C Placement Question

Leap Year Problem.....

2012-04-10T12:47:00.000-07:00

1-A year is called leap year if it is divisible by 400
2-If year is not divisible by 400 as well as 100 but it is divisible by 4 then that year are also leap year.

To determine a year is leap year or not...

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

printf("Enter any year: ");
scanf("%d",&year);

if(((year%4==0)&&(year%100!=0))||(year%400==0))
printf("%d is a leap year",year);
else
printf("%d is not a leap year",year);

return 0;
}

To find leap year in between a range of year....

#include<stdio.h>
int main(){
int year;
int min_year,max_year;

printf("Enter the lowest year: ");
scanf("%d",&min_year);

printf("Enter the heighest year: ");
scanf("%d",&max_year);

printf("Leap years in given range are: ");
for(year = min_year;year <= max_year; year++){
if(((year%4==0)&&(year%100!=0))||(year%400==0))
printf("%d ",year);
}

return 0;
}

Reverse a number using c

2012-04-10T12:32:00.000-07:00

1-Simple method of reversing an integer number with the help of while loop....

#include<stdio.h>

int main(){

int num,r,reverse=0;

printf("Enter any number: ");

scanf("%d",&num);

while(num){

r=num%10;

reverse=reverse*10+r;

num=num/10;

}

printf("Reversed of number: %d",reverse);

return 0;

}

2-With help of for loop....

#include<stdio.h>

int main(){

int num,r,reverse=0;

printf("Enter any number: ");

scanf("%d",&num);

for(;num!=0;num=num/10){

r=num%10;

reverse=reverse*10+r;

}

printf("Reversed of number: %d",reverse);

return 0;

}

3-With help of recursive function....

#include<stdio.h>

int main(){

int num,reverse;

printf("Enter any number: ");

scanf("%d",&num);

reverse=rev(num);

printf("Reverse of number: %d",reverse);

return 0;

}

int rev(int num){

static sum,r;

if(num){

r=num%10;

sum=sum*10+r;

rev(num/10);

}

else

return 0;

return sum;

}

4-And finally method for reversing a very large number which is beyond the range of int and long int....

#include<stdio.h>

#define MAX 1000

int main(){

char num[MAX];

int i=0,j,flag=0;

printf("Enter any positive integer: ");

scanf("%s",num);

while(num[i]){

if(num[i] < 48 || num[i] > 57){

printf("Invalid integer number");

return 0;

}

i++;

}

printf("Reverse: ");

for(j=i-1;j>=0;j--)

if(flag==0 && num[j] ==48){

}

else{

printf("%c",num[j]);

flag =1;

}

return 0;

preprocessor

2012-03-21T08:29:00.004-07:00

1) What will be output of following code?

#define max 10
void main()
{
int i;
i=++max;
clrscr();
printf("%d",i);
getch();
}

output: compiler error.
Explanation: max is preprocessor macro which process first before the actual compilation. Preprocessor paste the symbol to its constant value in entire the program before the compilation. so in this program max will be replaced by 10 before compilation. Thus program will be converted as:

void main()
{
int i;
i=++10;
clrscr();
printf("%d",i);
getch();
}

To know how we can see intermediate file click here

In this program we are trying to increment a constant symbol.

Meaning of ++10 IS:-
10=10+1
or 10=11

Which is error because we cannot assign constant value to another constant value .Hence compiler will give error.

(2) What will be output of following code?

#define max 10+2
void main()
{
int i;
i=max*max;
clrscr();
printf("%d",i);
getch();
}

Output:32

Explanation:

max is preprocessor macro which process first before the actual compilation. Preprocessor paste the symbol to the its constant value without any calculation in entire the program before the compilation. So in this program max will be replaced by 10+2 before compilation. Thus program will be converted as:

void main()
{
int i;
i=10+2*10+2;
clrscr();
printf("%d",i);
getch();
}

now i=10+2*10+2
i=10+20+2
i=32

(3) What will be output of following code?

#define A 4-2
#define B 3-1
void main()
{
int ratio=A/B;
printf("%d ",ratio);
getch();
}

Output:3
Explanation: A and B preprocessor macro which process first before the actual compilation. Preprocessor paste the symbol to its constant value without any calculation in entire the program before the compilation. So in this program A and B will be replaced by 4-2 and 3-1 respectively before compilation. Thus program will be converted as:

void main()
{
int ratio=4-2/3-1;
printf("%d ",ratio);
getch();
}

Here ratio=4-2/3-1
ratio=4-0-1
ratio=3

(4) What will be output of following code?

#define MAN(x,y) (x)>(y)?(x):(y)
void main()
{
int i=10,j=9,k=0;
k=MAN(i++,++j);
printf("%d %d %d",i,j,k);
getch();
}

Output: 11 11 11

Explanation: Preprocessor’s macro which process first before the actual compilation. Preprocessor paste the symbol to its constant value without any calculation in entire the program before the compilation. Thus program will be converted as:

void main()
{
int i=10,j=9,k=0;
k=(i++)>(++j)?(i++):(++j);
printf("%d %d %d",i,j,k);
getch();
}

now k=(i++)>(++j)?(i++):(++j);
first it will check the condition
(i++)>(++j)

i++ i.e. when postfix is used with variable in expression then expression is evaluated first with original value then variable is incremented

Or 10>10

This condition is false.
Now i = 10+1 = 11
There is rule, only false part will execute after? i.e. ++j, i++ will be not execute.
So after ++j
j=10+1=11;

And k will assign value of j .so k=11;

(5) What will be output of following code?

#define START main() {
#define PRINT printf("*******");
#define END }
START
PRINT
END

Output: *******
Explanation:

Preprocessor macro which process first before the actual compilation. Preprocessor paste the symbol to its constant value i.e. symbols without any calculation in entire the program before the compilation. Thus program will be converted as:

main()
{
printf("*******");
}

(6) What will be output of following code?
#define CUBE(x) (x*x*x)
#define M 5
#define N M+1
#define PRINT printf("RITESH");
void main()
{
int volume =CUBE(3+2);
clrscr();
printf("%d %d ",volume,N);
PRINT
getch();
}

Output: 17 6
Explanation:

Preprocessor macro which process first before the actual compilation. Preprocessor paste the symbol to its constant value without any calculation in entire the program before the compilation. Thus program will be converted as:

void main()
{
int volume =(3+2*3+2*3+2);
clrscr();
printf("%d %d ",volume,5+1);
PRINT
getch();
}

