PROBLEM WITH ANSWERS

Though no operator= function taking complex, double is defined, the double on the rhs is converted into a temporary object using the single argument constructor taking double and assigned to the lvalue.

2.void main()

{

int a, *pa, &ra;

pa = &a;

ra = a;

cout <<"a="<<<"*pa="<<*pa <<"ra"<

}

Answer :

Compiler Error: 'ra',reference must be initialized

Explanation :

Pointers are different from references. One of the main

differences is that the pointers can be both initialized and assigned,

whereas references can only be initialized. So this code issues an error.

placement c++ programming language Question

noreply@blogger.com (Anonymous) — Mon, 1 Dec 2008 16:51:00 -0800

1) Determine the output of the 'C++' Codelet.

class base

{

public :

out()

{

cout<<"base ";

}

};

class deri{

public : out()

{

cout<<"deri ";

}

};

void main()

{ deri dp[3];

base *bp = (base*)dp;

for (int i=0; i<3;i++)

(bp++)->out();

}

2) Justify the use of virtual constructors and destructors in C++.

3) Each C++ object possesses the 4 member fns,(which can be declared by the programmer explicitly or by the implementation if they are not available). What are those 4 functions?

4) What is wrong with this class declaration?

class something

{

char *str;

public:

something(){

st = new char[10]; }

~something()

{

delete str;

}

};

5) Inheritance is also known as -------- relationship. Containership as ________ relationship.

6) When is it necessary to use member-wise initialization list (also known as header initialization list) in C++?

7) Which is the only operator in C++ which can be overloaded but NOT inherited.

8) Is there anything wrong with this C++ class declaration?

class temp

{

int value1;

mutable int value2;

public :

void fun(int val)

const{

((temp*) this)->value1 = 10;

value2 = 10;

}

};

Technical /logical interview questions and solution in C language

noreply@blogger.com (Anonymous) — Mon, 1 Dec 2008 16:50:00 -0800

1.)

void main()

{

int const * p=5;

printf("%d",++(*p));

}

Answer:

Compiler error: Cannot modify a constant value.

Explanation:-

p is a pointer to a "constant integer". But we tried to change the value of the "constant integer".

2.)

main()

{

char s[ ]="man";

int i;

for(i=0;s[ i ];i++)

printf("\n%c%c%c%c",s[ i ],*(s+i),*(i+s),i[s]);

}

Answer:

mmmm

aaaa

nnnn

Explanation:-

s[i], *(i+s), *(s+i), i[s] are all different ways of expressing the same idea. Generally array name is the base address for that array. Here s is the base address. i is the index number/displacement from the base address. So, indirecting it with * is same as s[i]. i[s] may be surprising. But in the case of C it is same as s[i].

3.)

main()

{

float me = 1.1;

double you = 1.1;

if(me==you)

printf("I love U");

else

printf("I hate U");

}

Answer:

I hate U

Explanation:-

For floating point numbers (float, double, long double) the values cannot be predicted exactly. Depending on the number of bytes, the precession with of the value represented varies. Float takes 4 bytes and long double takes 10 bytes. So float stores 0.9 with less precision than long double.

Rule of Thumb:

Never compare or at-least be cautious when using floating point numbers with relational operators (== , >, <, <=, >=,!= ) .

4.)

main()

{

static int var = 5;

printf("%d ",var--);

if(var)

main();

}

Answer:

5 4 3 2 1

Explanation:

When static storage class is given, it is initialized once. The change in the value of a static variable is retained even between the function calls. Main is also treated like any other ordinary function, which can be called recursively.

5.)

main()

{

int c[ ]={2.8,3.4,4,6.7,5};

int j,*p=c,*q=c;

for(j=0;j<5;j++)>

printf(" %d ",*c);

++q;

}

for(j=0;j<5;j++)

{

printf(" %d ",*p);

++p;

}

Answer:

2 2 2 2 2 2 3 4 6 5

Explanation:

Initially pointer c is assigned to both p and q. In the first loop, since only q is incremented and not c , the value 2 will be printed 5 times. In second loop p itself is incremented. So the values 2 3 4 6 5 will be printed.

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Technical / logical interview questions in C language

noreply@blogger.com (Anonymous) — Mon, 1 Dec 2008 16:49:00 -0800

1.)

main()

{

extern int i;

i=20;

printf("%d",i);

}

Answer:

Linker Error : Undefined symbol '_i'

Explanation:

extern storage class in the following declaration,

extern int i;

specifies to the compiler that the memory for i is allocated in some other program and that address will be given to the current program at the time of linking. But linker finds that no other variable of name i is available in any other program with memory space allocated for it. Hence a linker error has occurred .

2.)

main()

{

int i=-1,j=-1,k=0,l=2,m;

m=i++&&j++&&k++||l++;

printf("%d %d %d %d %d",i,j,k,l,m);

}

Answer:

0 0 1 3 1

Explanation :

Logical operations always give a result of 1 or 0 . And also the logical AND (&&) operator has higher priority over the logical OR (||) operator. So the expression ‘i++ && j++ && k++’ is executed first. The result of this expression is 0 (-1 && -1 && 0 = 0). Now the expression is 0 || 2 which evaluates to 1 (because OR operator always gives 1 except for ‘0 || 0’ combination- for which it gives 0). So the value of m is 1. The values of other variables are also incremented by 1.

