that are initialized to zero 
by default. 
             The Heap area is managed by malloc, realloc, and free,
  2. Stack
             he stack is a LIFO structure, typically located 
in the higher parts of memory. It usually "grows down" with 
every register, immediate value or stack frame being added 
to it. A stack frame consists at minimum of a return address.
  3.Code segment

When we malloc sm size where does it reside.
Where is a.out stored

Stack increses in which direction ?
Ans:
Sack starts from highe memory locations and it increases in lower memory locations

Representation of stack and queue using linked list and then perform insertion and deletion

Models in software Eng?.Explain extreme programming
Ans:
Waterfall Model
Iterative Model
Spiral Model

What are triggers?example
Ans:
A Trigger is a named database object which defines some action that the database should take when some databases related event occurs. Triggers are executed when you issues a data manipulation command like INSERT, DELETE, UPDATE on a table for which the trigger has been created. They are automatically executed and also transparent to the user. But for creating the trigger the user must have the CREATE TRIGGER privilege.

3 properties of object oriented analysis and design
Ans:
Data Abstraction
Re usability
Polymorphism

Normal forms in dbms?
Ans:
1 NF
2 NF
3 NF
BCNF
4 NF
and 5 NF