Technical set 2

2012-03-07T21:50:00.001-08:00

1.
What will be output when you will execute following c code?
#include<stdio.h>
void main(){
switch(5/2*6+3.0){
case 3:printf("David Beckham");
break;
case 15:printf("Ronaldinho");
break;
case 0:printf("Lionel Messi");
break;
default:printf("Ronaldo");
}
}
Choose all that apply:
(A) David Beckham
(B) Ronaldinho
(C) Lionel Messi
(D) Ronaldo
(E) Compilation error
Explanation:e
2.
What will be output when you will execute following c code?
#include<stdio.h>
void main(){
unsigned char c=280;
switch(c){
printf("Start\t");
case 280:printf("David Beckham\t");
case 24: printf("Ronaldinho\t");
default: printf("Ronaldo\t");
printf("End");
}
}
Choose all that apply:
(A)Start David Beckham Ronaldinho Ronaldo End
(B) Start David Beckham Ronaldinho Ronaldo
(C) Start Ronaldinho Ronaldo End
(D)Ronaldinho Ronaldo End
(E) Compilation error
Explanation:d
3.
What will be output when you will execute following c code?
#include<stdio.h>
#define TRUE 1
void main(){
switch(TRUE){
printf("cquestionbank.blogspot.com");
}
}
Choose all that apply:
(A) cquestionbank.blogspot.com
(B) It will print nothing
(C) Runtime error
(D) Compilation error
(E) None of the above
Explanation:b
4.
What will be output when you will execute following c code?
#include<stdio.h>
void main(){
static int i;
int j=1;
int arr[5]={1,2,3,4};
switch(arr[j]){
case 1: i++;break;
case 2: i+=2;j=3;continue;
case 3: i%=2;j=4;continue;
default: --i;
}
printf("%d",i);
}
Choose all that apply:
(A) 0
(B) 1
(C) 2
(D) Compilation error
(E) None of the above
Explanation:d
5.
What will be output when you will execute following c code?
#include<stdio.h>
void main(){
static int i;
int j;
for(j=0;j<=5;j+=2)
switch(j){
case 1: i++;break;
case 2: i+=2;
case 4: i%=2;j=-1;continue;
default: --i;continue;
}
printf("%d",i);
}
Choose all that apply:
(A) 0
(B) 1
(C) 2
(D) Compilation error
(E) None of the above
Explanation:a
6.
What will be output when you will execute following c code?
#include<stdio.h>
void main(){
int x=3;
while(1){
switch(x){
case 5/2: x+=x+++x;
case 3%4: x+=x---x;continue;
case 3>=3: x+=!!!x;break;
case 5&&0:x+=~~~x;continue;
default: x+=-x--;
}
break;
}
printf("%d",x);
}
Choose all that apply:
(A) 3
(B) -1
(C) 5
(D) Compilation error
(E) None of the above
Explanation:b
7.
What will be output when you will execute following c code?
#include<stdio.h>
void main(){
char *str="cquestionbank.blogspot.com";
int a=2;
switch('A'){
case 97:
switch(97){
default: str+=1;
}
case 65:
switch(97){
case 'A':str+=2;
case 'a':str+=4;
}
default:
for(;a;a--)
str+=8;
}
printf("%s",str);
}
Choose all that apply:
(A) cquestionbank.blogspot.com
(B) blogspot.com
(C) com
(D) Compilation error
(E) None of the above
Explanation:e
8.
What will be output when you will execute following c code?
#include<stdio.h>
void main(){
switch(2){
case 1L:printf("No");
case 2L:printf("%s","I");
goto Love;
case 3L:printf("Please");
case 4L:Love:printf("Hi");
}
}
Choose all that apply:
(A) I
(B) IPleaseHi
(C) IHi
(D) Compilation error
(E) None of the above
Explanation:c
9.
What will be output when you will execute following c code?
#include<stdio.h>
void main(){
int a=5;
a=a>=4;
switch(2){
case 0:int a=8;
case 1:int a=10;
case 2:++a;
case 3:printf("%d",a);
}
}
Choose all that apply:
(A) 8
(B) 11
(C) 10
(D) Compilation error
(E) None of the above
Explanation:e
10.
What will be output when you will execute following c code?
#include<stdio.h>
void main(){
int a=3,b=2;
a=a==b==0;
switch(1){
a=a+10;
}
sizeof(a++);
printf("%d",a);
}
Choose all that apply:
(A) 10
(B) 11
(C) 12
(D)1
(E) Compilation error
Explanation:d

Technical set 1

2012-03-07T21:40:00.001-08:00

1.
What will be output when you will execute following c code?#include<stdio.h>
void main(){
     int check=2;
     switch(check){
        case 1: printf("D.W.Steyn");
        case 2: printf(" M.G.Johnson");
        case 3: printf(" Mohammad Asif");
        default: printf(" M.Muralidaran");
     }
}
Choose all that apply:
(A)    M.G.Johnson
(B)    M.Muralidaran
(C)
M.G.Johnson Mohammad Asif M.Muralidaran
(D)    Compilation error
(E)    None of the above
Explanation:c
2.
What will be output when you will execute following c code?#include<stdio.h>
void main(){
     int movie=1;
     switch(movie<<2+movie){
        default:printf("3 Idiots");
        case 4: printf(" Ghajini");
        case 5: printf(" Krrish");
        case 8: printf(" Race");
     }
}
Choose all that apply:
(A)    3 Idiots Ghajini Krrish Race
(B)    Race
(C)    Krrish
(D)    Ghajini Krrish Race
(E)    Compilation error
Explanation:b
3.
What will be output when you will execute following c code?#include<stdio.h>
#define L 10
void main(){
     auto money=10;
     switch(money,money*2){
        case L: printf("Willian");
                  break;
        case L*2:printf("Warren");
                  break;
        case L*3:printf("Carlos");
                  break;
        default: printf("Lawrence");
        case L*4:printf("Inqvar");
                  break;
     }
}
Choose all that apply:
(A)    Willian
(B)    Warren
(C)    Lawrence Inqvar
(D)    Compilation error: Misplaced default
(E)    None of the above
Explanation:b
4.
What will be output when you will execute following c code?#include<stdio.h>
void main(){
     int const X=0;
     switch(5/4/3){
        case X: printf("Clinton");
                  break;
        case X+1:printf("Gandhi");
                  break;
        case X+2:printf("Gates");
                  break;
        default: printf("Brown");
     }
}
Choose all that apply:
(A)    Clinton
(B)    Gandhi
(C)    Gates
(D)    Brown
(E)    Compilation error
Explanation:e
5.
What will be output when you will execute following c code?#include<stdio.h>
enum actor{
    SeanPenn=5,
    AlPacino=-2,
    GaryOldman,
    EdNorton
};
void main(){
     enum actor a=0;
     switch(a){
        case SeanPenn: printf("Kevin Spacey");
                         break;
        case AlPacino: printf("Paul Giamatti");
                         break;
        case GaryOldman:printf("Donald Shuterland");
                         break;
        case EdNorton: printf("Johnny Depp");
     }
}
Choose all that apply:
(A)    Kevin Spacey
(B)    Paul Giamatti
(C)    Donald Shuterland
(D)    Johnny Depp
(E)    Compilation error
Explanation:d
6.
What will be output when you will execute following c code?
#include<stdio.h>
void main(){
     switch(*(1+"AB" "CD"+1)){
        case 'A':printf("Pulp Fiction");
                  break;
        case 'B':printf("12 Angry Man");
                  break;
        case 'C':printf("Casabance");
                  break;
        case 'D':printf("Blood Diamond");
     }
}
Choose all that apply:
(A)    Pulp Fiction
(B)    12 Angry Man
(C)    Casabance
(D)    Blood Diamond
(E)    Compilation error
Explanation:c
7.
What will be output when you will execute following c code?#include<stdio.h>
void main(){
    char *str="ONE"
    str++;
    switch(str){
        case "ONE":printf("Brazil");
                break;
        case "NE": printf("Toy story");
                break;
        case "N": printf("His Girl Friday");
                break;
        case "E": printf("Casino Royale");
     }
}
Choose all that apply:
(A)    Brazil
(B)    Toy story
(C)    His Girl Friday
(D)    Casino Royale
(E)    Compilation error
Explanation:e
8.
What will be output when you will execute following c code?#include<stdio.h>
void main(){
    switch(5||2|1){
        case 3&2:printf("Anatomy of a Murder");
              break;
        case -~11:printf("Planet of Apes");
               break;
        case 6-3<<2:printf("The conversation");
               break;
    case 5>=5:printf("Shaun of the Dead");
     }
}
Choose all that apply:
(A)    Anatomy of a Murder
(B)    Planet of Apes
(C)    The conversation
(D)    Shaun of the Dead
(E)    Compilation error
Explanation:e
9.
What will be output when you will execute following c code?#include<stdio.h>
void main(){
    switch(6){
        case 6.0f:printf("Sangakkara");
               break;
        case 6.0: printf("Sehwag");
               break;
        case 6.0L:printf("Steyn");
               break;
        default: printf("Smith");
    }
}
Choose all that apply:
(A)    Sangakkara
(B)    Sehwag
(C)    Steyn
(D)    Smith
(E)    Compilation error
Explanation:e
10.
What will be output when you will execute following c code?
#include<stdio.h>
void main(){
    switch(0X0){
        case NULL:printf("Thierry Henry");
             break;
        case '\0':printf("Steven Gerrend");
             break;
        case 0: printf("Kaka");
             break;
        default: printf("Michael Ballack");
    }
}
Choose all that apply:
(A)    Thierry Henry
(B)    Steven Gerrend
(C)    Kaka
(D)    Michael Ballack
(E)    Compilation error
Explanation:e