3.)

main()

{

char *p;

printf("%d %d ",sizeof(*p),sizeof(p));

}

Answer:

1 2

Explanation:

The sizeof() operator gives the number of bytes taken by its operand. P is a character pointer, which needs one byte for storing its value (a character). Hence sizeof(*p) gives a value of 1. Since it needs two bytes to store the address of the character pointer sizeof(p) gives 2.

4.)

main()

{

int i=3;

switch(i)

{

default:printf("zero");

case 1: printf("one");

break;

case 2:printf("two");

break;

case 3: printf("three");

break;

}

Answer :-

three

Explanation :

The default case can be placed anywhere inside the loop. It is executed only when all other cases doesn't match.

5.)

main()

{

printf("%x",-1<<4);

}

Answer:

fff0

Explanation :

-1 is internally represented as all 1's. When left shifted four times the least significant 4 bits are filled with 0's.The %x format specifier specifies that the integer value be printed as a hexadecimal value.

ALL PLACEMENT TECHNICAL INTERVIEW C PROGRAMMING LANGUAGE QUESTION BANCK

C programming Language Technical Questions And Answers

noreply@blogger.com (Anonymous) — Mon, 1 Dec 2008 16:48:00 -0800

1.)

main()

{

char string[]="Hello World";

display(string);

}

void display(char *string)

{

printf("%s",string);

}

Answer:

Compiler Error : Type mismatch in redeclaration of function display

Explanation :

In third line, when the function display is encountered, the compiler doesn't know anything about the function display. It assumes the arguments and return types to be integers, (which is the default type). When it sees the actual function display, the arguments and type contradicts with what it has assumed previously. Hence a compile time error occurs.

2.)

main()

{

int c=- -2;

printf("c=%d",c);

}

Answer:

c=2;

Explanation:

Here unary minus (or negation) operator is used twice. Same maths rules applies, ie. minus * minus= plus.

Note:

However you cannot give like --2. Because -- operator can only be applied to variables as a decrement operator (eg., i--). 2 is a constant and not a variable.

3.)

#define int char

main()

{

int i=65;

printf("sizeof(i)=%d",sizeof(i));

}

Answer:

sizeof(i)=1

Explanation:

Since the #define replaces the string int by the macro char

4.)

main()

{

int i=10;

i=!i>14;

printf("i=%d",i);

}

Answer:

i=0

Explanation:

In the expression !i>14 , NOT (!) operator has more precedence than ‘ >’ symbol. ! is a unary logical operator. !i (!10) is 0 (not of true is false). 0>14 is false (zero).

5.)

#include

main()

{

char s[]={'a','b','c','\n','c','\0'};

char *p,*str,*str1;

p=&s[3];

str=p;

str1=s;

printf("%d",++*p + ++*str1-32);

}

Answer:

Explanation:

p is pointing to character '\n'. str1 is pointing to character 'a' ++*p. "p is pointing to '\n' and that is incremented by one." the ASCII value of '\n' is 10, which is then incremented to 11. The value of ++*p is 11. ++*str1, str1 is pointing to 'a' that is incremented by 1 and it becomes 'b'. ASCII value of 'b' is 98.

Now performing (11 + 98 – 32), we get 77("M");

So we get the output 77 :: "M" (Ascii is 77).

ALL PLACEMENT TECHNICAL INTERVIEW C PROGRAMMING LANGUAGE QUESTION BANCK

C Language Technical Questions And Answers

noreply@blogger.com (Anonymous) — Mon, 1 Dec 2008 16:47:00 -0800

1.)

main()

{

char string[]="Hello World";

display(string);

}

void display(char *string)

{

printf("%s",string);

}

Answer:

Compiler Error : Type mismatch in redeclaration of function display

Explanation :

2.)

main()

{

int c=- -2;

printf("c=%d",c);

}

Answer:

c=2;

Explanation:

Here unary minus (or negation) operator is used twice. Same maths rules applies, ie. minus * minus= plus.

Note:

However you cannot give like --2. Because -- operator can only be applied to variables as a decrement operator (eg., i--). 2 is a constant and not a variable.

3.)

#define int char

main()

{

int i=65;

printf("sizeof(i)=%d",sizeof(i));

}

Answer:

sizeof(i)=1

Explanation:

Since the #define replaces the string int by the macro char

4.)

main()

{

int i=10;

i=!i>14;

printf("i=%d",i);

}

Answer:

i=0

Explanation:

In the expression !i>14 , NOT (!) operator has more precedence than ‘ >’ symbol. ! is a unary logical operator. !i (!10) is 0 (not of true is false). 0>14 is false (zero).

5.)

#include

main()

{

char s[]={'a','b','c','\n','c','\0'};

char *p,*str,*str1;

p=&s[3];

str=p;

str1=s;

printf("%d",++*p + ++*str1-32);

}

Answer:

Explanation:

Now performing (11 + 98 – 32), we get 77("M");

So we get the output 77 :: "M" (Ascii is 77).

ALL PLACEMENT TECHNICAL INTERVIEW QUESTION BANCK

C Language Technical Interview Questions And solution

noreply@blogger.com (Anonymous) — Mon, 1 Dec 2008 16:46:00 -0800

1.)

#include

main()

{

int a[2][2][2] = { {10,2,3,4}, {5,6,7,8} };

int *p,*q;

p=&a[2][2][2];

*q=***a;

printf("%d----%d",*p,*q);

}

Answer:

SomeGarbageValue---1

Explanation:

p=&a[2][2][2] you declare only two 2D arrays, but you are trying to access the third 2D(which you are not declared) it will print garbage values. *q=***a starting address of a is assigned integer pointer. Now q is pointing to starting address of a. If you print *q, it will print first element of 3D array.

2.)

#include

main()

{

struct xx

{

int x=3;

char name[]="hello";

};

struct xx *s;

printf("%d",s->x);

printf("%s",s->name);

}

Answer:

Compiler Error

Explanation:

You should not initialize variables in declaration

3.)

#include

main()

{

struct xx

{

int x;

struct yy

{

char s;

struct xx *p;

};

struct yy *q;

};

}

Answer:

Compiler Error

Explanation:

The structure yy is nested within structure xx. Hence, the elements are of yy are to be accessed through the instance of structure xx, which needs an instance of yy to be known. If the instance is created after defining the structure the compiler will not know about the instance relative to xx. Hence for nested structure yy you have to declare member.

4.)

main()

{

printf("\nab");

printf("\bsi");

printf("\rha");

}

Answer:

hai

Explanation:

\n - newline

\b - backspace

\r - linefeed

5.)

main()

{

int i=5;

printf("%d%d%d%d%d%d",i++,i--,++i,--i,i);

}

Answer:-

45545

Explanation:

The arguments in a function call are pushed into the stack from left to right. The evaluation is by popping out from the stack. and the evaluation is from right to left, hence the result.

ALL PLACEMENT TECHNICAL INTERVIEW C PROGRAMMING LANGUAGE QUESTION BANCK

C Language Technical Interview Questions And Answers

noreply@blogger.com (Anonymous) — Mon, 1 Dec 2008 16:45:00 -0800

1.)

#define square(x) x*x

main()

{

int i;

i = 64/square(4);

printf("%d",i);

}

Answer:-

Explanation:

the macro call square(4) will substituted by 4*4 so the expression becomes i = 64/4*4 . Since / and * has equal priority the expression will be evaluated as (64/4)*4 i.e. 16*4 = 64

2.)

main()

{

char *p="hai friends",*p1;

p1=p;

while(*p!='\0') ++*p++;

printf("%s %s",p,p1);

}

Answer:

ibj!gsjfoet

Explanation:

++*p++ will be parse in the given order

->> *p that is value at the location currently pointed by p will be taken

->> ++*p the retrieved value will be incremented

->> when ; is encountered the location will be incremented that is p++ will be executed

Hence, in the while loop initial value pointed by p is ‘h’, which is changed to ‘i’ by executing ++*p and pointer moves to point, ‘a’ which is similarly changed to ‘b’ and so on. Similarly blank space is converted to ‘!’. Thus, we obtain value in p becomes “ibj!gsjfoet” and since p reaches ‘\0’ and p1 points to p thus p1doesnot print anything.

3.)

#include

#define a 10

main()

{

#define a 50

printf("%d",a);

}

Answer:

Explanation:

The preprocessor directives can be redefined anywhere in the program. So the most recently assigned value will be taken.

4.)

#define clrscr() 100

main()

{

clrscr();

printf("%d\n",clrscr());

}

Answer:

100

Explanation:

Preprocessor executes as a seperate pass before the execution of the compiler. So textual replacement of clrscr() to 100 occurs.The input program to compiler looks like this :

main()

{

100;

printf("%d\n",100);

}

Note:

100; is an executable statement but with no action. So it doesn't give any problem

5.)

main()

{

printf("%p",main);

}

Answer:

Some address will be printed.

Explanation:

Function names are just addresses (just like array names are addresses).

main() is also a function. So the address of function main will be printed. %p in printf specifies that the argument is an address. They are printed as hexadecimal numbers.

ALL PLACEMENT TECHNICAL INTERVIEW C PROGRAMMING LANGUAGE QUESTION BANCK

C Language Interview Questions And Solution

noreply@blogger.com (Anonymous) — Mon, 1 Dec 2008 16:44:00 -0800

main()

{

clrscr();

}

clrscr();

Answer:

No output/error

Explanation:

The first clrscr() occurs inside a function. So it becomes a function call. In the second clrscr(); is a function declaration (because it is not inside any function).

enum colors {BLACK,BLUE,GREEN}

main()

{

printf("%d..%d..%d",BLACK,BLUE,GREEN);

return(1);

}

Answer:

0..1..2

Explanation:

enum assigns numbers starting from 0, if not explicitly defined.

void main()

{

char far *farther,*farthest;

printf("%d..%d",sizeof(farther),sizeof(farthest));

}

Answer:

4..2

Explanation:

the second pointer is of char type and not a far pointer

4) main()

{

int i=400,j=300;

printf("%d..%d");

}

Answer:

400..300

Explanation:

printf takes the values of the first two assignments of the program. Any number of printf's may be given. All of them take only the first two values. If more number of assignments given in the program,then printf will take garbage values.