C Placement Question

2012-03-04T08:01:00.001-08:00

C Question set

c question set 1	c question set 2	c question set 3
c question set 4	c question set 5	c question set 6
pointers	array	data types
string	if-else set 1	if-else set 2
printf

printf

2012-03-04T06:44:00.001-08:00

(1)
#include<stdio.h>
#include<conio.h>
void main()
{
int a=5,b=6,c=11;
clrscr();
printf("%d %d %d");
getch();
}
What will output when you compile and run the above code?
(a)Garbage value garbage value garbage value
(b)5 6 11
(c)11 6 5
(d)Compiler error
Answer: (c)

(2)
#include<stdio.h>
void main()
{
char *str="CQUESTIONBANK";
clrscr();
printf(str+9);
getch();
}
What will output when you compile and run the above code?
(a)CQESTIONBANK
(b)CQUESTION
(c)BANK
(d)Compiler error
Answer: (c)
(3)
#include<stdio.h>
void main()
{
clrscr();
printf("%d",printf("CQUESTIONBANK"));
getch();
}
What will output when you compile and run the above code?
(a)13CQUESTIONBANK
(b)CQUESTIONBANK13
(c)Garbage CQUESTIONBANK
(d)Compiler error
Answer: (b)
(4)#include<stdio.h>
#include<conio.h>
void main()
{
short int a=5;
clrscr();
printf("%d"+1,a);
getch();
}
What will output when you compile and run the above code?(a)6
(b)51
(c)d
(d)Compiler error
Answer: (c)
(5)
#include<stdio.h>
void main()
{
int i=85;
clrscr();
printf("%p %Fp",i,i);
getch();
}
What will output when you compile and run the above code?
(a)85 85
(b)0055 034E:0055
(c)0055 FFFF:0055
(d)Compiler error
Answer: (b)

(6)#include<stdio.h>
static struct student
{
int a;
    int b;
    int c;
int d;
}s1={6,7,8,9},s2={4,3,2,1},s3;
void main()
{
s3=s1+s2;
clrscr();
printf("%d %d %d %d",s3.a,s3.b,s3.c,s3.d);
getch();
}
What will output when you compile and run the above code?(a)6789
(b)4321
(c)10101010
(d)Compiler error
Answer: (d)
(7)#include<stdio.h>
extern struct student
{
int a;
    int b;
    int c;
int d;
}s={6,7,8,9};
void main()
{
clrscr();
printf("%d %d %d %d",s.a,s.b,s.c,s.d);
getch();
}
What will output when you compile and run the above code?(a)6789
(b)9876
(c)0000
(d)Compiler error
Answer: (a)
(8)#include<stdio.h>
struct student
{
static int a;
register int b;
auto int c;
extern int d;
}s={6,7,8,9};
void main()
{
printf("%d %d % %d",s.a,s.b,s.c,s.d);
}
What will output when you compile and run the above code?(a)6789
(b)9876
(c)0000
(d)Compiler error
Answer: (d)
(9)#include<stdio.h>
struct student
{
int roll;
int cgpa;
int sgpa[8];
};
void main()
{
struct student s={12,8,7,2,5,9};
int *ptr;
ptr=(int *)&s;
clrscr();
printf("%d",*(ptr+3));
getch();
}
What will output when you compile and run the above code?(a)8
(b)7
(c)2
(d)Compiler error
Answer: (c)
(10)#include<stdio.h>
struct game
{
int level;
int score;
struct player
{
char *name;
}g2={"anil"};
}g3={10,200};
void main()
{
struct game g1=g3;
clrscr();
printf("%d %d %s",g1.level,g1.score,g1.g2.name);
getch();
}
What will output when you compile and run the above code?(a)10 200 anil
(b)200 10 anil
(c)10 200 null
(d)Compiler error
Answer: (d)
(11)#include<stdio.h>
struct game
{
int level;
int score;
struct player
{
char *name;
}g2;
}g1;
void main()
{
clrscr();
printf("%d %d %s",g1.level,g1.score,g1.g2.name);
getch();
}
What will output when you compile and run the above code?(a)Garbage_value garbage_value garbage_value
(b)0 0 (null)
(c)Run time error
(d)Compiler error
Answer: (b)

if-else set 2

2012-03-01T03:56:00.001-08:00

1.
What will be output when you will execute following c code?
#include<stdio.h>
void main(){
int a=10;
if(printf("%d",a>=10)-10)
for(;;)
break;
else;
}
Choose all that apply:
(A) It will print nothing
(B) 0
(C) 1
(D) Compilation error: Misplaced else
(E) Infinite loop
Explanation:C
Return type of printf function is int. This function return a integral value which is equal to number of charcters printf function will print on console.
Operator >= will return 1 if both operands are either equal or first operand is grater than second operand. So a>=10 will return 1 since a is equal to 10.Thus printf function will print 1. Since this function is printing only one character so it will also return 1. So, printf("%d",a>=10) - 10
= 1 - 10
= -9
Since -9 is non-zero number so if(-9) is true condition and if clause will execute.

2.
What will be output when you will execute following c code?
#include<stdio.h>
void main(){
int a=5,b=10;
if(++a||++b)
printf("%d %d",a,b);
else
printf("John Terry");
}
Choose all that apply:
(A) 5 10
(B) 6 11
(C) 6 10
(D) 5 11
(E) John Terry
Explanation:c
Consider the following expression:
++a || ++b
In this expression || is Logical OR operator. Two important properties of this operator are:
Property 1:
(Expression1) || (Expression2)
|| operator returns 0 if and only if both expressions return a zero otherwise it || operator returns 1.
Property 2:
To optimize the execution time there is rule, Expression2 will only evaluate if and only if Expression1 return zero.
In this program initial value of a is 5. So ++a will be 6. Since ++a is returning a non-zero so ++b will not execute and if condition will be true and if clause will be executed.

3.
What will be output when you will execute following c code?
#include<stdio.h>
void main(){
static int i;
for(;;)
if(i+++"The Matrix")
printf("Memento");
else
break;
}
Choose all that apply:
(A) It will print Memento at one time
(B) It will print Memento at three times
(C) It will print Memento at 65535 times
(D) It will print Memento at infinite times
(E) Compilation error: Unknown operator +++
Explanation:C
think urself

4.
What will be output when you will execute following c code?
#include<stdio.h>
void main(){
int x=1;
if(x--)
printf("The Godfather");
--x;
else
printf("%d",x);
}
Choose all that apply:
(A) The Godfather
(B) 1
(C) 0
(D) Compilation error
(E) None of the above
Explanation:d
If you are not using { and } in if clause then you can write only one statement. Otherwise it will cause of compilation error: Misplace else

5.
What will be output when you will execute following c code?
#include<stdio.h>
void main(){
if('\0');
else if(NULL)
printf("cquestionbank");
else;
}
Choose all that apply:
(A) cquestionbank
(B) It will print nothing
(C) Warning: Condition is always true
(D) Warning: Unreachable code
(E) Compilation error: if statement without any body
Explanation:b,d
‘\0’ is null character constant. Its ASCII value is zero. if(0) means false so program control will check it else if clause.
NULL is macro constant which has been defined in stdio.h which also returns zero.