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C Language Interview Questions And Answers

noreply@blogger.com (Anonymous) — Mon, 1 Dec 2008 16:44:00 -0800

main()

{

char *p;

p="Hello";

printf("%c\n",*&*p);

}

Answer:

Explanation:

* is a dereference operator & is a reference operator. They can be applied any number of times provided it is meaningful. Here p points to the first character in the string "Hello". *p dereferences it and so its value is H. Again & references it to an address and * dereferences it to the value H.

2) main()

{

int i=1;

while (i<=5)

{

printf("%d",i);

if (i>2)

goto here;

i++;

}

fun()

{

here:

printf("PP");

}

Answer:

Compiler error: Undefined label 'here' in function main

Explanation:

Labels have functions scope, in other words The scope of the labels is limited to functions . The label 'here' is available in function fun() Hence it is not visible in function main.

main()

{

static char names[5][20]={"pascal","ada","cobol","fortran","perl"};

int i;

char *t;

t=names[3];

names[3]=names[4];

names[4]=t;

for (i=0;i<=4;i++)

printf("%s",names[i]);

}

Answer:

Compiler error: Lvalue required in function main

Explanation:

Array names are pointer constants. So it cannot be modified.

4) void main()

{

int i=5;

printf("%d",i++ + ++i);

}

Answer:

Output Cannot be predicted exactly.

Explanation:

Side effects are involved in the evaluation of i

5) void main()

{

int i=5;

printf("%d",i+++++i);

}

Answer:

Compiler Error

Explanation:

The expression i+++++i is parsed as i ++ ++ + i which is an illegal combination of operators.

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online Placement Question of c programming languge

noreply@blogger.com (Anonymous) — Mon, 1 Dec 2008 16:43:00 -0800

1) #include

main()

{

int i=1,j=2;

switch(i)

{

case 1: printf("GOOD");

break;

case j: printf("BAD");

break;

}

Answer:-

Compiler Error: Constant expression required in function main.

Explanation:

The case statement can have only constant expressions (this implies that we cannot use variable names directly so an error).

Note:

Enumerated types can be used in case statements.

main()

{

int i;

printf("%d",scanf("%d",&i)); // value 10 is given as input here

}

Answer:-

Explanation:

Scanf returns number of items successfully read and not 1/0. Here 10 is given as input which should have been scanned successfully. So number of items read is 1.

#define f(g,g2) g##g2

main()

{

int var12=100;

printf("%d",f(var,12));

}

Answer:-

100

main()

{

int i=0;

for(;i++;printf("%d",i)) ;

printf("%d",i);

}

Answer:-

Explanation:

before entering into the for loop the checking condition is "evaluated". Here it evaluates to 0 (false) and comes out of the loop, and i is incremented (note the semicolon after the for loop).

#include

main()

{

char s[]={'a','b','c','\n','c','\0'};

char *p,*str,*str1;

p=&s[3];

str=p;

str1=s;

printf("%d",++*p + ++*str1-32);

}

Answer:-

Explanation:

p is pointing to character '\n'.str1 is pointing to character 'a' ++*p meAnswer:"p is pointing to '\n' and that is incremented by one." the ASCII value of '\n' is 10. then it is incremented to 11. the value of ++*p is 11. ++*str1 meAnswer:"str1 is pointing to 'a' that is incremented by 1 and it becomes 'b'. ASCII value of 'b' is 98. both 11 and 98 is added and result is subtracted from 32.

i.e. (11+98-32)=77("M");

ALL PLACEMENT TECHNICAL INTERVIEW QUESTION BANCK

online placement c programming question

noreply@blogger.com (Anonymous) — Mon, 1 Dec 2008 16:42:00 -0800

1) #include

main()

{

struct xx

{

int x=3;

char name[]="hello";

};

struct xx *s=malloc(sizeof(struct xx));

printf("%d",s->x);

printf("%s",s->name);

}

Answer:

Compiler Error

Explanation:

Initialization should not be done for structure members inside the structure declaration

2) #include

main()

{

struct xx

{

int x;

struct yy

{

char s;

struct xx *p;

};

struct yy *q;

};

}

Answer:

Compiler Error

Explanation:

in the end of nested structure yy a member have to be declared.

3) main()

{

extern int i;

i=20;

printf("%d",sizeof(i));

}

Answer:

Linker error: undefined symbol '_i'.

Explanation:

extern declaration specifies that the variable i is defined somewhere else. The compiler passes the external variable to be resolved by the linker. So compiler doesn't find an error. During linking the linker searches for the definition of i. Since it is not found the linker flags an error.

main()

{

printf("%d", out);

}

int out=100;

Answer:

Compiler error: undefined symbol out in function main.

Explanation:

The rule is that a variable is available for use from the point of declaration. Even though a is a global variable, it is not available for main. Hence an error.

5) main()

{

extern out;

printf("%d", out);

}

int out=100;

Answer:-

100

Explanation:

This is the correct way of writing the previous program.

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online test of c programming languge

noreply@blogger.com (Anonymous) — Mon, 1 Dec 2008 16:41:00 -0800

1) main()

{

show();

}

void show()

{

printf("I'm the greatest");

}

Answer:

Compier error: Type mismatch in redeclaration of show.

Explanation:

When the compiler sees the function show it doesn't know anything about it. So the default return type (ie, int) is assumed. But when compiler sees the actual definition of show mismatch occurs since it is declared as void. Hence the error.

The solutions are as follows:

1. declare void show() in main() .

2. define show() before main().

3. declare extern void show() before the use of show().

2) main( )

{

int a[2][3][2] = {{{2,4},{7,8},{3,4}},{{2,2},{2,3},{3,4}}};

printf(“%u %u %u %d \n”,a,*a,**a,***a);

printf(“%u %u %u %d \n”,a+1,*a+1,**a+1,***a+1);

}

Answer:

100, 100, 100, 2

114, 104, 102, 3

Explanation:

The given array is a 3-D one. It can also be viewed as a 1-D array.

100 102 104 106 108 110 112 114 116 118 120 122

thus, for the first printf statement a, *a, **a give address of first element . since the indirection ***a gives the value. Hence, the first line of the output.

for the second printf a+1 increases in the third dimension thus points to value at 114, *a+1 increments in second dimension thus points to 104, **a +1 increments the first dimension thus points to 102 and ***a+1 first gets the value at first location and then increments it by 1. Hence, the output.

3) main( )

{

int a[ ] = {10,20,30,40,50},j,*p;

for(j=0; j<5;>

{

printf(“%d” ,*a);

a++;

}

p = a;

for(j=0; j<5;>

{

printf(“%d ” ,*p);

p++;

}

Answer:

Compiler error: lvalue required.

Explanation:

Error is in line with statement a++. The operand must be an lvalue and may be of any of scalar type for the any operator, array name only when subscripted is an lvalue. Simply array name is a non-modifiable lvalue.

4) main( )

{

static int a[ ] = {0,1,2,3,4};

int *p[ ] = {a,a+1,a+2,a+3,a+4};

int **ptr = p;

ptr++;

printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);

*ptr++;

printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);

*++ptr;

printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);

++*ptr;

printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);

}

Answer:

111

222

333

344

Explanation:

Let us consider the array and the two pointers with some address

After execution of the instruction ptr++ value in ptr becomes 1002, if scaling factor for integer is 2 bytes. Now ptr – p is value in ptr – starting location of array p, (1002 – 1000) / (scaling factor) = 1, *ptr – a = value at address pointed by ptr – starting value of array a, 1002 has a value 102 so the value is (102 – 100)/(scaling factor) = 1, **ptr is the value stored in the location pointed by the pointer of ptr = value pointed by value pointed by 1002 = value pointed by 102 = 1. Hence the output of the firs printf is 1, 1, 1.

After execution of *ptr++ increments value of the value in ptr by scaling factor, so it becomes1004. Hence, the outputs for the second printf are ptr – p = 2, *ptr – a = 2, **ptr = 2.

After execution of *++ptr increments value of the value in ptr by scaling factor, so it becomes1004. Hence, the outputs for the third printf are ptr – p = 3, *ptr – a = 3, **ptr = 3.

After execution of ++*ptr value in ptr remains the same, the value pointed by the value is incremented by the scaling factor. So the value in array p at location 1006 changes from 106 10 108,. Hence, the outputs for the fourth printf are ptr – p = 1006 – 1000 = 3, *ptr – a = 108 – 100 = 4, **ptr = 4.

5) main( )

{

char *q;

int j;

for (j=0; j<3;>

}

Explanation:

Here we have only one pointer to type char and since we take input in the same pointer thus we keep writing over in the same location, each time shifting the pointer value by 1. Suppose the inputs are MOUSE, TRACK and VIRTUAL. Then for the first input suppose the pointer starts at location 100 then the input one is stored as

M

O

U

When the second input is given the pointer is incremented as j value becomes 1, so the input is filled in memory starting from 101.

M

T

R

A

C

K

\0

The third input starts filling from the location 102

M

T

V

I

R

T

U

A

L

This is the final value stored .

The first printf prints the values at the position q, q+1 and q+2 = M T V

The second printf prints three strings starting from locations q, q+1, q+2

i.e MTVIRTUAL, TVIRTUAL and VIRTUAL.