6.
What will be output when you will execute following c code?
#include<stdio.h>
void main(){
int a=5,b=10;
clrscr();
if(a<++a||b<++b)
printf("%d %d",a,b);
else
printf("John Terry");
}
Choose all that apply:
(A) 5 10
(B) 6 11
(C) 6 10
(D) Compilation error
(E) John Terry
Explanation:e
Consider the following expression:
a<++a||b<++b
In the above expression || is logical OR operator. It divides any expression in the sub expressions. In this way we have two sub expressions:
(1) a<++a
(2) b<++b
In the expression: a< ++a
There are two operators. There precedence and associate are:
Precedence
Operator
Associate
1
++
Right to left
2
<
Left to right
From table it is clear first ++ operator will perform the operation then < operator.
One important property of pre-increment (++) operator is: In any expression first pre-increment increments the value of variable then it assigns same final value of the variable to all that variables. So in the expression: a < ++a
Initial value of variable a is 5.
Step 1: Increment the value of variable a in whole expression. Final value of a is 6.
Step 2: Now start assigning value to all a in the expression. After assigning 6 expression will be:
6 < 6
Since condition is false .So second expression i.e. b<++b will be evaluated. Again 11 < 11 is false. So || will operator will return zero and else clause will execute.

7.
What will be output when you will execute following c code?
#include<stdio.h>
void main(){
int x=1,y=2;
if(--x && --y)
printf("x=%d y=%d",x,y);
else
printf("%d %d",x,y);
}
Choose all that apply:
(A) 1 2
(B) x=1 y=2
(C) 0 2
(D) x=0 y=1
(E) 0 1
Explanation:c
Consider the following expression:
--x && --y
In this expression && is Logical AND operator. Two important properties of this operator are:
Property 1:
(Expression1) && (Expression2)
&& operator returns 1 if and only if both expressions return a non-zero value other wise it && operator returns 0.
Property 2:
To optimize the execution time there is rule, Expression2 will only evaluate if and only if Expression1 return a non-zero value.
In this program initial value of x is 1. So –x will be zero. Since -–x is returning zero so -–y will not execute and if condition will be false. Hence else part will be executed.

8.
What will be output when you will execute following c code?
#include<stdio.h>
void main(){
signed int a=-1;
unsigned int b=-1u;
if(a==b)
printf("The Lord of the Rings");
else
printf("American Beauty");
}
Choose all that apply:
(A) The Lord of the Rings
(B) American Beauty
(C) Compilation error: Cannot compare signed number with unsigned number
(D) Compilation error: Undefined symbol -1u
(E) Warning: Illegal operation
Explanation:a

9.
What will be output when you will execute following c code?
#include<stdio.h>
void main(){
char c=256;
char *ptr="Leon";
if(c==0)
while(!c)
if(*ptr++)
printf("+%u",c);
else
break;
}
Choose all that apply:
(A) +256+256+256+256
(B) 0000
(C) +0+0+0+0
(D) It will print +256 at infinite times
(E) Compilation error
Explanation:c
In the above program c is signed (default) char variable. Range of signed char variable in Turbo c is from -128 to 127. But we are assigning 256 which is beyond the range of variable c. Hence variable c will store corresponding cyclic value according to following diagram:
Since 256 is positive number move from zero in clock wise direction. You will get final value of c is zero.
if(c==0)
It is true since value of c is zero.
Negation operator i.e. ! is always return either zero or one according to following rule:
!0 = 1
!(Non-zero number) = 0
So,
!c = !0 =1
As we know in c zero represents false and any non-zero number represents true. So
while(!c) i.e. while(1) is always true.
In the above program prt is character pointer. It is pointing to first character of string “Leon” according to following diagram:
In the above figure value in circle represents ASCII value of corresponding character.
Initially *ptr means ‘L’. So *ptr will return ASCII value of character constant ‘L’ i.e. 76
if(*ptr++) is equivalent to : if(‘L’) is equivalent to: if(76) . It is true so in first iteration it will print +0. Due to ++ operation in second iteration ptr will point to character constant ‘e’ and so on. When ptr will point ‘\0’ i.e. null character .It will return its ASCII value i.e. 0. So if(0) is false. Hence else part will execute.

10.
What will be output when you will execute following c code?
#include<stdio.h>
void main(){
int a=2;
if(a--,--a,a)
printf("The Dalai Lama");
else
printf("Jim Rogers");
}
Choose all that apply:
(A) The Dalai Lama
(B) Jim Rogers
(C) Run time error
(D) Compilation error: Multiple parameters in if statement
(E) None of the above
Explanation:b
Consider the following expression:
a-- , --a , a
In c comma is behaves as separator as well as operator.In the above expression comma is behaving as operator.Comma operator enjoy lest precedence in precedence table andits associatively is left to right. So first of all left most comma operator will perform operation then right most comma will operator in the above expression.
After performing a-- : a will be 2
After performing --a : a will be 0
a=0
As we know in c zero represents false and any non-zero number represents true. Hence else part will execute.

if-else set 1

2012-03-01T03:25:00.001-08:00

(1).

What will be output when you will execute following c code?

#include<stdio.h>
void main(){
    int a=5,b=10,c=1;
    if(a&&b>c){
         printf("cquestionbank");
    }
    else{
         break;
    }
}

Choose all that apply:

(A) cquestionbank
(B) It will print nothing
(C) Run time error
(D) Compilation error
(E) None of the above

Explanation: D

Keyword break is not syntactical part of if-else statement. So we cannot use break keyword in if-else statement. This keyword can be use in case of loop or switch case statement.
Hence when you will compile above code compiler will show an error message: Misplaced break.

(2).

What will be output when you will execute following c code?

#define PRINT printf("Star Wars");printf(" Psycho");
#include<stdio.h>
void main(){
    int x=1;
    if(x--)
         PRINT
    else
         printf("The Shawshank Redemption");
}

Choose all that apply:

(A) Stars Wars Psycho
(B) The Shawshank Redemption
(C) Warning: Condition is always true
(D) Warning: Condition is always false
(E) Compilation error

Explanation:E

PRINT is macro constant. Macro PRINT will be replaced by its defined statement just before the actual compilation starts. Above code is converted as:

void main(){
    int x=1;
    if(x--)
         printf("Star Wars");
printf(" Psycho");
    else
         printf("The Shawshank Redemption");
}
If you are not using opening and closing curly bracket in if clause, then you can write only one statement in the if clause. So compiler will think:
(i)
if(x--)
    printf("Star Wars");

It is if statement without any else. It is ok.
(ii)
printf(" Psycho");

It is a function call. It is also ok
(iii)
else
printf("The Shawshank Redemption");

You cannot write else clause without any if clause. It is cause of compilation error. Hence compiler will show an error message: Misplaced else

(3).

What will be output when you will execute following c code?