ALL PLACEMENT TECHNICAL INTERVIEW QUESTION BANCK

placement c programming quiz with solution

noreply@blogger.com (Anonymous) — Mon, 1 Dec 2008 16:40:00 -0800

main( )

{

void *vp;

char ch = ‘g’, *cp = “goofy”;

int j = 20;

vp = &ch;

printf(“%c”, *(char *)vp);

vp = &j;

printf(“%d”,*(int *)vp);

vp = cp;

printf(“%s”,(char *)vp + 3);

}

Answer:

g20fy

Explanation:

Since a void pointer is used it can be type casted to any other type pointer. vp = &ch stores address of char ch and the next statement prints the value stored in vp after type casting it to the proper data type pointer. the output is ‘g’. Similarly the output from second printf is ‘20’. The third printf statement type casts it to print the string from the 4th value hence the output is ‘fy’.

main ( )

{

static char *s[ ] = {“black”, “white”, “yellow”, “violet”};

char **ptr[ ] = {s+3, s+2, s+1, s}, ***p;

p = ptr;

**++p;

printf(“%s”,*--*++p + 3);

}

Answer:

Explanation:

In this problem we have an array of char pointers pointing to start of 4 strings. Then we have ptr which is a pointer to a pointer of type char and a variable p which is a pointer to a pointer to a pointer of type char. p hold the initial value of ptr, i.e. p = s+3. The next statement increment value in p by 1 , thus now value of p = s+2. In the printf statement the expression is evaluated *++p causes gets value s+1 then the pre decrement is executed and we get s+1 – 1 = s . the indirection operator now gets the value from the array of s and adds 3 to the starting address. The string is printed starting from this position. Thus, the output is ‘ck’.

main()

{

int i, n;

char *x = “girl”;

n = strlen(x);

*x = x[n];

for(i=0; i

{

printf(“%s\n”,x);

x++;

}

Answer:

(blank space)

irl

Explanation:

Here a string (a pointer to char) is initialized with a value “girl”. The strlen function returns the length of the string, thus n has a value 4. The next statement assigns value at the nth location (‘\0’) to the first location. Now the string becomes “\0irl” . Now the printf statement prints the string after each iteration it increments it starting position. Loop starts from 0 to 4. The first time x[0] = ‘\0’ hence it prints nothing and pointer value is incremented. The second time it prints from x[1] i.e “irl” and the third time it prints “rl” and the last time it prints “l” and the loop terminates.

4) int i,j;

for(i=0;i<=10;i++)

{

j+=5;

assert(i<5);

}

Answer:

Runtime error: Abnormal program termination.

assert failed (i<5),>,

Explanation:

asserts are used during debugging to make sure that certain conditions are satisfied. If assertion fails, the program will terminate reporting the same. After debugging use,

#undef NDEBUG

and this will disable all the assertions from the source code. Assertion

is a good debugging tool to make use of.

5) main()

{

int i=-1;

+i;

printf("i = %d, +i = %d \n",i,+i);

}

Answer:

i = -1, +i = -1

Explanation:

Unary + is the only dummy operator in C. Where-ever it comes you can just ignore it just because it has no effect in the expressions (hence the name dummy operator).

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c programming quiz with solution

noreply@blogger.com (Anonymous) — Mon, 1 Dec 2008 16:39:00 -0800

1)What are the files which are automatically opened when a C file is executed?

Answer:

stdin, stdout, stderr (standard input,standard output,standard error).

2) what will be the position of the file marker?

a: fseek(ptr,0,SEEK_SET);

b: fseek(ptr,0,SEEK_CUR);

Answer :

a: The SEEK_SET sets the file position marker to the starting of the file.

b: The SEEK_CUR sets the file position marker to the current position

of the file.

3) main()

{

char name[10],s[12];

scanf(" \"%[^\"]\"",s);

}

How scanf will execute?

Answer:

First it checks for the leading white space and discards it.Then it matches with a quotation mark and then it reads all character upto another quotation mark.

4)What is the problem with the following code segment?

while ((fgets(receiving array,50,file_ptr)) != EOF);

Answer & Explanation:

fgets returns a pointer. So the correct end of file check is checking for != NULL.

5) main()

{

main();

}

Answer:

Runtime error : Stack overflow.

Explanation:

main function calls itself again and again. Each time the function is called its return address is stored in the call stack. Since there is no condition to terminate the function call, the call stack overflows at runtime. So it terminates the program and results in an error.

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C programming Language quiz with solution

noreply@blogger.com (Anonymous) — Mon, 1 Dec 2008 16:38:00 -0800

main()

{

char *cptr,c;

void *vptr,v;

c=10; v=0;

cptr=&c; vptr=&v;

printf("%c%v",c,v);

}

Answer:

Compiler error (at line number 4): size of v is Unknown.

Explanation:

You can create a variable of type void * but not of type void, since void is an empty type. In the second line you are creating variable vptr of type void * and v of type void hence an error.

2) main()

{

char *str1="abcd";

char str2[]="abcd";

printf("%d %d %d",sizeof(str1),sizeof(str2),sizeof("abcd"));

}

Answer:-

2 5 5

Explanation:

In first sizeof, str1 is a character pointer so it gives you the size of the pointer variable. In second sizeof the name str2 indicates the name of the array whose size is 5 (including the '\0' termination character). The third sizeof is similar to the second one.

3) main()

{

char not;

not=!2;

printf("%d",not);

}

Answer:

Explanation:

! is a logical operator. In C the value 0 is considered to be the boolean value FALSE, and any non-zero value is considered to be the boolean value TRUE. Here 2 is a non-zero value so TRUE. !TRUE is FALSE (0) so it prints 0.