#define True 5==5
#include<stdio.h>
void main(){
    if(.001-0.1f)
         printf("David Beckham");
    else if(True)
         printf("Ronaldinho");
    else
        printf("Cristiano Ronaldo");
}

Choose all that apply:

(A) David Beckham
(B) Ronaldinho
(C) Cristiano Ronaldo
(D) Warning: Condition is always true
(E) Warning: Unreachable code

Explanation:a,d,e

As we know in c zero represents false and any non-zero number represents true. So in the above code:

(0.001 – 0.1f) is not zero so it represents true. So only if clause will execute and it will print: David Beckham on console.
But it is bad programming practice to write constant as a condition in if clause. Hence compiler will show a warning message: Condition is always true

Since condition is always true, so else clause will never execute. Program control cannot reach at else part. So compiler will show another warning message:
Unreachable code

(4).

What will be output when you will execute following c code?

#include<stdio.h>
void main(){
    int a=100;
    if(a>10)
         printf("M.S. Dhoni");
    else if(a>20)
         printf("M.E.K Hussey");
    else if(a>30)
           printf("A.B. de villiers");
}

Choose all that apply:

(A) M.S. Dhoni
(B) A.B. de villiers
(C) M.S Dhoni
M.E.K Hussey
A.B. de Villiers
(D) Compilation error: More than one conditions are true
(E) None of the above

Explanation:a

In case of if – if else – if else … Statement if first if clause is true the compiler will never check rest of the if else clause and so on.

(5).

What will be output when you will execute following c code?

#include<stdio.h>
void main(){
    int x=-1,y=-1;
    if(++x=++y)
         printf("R.T. Ponting");
    else
         printf("C.H. Gayle");
}

Choose all that apply:

(A) R.T Ponting
(B) C.H. Gayle
(C) Warning: x and y are assigned a value that is never used
(D) Warning: Condition is always true
(E) Compilation error

Explanation:c,e

Consider following statement:
++x=++y
As we know ++ is pre increment operator in the above statement. This operator increments the value of any integral variable by one and return that value. After performing pre increments above statement will be:

0=0
In C language it is illegal to assign a constant value to another constant. Left side of = operator must be a container i.e. a variable. So compiler will show an error message: Lvalue required

In c if you assign any value to variable but you don’t perform any operator or perform operation only using unary operator on the variable the complier will show a warning message: Variable is assigned a value that is never

(6).

What will be output when you will execute following c code?

#include<stdio.h>
void main(){
    if(sizeof(void))
         printf("M. Muralilidaran");
    else
         printf("Harbhajan Singh");
}

Choose all that apply:

(A) M. Muralilidaran
(B) Harbhajan Singh
(C) Warning: Condition is always false
(D) Compilation error
(E) None of the above

Explanation:d

It illegal to find size of void data type using sizeof operator. Because size of void data type is meaning less.

(7).

What will be output when you will execute following c code?

#include<stdio.h>
void main(){
    int m=5,n=10,q=20;
    if(q/n*m)
         printf("William Gates");
    else
         printf(" Warren Buffet");
         printf(" Carlos Slim Helu");
}

Choose all that apply:
(A) William Gates
(B) Warren Buffet Carlos Slim Helu
(C) Run time error
(D) Compilation error
(E) None of the above

Explanation:e

Consider the following expression:
q / n * m

In this expression there are two operators. They are:
/: Division operator
*: Multiplication operator
Precedence and associate of each operator is as follow:

Precedence
Operator
Associate
1
/ , *
Left to right
Precedence of both operators is same. Hence associate will decide which operator will execute first. Since Associate is left to right. So / operator will execute then * operator will execute.
= q / n * m
= 20 / 10 * 5
= 2 * 5
=10

As we know in c zero represents false and any non-zero number represents true. Since 10 is non- zero number so if clause will execute and print: William Gates

Since in else clause there is not any opening and closing curly bracket. So compiler will treat only one statement as a else part. Hence last statement i.e.
printf(" Carlos Slim Helu");

is not part of if-else statement. So at the end compiler will also print: Carlos Slim Helu
So output of above code will be:
William Gates Carlos Slim Helu

(8).

What will be output when you will execute following c code?

#include<stdio.h>
void main(){
if(!printf("Mukesh Ambani"))
if(printf(" Lakashmi Mittal"))
}

Choose all that apply:

(A) Mukesh Ambani
(B) Lakashmi Mittal
(C) It will print nothing
(D) Mukesh Ambani Lakashmi Mittal
(E) Compilation error: if statement without body

Explanation:a

Return type of printf function is int. This function return a integral value which is equal to number of characters a printf function will print on console. First of all printf function will: Mukesh Ambani. Since it is printing 13 character so it will return 13. So,

!printf("Mukesh Ambani")
= !13
= 0
In c language zero represents false. So if(0) is false so next statement which inside the body of first if statement will not execute.

(9).

What will be output when you will execute following c code?

#include<stdio.h>
void main(){
    if("ABC") printf("Barack Obama\n");
    if(-1)    printf("Hu Jintao\n");
    if(.92L) printf("Nicolas Sarkozy\n");
    if(0)     printf("Ben Bernanke\n");
    if('W')   printf("Vladimir Putin\n");
}

Choose all that apply:

(A) It will print nothing

(B) Barack Obama
       Hu Jintao
       Nicolas Sarkozy
       Vladimir Putin

(D) Hu Jintao
Nicolas Sarkozy
Vladimir Putin

(E) Compilation error

Explanation:b

“ABC”: It is string constant and it will always return a non-zero memory address.
0.92L: It is long double constant.
‘W’: It is character constant and its ASCII value is

As we know in c language zero represents false and any non-zero number represents true. In this program condition of first, second, third and fifth if statements are true.

(10).

What will be output when you will execute following c code?

#include<stdio.h>
void main(){
    if(0xA)
         if(052)
             if('\xeb')
                 if('\012')
                      printf("Tom hanks");
                 else;
             else;
         else;
    else;
}

Choose all that apply:

(A) Tom hanks
(B) Compilation error: Misplaced else
(C) Compilation error: If without any body
(D) Compilation error: Undefined symbol
(E) Warning: Condition is always true

Explanation:a,e

oxA: It is hexadecimal integer constant.
052: It octal integer constant.
‘\xeb’: It is hexadecimal character constant.
‘\012’: It is octal character constant.

As we know in c zero represents false and any non-zero number represents true. All of the above constants return a non-zero value. So all if conditions in the above program are true.

In c it is possible to write else clause without any body.

string

2012-02-24T07:04:00.001-08:00

(1)

Without using any semicolon (;) in program write a c program which output is: HELLO WORLD?

Answer:
void main()
{
if(printf("HELLO WORLD"))
{
}
}

(2)

What will be output of following code?
void main()
{
char a[5];
a[0]='q';
a[1]='u';
a[2]='e';
clrscr();
printf("%s",a);
getch();
}
Output: garbage
Explanation: %s is used for string but a is not a string it is only array of character since its last character is not null character.
If you will write a[3]=’\0’; then output will be que.

(3)

Write a scanf statement which can store one line of string which includes white space.

Answer:
void main()
{
char a[30];
clrscr();
scanf("%[^\n]",a);
printf("%s",a);
getch();
}

(4)

Write a c program in which a scanf function can store a paragraph at time?

Answer:
void main()
{
char a[30];
clrscr();
scanf("%[^\t]",a);
printf("%s",a);
getch();
}
Note: Paragraph will end when you will press tab key.

(5)

In the following program
void main()
{
extern char a[30];
}
How much array a is reserving memory space?
Answer:
It is not reserving any memory space because array a has only declared it has not initialized .Any not initialized variable doesn’t reserve any memory space.

data types

2012-02-24T06:36:00.003-08:00

1.
What will be output when you will execute following c code?

#include<stdio.h>
void main(){
printf("%d\t",sizeof(6.5));
printf("%d\t",sizeof(90000));
printf("%d",sizeof('A'));
}

Choose all that apply:
(A) 4 2 1
(B) 8 2 1
(C) 4 4 1
(D) 8 4 1
(E) 8 4 2

Explanation: D

2.
Consider on following declaring of enum.
(i) enum cricket {Gambhir,Smith,Sehwag}c;
(ii) enum cricket {Gambhir,Smith,Sehwag};
(iii) enum {Gambhir,Smith=-5,Sehwag}c;
(iv) enum c {Gambhir,Smith,Sehwag};
Choose correct one:
(A)Only (i) is correct declaration
(B) Only (i) and (ii) is correct declaration
(C) Only (i) and (iii) are correct declaration
(D) Only (i),(ii) and are correct declaration
(E) All four are correct declaration

Explanation: E

3.
What will be output when you will execute following c code?

#include<stdio.h>
void main(){
signed x;
unsigned y;
x = 10 +- 10u + 10u +- 10;
y = x;
if(x==y)
printf("%d %d",x,y);
else if(x!=y)
printf("%u %u",x,y);
}
Choose all that apply:
(A) 0 0
(B) 65536 -10
(C) 0 65536
(D) 65536 0
(E) Compilation error

Explanation:A

4.
Which of the following is not modifier of data type in c?

(A) extern
(B) interrupt
(C) huge
(D) register
(E) All of these are modifiers of data type

Explanation:E

5.
What will be output when you will execute following c code?

#include<stdio.h>
void main(){
double num=5.2;
int var=5;
printf("%d\t",sizeof(!num));
printf("%d\t",sizeof(var=15/2));
printf("%d",var);
}

Choose all that apply:
(A) 4 2 7
(B) 4 4 5
(C) 2 2 5
(D) 2 4 7
(E) 8 2 7

Explanation: C

6.
What will be output when you will execute following c code?

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

Choose all that apply:
(A) 0
(B) 10
(C) Garbage value
(D) Any memory address
(E) Error: Cannot modify const object

Explanation: B

7.
Consider on following declaration:
(i) short i=10;
(ii) static i=10;
(iii) unsigned i=10;
(iv) const i=10;

Choose correct one:
(A)Only (iv) is incorrect
(B) Only (ii) and (iv) are incorrect
(C) Only (ii),(iii) and (iv) are correct
(D) Only (iii) is correct
(E) All are correct declaration

Explanation: E

8.
What will be output when you will execute following c code?

#include<stdio.h>
void main(){
int a= sizeof(signed) +sizeof(unsigned);
int b=sizeof(const)+sizeof(volatile);
printf("%d",a+++b);
}

Choose all that apply:
(A) 10
(B) 9
(C) 8
(D) Error: Cannot find size of modifiers
(E) Error: Undefined operator +++

Explanation: C

9.
What will be output when you will execute following c code?

#include<stdio.h>
void main(){
signed x,a;
unsigned y,b;
a=(signed)10u;
b=(unsigned)-10;
y = (signed)10u + (unsigned)-10;
x = y;
printf("%d %u\t",a,b);
if(x==y)
printf("%d %d",x,y);
else if(x!=y)
printf("%u %u",x,y);
}

Choose all that apply:
(A) 10 -10 0 0
(B) 10 -10 65516 -10
(C) 10 -10 10 -10
(D) 10 65526 0 0
(E) Compilation error

Explanation: D

10.
Which of the following is integral data type?

(A) void
(B) char
(C) float
(D) double
(E) None of these

Explanation: B

11.
What will be output when you will execute following c code?

#include<stdio.h>
void main(){
volatile int a=11;
printf("%d",a);
}

Choose all that apply:
(A) 11
(B) Garbage
(C) -2
(D) We cannot predict
(E) Compilation error

Explanation: D

12.
What is the range of signed int data type in that compiler in which size of int is two byte?

(A) -255 to 255
(B) -32767 to 32767
(C) -32768 to 32768
(D) -32767 to 32768
(E) -32768 to 32767

Explanation: E

13.
What will be output when you will execute following c code?

#include<stdio.h>
const enum Alpha{
X,
Y=5,
Z
}p=10;
void main(){
enum Alpha a,b;
a= X;
b= Z;

printf("%d",a+b-p);

}

Choose all that apply:
(A) -4
(B) -5
(C) 10
(D) 11
(E) Error: Cannot modify constant object

Explanation: A

14.
What will be output when you will execute following c code?

#include<stdio.h>
void main(){
char a=250;
int expr;
expr= a+ !a + ~a + ++a;
printf("%d",expr);
}

Choose all that apply:
(A) 249
(B) 250
(C) 0
(D) -6
(E) Compilation error

Explanation: D

15.
Consider on order of modifiers in following declaration:

(i)char volatile register unsigned c;
(ii)volatile register unsigned char c;
(iii)register volatile unsigned char c;
(iv)unsigned char volatile register c;

Choose correct one:
(A) Only (ii) is correct declaration
(B) Only (i) is correction declaration
(C) All are incorrect
(D) All are correct but they are different
(E) All are correct and same

Explanation: E

16.
What will be output when you will execute following c code?

#include<stdio.h>
void main(){
int a=-5;
unsigned int b=-5u;
if(a==b)
printf("Avatar");
else
printf("Alien");
}

Choose all that apply:
(A) Avatar
(B) Alien
(C) Run time error
(D) Error: Illegal assignment
(E) Error: Don’t compare signed no. with unsigned no.

Explanation: A

17.
What will be output when you will execute following c code?

#include<stdio.h>
extern enum cricket x;
void main(){
printf("%d",x);
}
const enum cricket{
Taylor,
Kallis=17,
Chanderpaul
}x=Taylor|Kallis&Chanderpaul;

Choose all that apply:
(A) 0
(B) 15
(C) 16
(D) 17
(E) Compilation error

Explanation: C

18.
Which of the following is not derived data type in c?

(A) Function
(B) Pointer
(C) Enumeration
(D) Array
(E) All are derived data type

Explanation: C

19.
What will be output when you will execute following c code?

#include<stdio.h>
enum A{
x,y=5,
enum B{
p=10,q
}varp;
}varx;

void main(){
printf("%d %d",x,varp.q);
}

Choose all that apply:
(A) 0 11
(B) 5 10
(C) 4 11
(D) 0 10
(E) Compilation error

Explanation: E

20.
Consider on following declaration in c:

(i)short const register i=10;
(ii)static volatile const int i=10;
(iii)unsigned auto long register i=10;
(iv)signed extern float i=10.0;

Choose correct one:
(A) Only (iv)is correct
(B) Only (ii) and (iv) is correct
(C) Only (i) and (ii) is correct
(D) Only (iii) correct
(E) All are correct declaration

Explanation: C

array

2012-02-23T07:18:00.000-08:00

What will be output if you will execute following c code?

#include<stdio.h>
void main(){
char arr[7]="Network";
printf("%s",arr);
}

(A) Network
(B) N
(C) network
(D) Garbage value
(E) Compilation error
Explanation: **d ,array size overflow....
2.
What will be output if you will execute following c code?