4) #define FALSE -1

#define TRUE 1

#define NULL 0

main() {

if(NULL)

puts("NULL");

else if(FALSE)

puts("TRUE");

else

puts("FALSE");

}

Answer:-

TRUE

Explanation:

The input program to the compiler after processing by the preprocessor is,

main(){

if(0)

puts("NULL");

else if(-1)

puts("TRUE");

else

puts("FALSE");

}

Preprocessor doesn't replace the values given inside the double quotes. The check by if condition is boolean value false so it goes to else. In second if -1 is boolean value true hence "TRUE" is printed.

5) main()

{

int k=1;

printf("%d==1 is ""%s",k,k==1?"TRUE":"FALSE");

}

Answer:

1==1 is TRUE

Explanation:

When two strings are placed together (or separated by white-space) they are concatenated (this is called as "stringization" operation). So the string is as if it is given as "%d==1 is %s". The conditional operator( ?: ) evaluates to "TRUE".

ALL PLACEMENT TECHNICAL INTERVIEW QUESTION BANCK

C quiz with solution

noreply@blogger.com (Anonymous) — Mon, 1 Dec 2008 16:37:00 -0800

1) main()

{

int y;

scanf("%d",&y); // input given is 2000

if( (y%4==0 && y%100 != 0) || y%100 == 0 )

printf("%d is a leap year");

else

printf("%d is not a leap year");

}

Answer:

2000 is a leap year

Explanation:

An ordinary program to check if leap year or not.

2) #define max 5

#define int arr1[max]

main()

{

typedef char arr2[max];

arr1 list={0,1,2,3,4};

arr2 name="name";

printf("%d %s",list[0],name);

}

Answer:

Compiler error (in the line arr1 list = {0,1,2,3,4})

Explanation:

arr2 is declared of type array of size 5 of characters. So it can be used to declare the variable name of the type arr2. But it is not the case of arr1. Hence an error.

Rule of Thumb:

#defines are used for textual replacement whereas typedefs are used for declaring new types.

3) int i=10;

main()

{

extern int i;

{

int i=20;

{

const volatile unsigned i=30;

printf("%d",i);

}

printf("%d",i);

}

printf("%d",i);

}

Answer:

30,20,10

Explanation:

'{' introduces new block and thus new scope. In the innermost block i is declared as,

const volatile unsigned

which is a valid declaration. i is assumed of type int. So printf prints 30. In the next block, i has value 20 and so printf prints 20. In the outermost block, i is declared as extern, so no storage space is allocated for it. After compilation is over the linker resolves it to global variable i (since it is the only variable visible there). So it prints i's value as 10.

4) main()

{

int *j;

{

int i=10;

j=&i;

}

printf("%d",*j);

}

Answer:

Explanation:

The variable i is a block level variable and the visibility is inside that block only. But the lifetime of i is lifetime of the function so it lives upto the exit of main function. Since the i is still allocated space, *j prints the value stored in i since j points i.

5) main()

{

int i=-1;

-i;

printf("i = %d, -i = %d \n",i,-i);

}

Answer:

i = -1, -i = 1

Explanation:

-i is executed and this execution doesn't affect the value of i. In printf first you just print the value of i. After that the value of the expression -i = -(-1) is printed.

ALL PLACEMENT TECHNICAL INTERVIEW QUESTION BANCK

C Language Quiz Question with solution

noreply@blogger.com (Anonymous) — Mon, 1 Dec 2008 16:36:00 -0800

1) main()

{

int y;

scanf("%d",&y); // input given is 2000

if( (y%4==0 && y%100 != 0) || y%100 == 0 )

printf("%d is a leap year");

else

printf("%d is not a leap year");

}

Answer:

2000 is a leap year

Explanation:

An ordinary program to check if leap year or not.

2) #define max 5

#define int arr1[max]

main()

{

typedef char arr2[max];

arr1 list={0,1,2,3,4};

arr2 name="name";

printf("%d %s",list[0],name);

}

Answer:-

Compiler error (in the line arr1 list = {0,1,2,3,4})

Explanation:

arr2 is declared of type array of size 5 of characters. So it can be used to declare the variable name of the type arr2. But it is not the case of arr1. Hence an error.

Rule of Thumb:

#defines are used for textual replacement whereas typedefs are used for declaring new types.

3) int i=10;

main()

{

extern int i;

{

int i=20;

{

const volatile unsigned i=30;

printf("%d",i);

}

printf("%d",i);

}

printf("%d",i);

}

Answer:-

30,20,10

Explanation:

'{' introduces new block and thus new scope. In the innermost block i is declared as,

const volatile unsigned

4) main()

{

int *j;

{

int i=10;

j=&i;

}

printf("%d",*j);

}

Answer:-

Explanation:

5) main()

{

int i=-1;

-i;

printf("i = %d, -i = %d \n",i,-i);

}

Answer:

i = -1, -i = 1

Explanation:

-i is executed and this execution doesn't affect the value of i. In printf first you just print the value of i. After that the value of the expression -i = -(-1) is printed.