#include<stdio.h>
void main(){
char arr[11]="The African Queen";
printf("%s",arr);
}
(A) The African Queen
(B) The
(C) Queen
(D) null
(E) Compilation error
Explanation: **d
3.
What will be output if you will execute following c code?
#include<stdio.h>
void main(){
char arr[20]="MysticRiver";
printf("%d",sizeof(arr));
}
(A) 20
(B) 11
(C) 12
(D) 22
(E) 24
Explanation: **a

4.
What will be output if you will execute following c code?
#include<stdio.h>
void main(){
int const SIZE=5;
int expr;
double value[SIZE]={2.0,4.0,6.0,8.0,10.0};
expr=1|2;
printf("%f",value[expr]);
}
(A) 2.000000
(B) 4.000000
(C) 6.000000
(D) 8.000000
(E) Compilation error
Explanation: **d ,as precedence of bitwise inclusive OR ( | ) is more than assignment operator so   1|2=3  expr=3;

5.
What will be output if you will execute following c code?

#include<stdio.h>
enum power{
Dalai,
Vladimir=3,
Barack,
Hillary
};
void main(){
float leader[Dalai+Hillary]={1.f,2.f,3.f,4.f,5.f};
enum power p=Barack;
printf("%0.f",leader[p>>1+1]);
}
(A) 1
(B) 2
(C) 3
(D) 5
(E) Compilation error
Explanation: **b , initial value of p is 4 right shift by 2 make p=1 leader[1]=2

6.
What will be output if you will execute following c code?

#include<stdio.h>
#define var 3
void main(){
char *cricket[var+~0]={"clarke","kallis"};
char *ptr=cricket[1+~0];
printf("%c",*++ptr);
}
(A) k
(B) a
(C) l
(D) i
(E) Compilation error
Explanation: **c

7.
What will be output if you will execute following c code?

#include<stdio.h>
#define var 3
void main(){
char data[2][3][2]={0,1,2,3,4,5,6,7,8,9,10,11};
printf("%o",data[0][2][1]);
}
(A) 5
(B) 6
(C) 7
(D) 8
(E) Compilation error
Explanation: **a

8.
What will be output if you will execute following c code?

#include<stdio.h>
#define var 3
void main(){
short num[3][2]={3,6,9,12,15,18};
printf("%d %d",*(num+1)[1],**(num+2));
}
(A) 12 15
(B) 12 12
(C) 15 15
(D) 15 12
(E) Compilation error
Explanation: **c

9.
What will be output if you will execute following c code?

#include<stdio.h>
#define var 3
void main(){
char *ptr="cquestionbank";
printf("%d",-3[ptr]);
}
(A) 100
(B) -100
(C) 101
(D) -101
(E) Compilation error
Explanation: **d

10.
What will be output if you will execute following c code?

#include<stdio.h>
void main(){
long myarr[2][4]={0l,1l,2l,3l,4l,5l,6l,7l};
printf("%ld\t",myarr[1][2]);
printf("%ld%ld\t",*(myarr[1]+3),3[myarr[1]]);
printf("%ld%ld%ld\t" ,*(*(myarr+1)+2),*(1[myarr]+2),3[1[myarr]]);
}
(A) 6 66 777
(B) 6 77 667
(C) 5 66 777
(D) 7 77 666
(E) 6 67 667
Explanation: **b

11.
What will be output if you will execute following c code?

#include<stdio.h>
void main(){
int array[2][3]={5,10,15,20,25,30};
int (*ptr)[2][3]=&array;
printf("%d\t",***ptr);
printf("%d\t",***(ptr+1));
printf("%d\t",**(*ptr+1));
printf("%d\t",*(*(*ptr+1)+2));
}
(A) 5 Garbage value 20 30
(B)   5 15 20 25
(C) 10 20 30 30
(D) 5 15 20 30
(E) Compilation error

Explanation: **a

12.
What will be output if you will execute following c code?

#include<stdio.h>
void main(){
static int a=2,b=4,c=8;
static int *arr1[2]={&a,&b};
static int *arr2[2]={&b,&c};
int* (*arr[2])[2]={&arr1,&arr2};
printf("%d %d\t",*(*arr[0])[1], *(*(**(arr+1)+1)));
}
(A) 2 4
(B) 4 2
(C) 4 8
(D) 8 4
(E) 8 2
Explanation: **c

13.
What will be output if you will execute following c code?

#include<stdio.h>
#include<math.h>
double myfun(double);
void main(){
double(*array[3])(double);
array[0]=exp;
array[1]=sqrt;
array[2]=myfun;
printf("%.1f\t",(*array)((*array[2])((**(array+1))(4))));
}
double myfun(double d){
d-=1;
return d;
}
(A) 1.4
(B) 2.8
(C) 4.2
(D) 3.0
(E) 2.7
Explanation: **e

14.
What will be output if you will execute following c code?

#include<stdio.h>
#include<math.h>
typedef struct{
char *name;
double salary;
}job;
void main(){
static job a={"TCS",15000.0};
static job b={"IBM",25000.0};
static job c={"Google",35000.0};
int x=5;
job * arr[3]={&a,&b,&c};
printf("%s %f\t",(3,x>>5-4)[*arr]);
}
double myfun(double d){
d-=1;
return d;
}
(A) Google 35000.000000
(B) TCS 15000.000000
(C) IBM 25000.000000
(D) null 15000.000000
(E) Google null
Explanation: **a

15.
What will be output if you will execute following c code?

#include<stdio.h>
union group{
char xarr[2][2];
char yarr[4];
};
void main(){
union group x={'A','B','C','D'};
printf("%c",x.xarr[x.yarr[2]-67][x.yarr[3]-67]);
}
(A) A
(B) B
(C) C
(D) D
(E) null
Explanation: **b

16.
What will be output if you will execute following c code?

#include<stdio.h>
void main(){
int a=5,b=10,c=15;
int *arr[3]={&a,&b,&c};
printf("%d",*arr[*arr[1]-8]);
}
(A) 5
(B) 10
(C) 18
(D) Garbage value
(E)   Compilation error

Explanation: **e

17.
What will be output if you will execute following c code?

#include<stdio.h>
void main(){
int arr[][3]={{1,2},{3,4,5},{5}};
printf("%d %d %d",sizeof(arr),arr[0][2],arr[1][2]);
}
(A)   6 0 4
(B) 6 1 5
(C) 18 0 5
(D) 18 1 5
(E) Compilation error
Explanation: **c

18.
What will be output if you will execute following c code?

#include<stdio.h>
void main(){
int xxx[10]={5};
printf("%d %d",xxx[1],xxx[9]);
}
(A) 0 5
(B) 5 5
(C) 5 0
(D) 0 0
(E) Compilation error
Explanation: **d
19.
What will be output if you will execute following c code?

#include<stdio.h>
#define WWW -1
enum {cat,rat};
void main(){
int Dhoni[]={2,'b',0x3,01001,'\x1d','\111',rat,WWW};
int i;
for(i=0;i<8;i++)
printf(" %d",Dhoni[i]);
}
(A) 2 98 3 513 29 73 0 -1
(B) 2 98 3 513 30 73 1 -1
(C) 2 99 3 513 29 73 1 -1
(D) 2 98 3 513 29 73 1 -1
(E) Compilation error
Explanation: **d

20.
What will be output if you will execute following c code?

#include<stdio.h>
void main(){
long double a;
signed char b;
int arr[sizeof(!a+b)];
printf("%d",sizeof(arr))
}
(A) 8
(B) 9
(C) 1
(D) 4
(E) Compilation error
Explanation: **d

pointers

2012-02-22T07:04:00.001-08:00

1.

What will be output of following program?
#include<stdio.h>
void main(){
int a = 320;
char *ptr;
ptr =( char *)&a;
printf("%d ",*ptr);
getch();
}

(A) 2
(B) 320
(C) 64
(D) Compilation error
(E) None of above

Explanation: c

As we know int is two byte data byte while char is one byte data byte. char pointer can keep the address one byte at time.
Binary value of 320 is 00000001 01000000 (In 16 bit)
Memory representation of int a = 320 is:
So ptr is pointing only first 8 bit which color is green and Decimal value is 64.