ALL PLACEMENT TECHNICAL INTERVIEW QUESTION BANCK

C programming Quiz Question with solution

noreply@blogger.com (Anonymous) — Mon, 1 Dec 2008 16:35:00 -0800

1) #include

main()

{

const int i=4;

float j;

j = ++i;

printf("%d %f", i,++j);

}

Answer:

Compiler error

Explanation:

i is a constant. you cannot change the value of constant

#include

main()

{

int a[2][2][2] = { {10,2,3,4}, {5,6,7,8} };

int *p,*q;

p=&a[2][2][2];

*q=***a;

printf("%d..%d",*p,*q);

}

Answer:

garbagevalue..1

Explanation:

p=&a[2][2][2] you declare only two 2D arrays. but you are trying to access the third 2D(which you are not declared) it will print garbage values. *q=***a starting address of a is assigned integer pointer. now q is pointing to starting address of a.if you print *q meAnswer:it will print first element of 3D array.

#include

main()

{

char j[ ]= "hello";

printf("%s %d",j,i);

}

Answer:-

hello 5

Explanation:

if you declare i as register compiler will treat it as ordinary integer and it will take integer value. i value may be stored either in register or in memory.

main()

{

int i=5,j=6,z;

printf("%d",i+++j);

}

Answer:

Explanation:

the expression i+++j is treated as (i++ + j)

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questions with answer in C programming language

noreply@blogger.com (Anonymous) — Mon, 1 Dec 2008 16:34:00 -0800

struct aaa

{

struct aaa *prev;

int i;

struct aaa *next;

};

main()

{

struct aaa abc,def,ghi,jkl;

int x=100;

abc.i=0;abc.prev=&jkl;

abc.next=&def;

def.i=1;def.prev=&abc;def.next=&ghi;

ghi.i=2;ghi.prev=&def;

ghi.next=&jkl;

jkl.i=3;jkl.prev=&ghi;jkl.next=&abc;

x=abc.next->next->prev->next->i;

printf("%d",x);

}

Answer:

Explanation:

above all statements form a double circular linked list;

abc.next->next->prev->next->i

this one points to "ghi" node the value of at particular node is 2.

struct point

{

int x;

int y;

};

struct point origin,*pp;

main()

{

pp=&origin;

printf("origin is(%d%d)\n",(*pp).x,(*pp).y);

printf("origin is (%d%d)\n",pp->x,pp->y);

}

Answer:

origin is(0,0)

Explanation:

pp is a pointer to structure. we can access the elements of the structure either with arrow mark or with indirection operator.

Note:

Since structure point is globally declared x & y are initialized as zeroes

main()

{

int i=_l_abc(10);

printf("%d\n",--i);

}

int _l_abc(int i)

{

return(i++);

}

Answer:

Explanation:

return(i++) it will first return i and then increments. i.e. 10 will be returned.

main()

{

char *p;

int *q;

long *r;

p=q=r=0;

p++;

q++;

r++;

printf("%p...%p...%p",p,q,r);

}

Answer:

0001...0002...0004

Explanation:

++ operator when applied to pointers increments address according to their corresponding data-types.

main()

{

char c=' ',x,convert(z);

getc(c);

if((c>='a') && (c<='z'))

x=convert(c);

printf("%c",x);

}

convert(z)

{

return z-32;

}

Answer:

Compiler error

Explanation:

declaration of convert and format of getc() are wrong.

ALL PLACEMENT TECHNICAL INTERVIEW QUESTION BANCK

Technical questions with solution in C programming language

noreply@blogger.com (Anonymous) — Mon, 1 Dec 2008 16:33:00 -0800

main(int argc, char **argv)

{

printf("enter the character");

getchar();

sum(argv[1],argv[2]);

}

sum(num1,num2)

int num1,num2;

{

return num1+num2;

}

Answer:

Compiler error.

Explanation:

argv[1] & argv[2] are strings. They are passed to the function sum without converting it to integer values.

# include

int one_d[]={1,2,3};

main()

{

int *ptr;

ptr=one_d;

ptr+=3;

printf("%d",*ptr);

}

Answer:-

garbage value

Explanation:

ptr pointer is pointing to out of the array range of one_d.

# include

aaa() {

printf("hi");

}

bbb(){

printf("hello");

}

ccc(){

printf("bye");

}

main()

{

int (*ptr[3])();

ptr[0]=aaa;

ptr[1]=bbb;

ptr[2]=ccc;

ptr[2]();

}

Answer:-

bye

Explanation:

ptr is array of pointers to functions of return type int.ptr[0] is assigned to address of the function aaa. Similarly ptr[1] and ptr[2] for bbb and ccc respectively. ptr[2]() is in effect of writing ccc(), since ptr[2] points to ccc.

#include

main()

{

FILE *ptr;

char i;

ptr=fopen("zzz.c","r");

while((i=fgetch(ptr))!=EOF)

printf("%c",i);

}

Answer:

contents of zzz.c followed by an infinite loop

Explanation:

The condition is checked against EOF, it should be checked against NULL.

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