2.

What will be output of following program?

#include<stdio.h>
#include<conio.h>
void main(){
void (*p)();
int (*q)();
int (*r)();
p = clrscr;
q = getch;
r = puts;
(*p)();
(*r)("cquestionbank.blogspot.com");
(*q)();
}

(A) NULL
(B) cquestionbank.blogspot.com
(C) c
(D) Compilation error
(E) None of above

Explanation: b
p is pointer to function whose parameter is void and return type is also void. r and q is pointer to function whose parameter is void and return type is int . So they can hold the address of such function.

3.

What will be output of following program?

#include<stdio.h>
void main(){
int i = 3;
int *j;
int **k;
j=&i;
k=&j;
printf(“%u %u %d ”,k,*k,**k);
}

(A) Address, Address, 3
(B) Address, 3, 3
(C) 3, 3, 3
(D) Compilation error
(E) None of above

Explanation: a
Memory representation
Here 6024, 8085, 9091 is any arbitrary address, it may be different.
Value of k is content of k in memory which is 8085
Value of *k means content of memory location which address k keeps.
k keeps address 8085 .
Content of at memory location 8085 is 6024
In the same way **k will equal to 3.
Short cut way to calculate:
Rule: * and & always cancel to each other
i.e. *&a = a
So *k = *(&j) since k = &j
*&j = j = 6024
And
**k = **(&j) = *(*&j) = *j = *(&i) = *&i = i = 3

4.

What will be output of following program?

#include<stdio.h>
void main(){
char far *p =(char far *)0x55550005;
char far *q =(char far *)0x53332225;
*p = 80;
(*p)++;
printf("%d",*q);
getch();
}

(A) 80
(B) 81
(C) 82
(D) Compilation error
(E) None of above

Explanation:b
Far address of p and q are representing same physical address.
Physical address of 0x55550005 = (0x5555) * (0x10) + (0x0005) = 0x55555
Physical address of 0x53332225 = (0x5333 * 0x10) + (0x2225) = 0x55555
*p = 80, means content at memory location 0x55555 is assigning value 25
(*p)++ means increase the content by one at memory location 0x5555 so now content at memory location 0x55555 is 81
*q also means content at memory location 0x55555 which is 26

5.

What will be output of following program?

#include<stdio.h>
#include<string.h>
void main(){
char *ptr1 = NULL;
char *ptr2 = 0;
strcpy(ptr1," c");
strcpy(ptr2,"questions");
printf("\n%s %s",ptr1,ptr2);
getch();
}

(A) c questions
(B) c (null)
(C) (null) (null)
(D) Compilation error
(E) None of above

Explanation: c
We cannot assign any string constant in null pointer by strcpy function.

6.

What will be output of following program?

#include<stdio.h>
void main(){
int huge *a =(int huge *)0x59990005;
int huge *b =(int huge *)0x59980015;
if(a == b)
printf("power of pointer");
else
printf("power of c");
getch();
}

(A) power of pointer
(B) power of c
(C) power of c power of c
(D) Compilation error
(E) None of above

Explanation: a
Here we are performing relational operation between two huge addresses. So first of all both a and b will normalize as:
a= (0x5999)* (0x10) + (0x0005) =0x9990+0x0005=0x9995
b= (0x5998)* (0x10) + (0x0015) =0x9980+0x0015=0x9995
Here both huge addresses are representing same physical address. So a==b is true.

7.

What will be output of following program?

#include<stdio.h>
#include<string.h>
void main(){
register a = 25;
int far *p;
p=&a;
printf("%d ",*p);
getch();
}

(A) 25
(B) 4
(C) Address
(D) Compilation error
(E) None of above

Explanation: d
Register data type stores in CPU. So it has not any memory address. Hence we cannot write &a.

8.

What will be output of following program?

#include<stdio.h>
#include<string.h>
void main(){
char far *p,*q;
printf("%d %d",sizeof(p),sizeof(q));
getch();
}

(A) 2 2
(B) 4 4
(C) 4 2
(D) 2 4
(E) None of above

Explanation: c
p is far pointer which size is 4 byte.
By default q is near pointer which size is 2 byte.

9.

What will be output of following program?

#include<stdio.h>
void main(){
int a = 10;
void *p = &a;
int *ptr = p;
printf("%u",*ptr);
getch();
}

(A) 10
(B) Address
(C) 2
(D) Compilation error
(E) None of above

Explanation: a
Void pointer can hold address of any data type without type casting. Any pointer can hold void pointer without type casting.

10.

What will be output of following program?

#include<stdio.h>
#include<string.h>
void main(){
int register a;
scanf("%d",&a);
printf("%d",a);
getch();
}

//if a=25

(A) 25
(B) Address
(C) 0
(D) Compilation error
(E) None of above

Explanation: d
Register data type stores in CPU. So it has not any memory address. Hence we cannot write &a.

11.

What will be output of following program?

#include<stdio.h>
void main(){
char arr[10];
arr = "world";
printf("%s",arr);
getch();
}

(A) world
(B) w
(C) Null
(D) Compilation error
(E) None of above

Explanation: d
Compilation error Lvalue required
Array name is constant pointer and we cannot assign any value in constant data type after declaration.

12.

What will be output of following program?

#include<stdio.h>
#include<string.h>
void main(){
int a,b,c,d;
char *p = ( char *)0;
int *q = ( int *q)0;
float *r = ( float *)0;
double *s = 0;
a = (int)(p+1);
b = (int)(q+1);
c = (int)(r+1);
d = (int)(s+1);
printf("%d %d %d %d",a,b,c,d);
}

(A) 2 2 2 2
(B) 1 2 4 8
(C) 1 2 2 4
(D) Compilation error
(E) None of above

Explanation: b
Address = next address
Since initial address of all data type is zero. So its
next address will be size of data type.

13.

What will be output of following program?

#include<stdio.h>
#include<string.h>
void main(){
int a = 5,b = 10,c;
int *p = &a,*q = &b;
c = p - q;
printf("%d" , c);
getch();
}

(A) 1
(B) 5
(C) -5
(D) Compilation error
(E) None of above

Explanation: a
Difference of two same type of pointer is always one.

14.

What will be output of following program?

#include<stdio.h>
unsigned long int (* avg())[3]{
static unsigned long int arr[3] = {1,2,3};
return &arr;
}
void main(){
unsigned long int (*ptr)[3];
ptr = avg();
printf("%d" , *(*ptr+2));
getch();
}

(A) 1
(B) 2
(C) 3
(D) Compilation error
(E) None of above

Explanation:c

15.

What will be output of following program?

#include<stdio.h>
void main(){
int * p , b;
b = sizeof(p);
printf(“%d” , b);
}

(A) 2
(B) 4
(C) 8
(D) Compilation error
(E) None of above

Explanation: e

Output: 2 or 4
since in this question it has not written p is which type of pointer. So its output will depend upon which memory model has selected. Default memory model is small.

16.

What will be output of following program?

#include<stdio.h>
void main(){
int i = 5 , j;
int *p , *q;
p = &i;
q = &j;
j = 5;
printf("value of i : %d value of j : %d",*p,*q);
getch();
}

(A) 5 5
(B) Address Address
(C) 5 Address
(D) Compilation error
(E) None of above

Explanation: a

17.

What will be output of following program?

#include<stdio.h>
void main(){
int i = 5;
int *p;
p = &i;
printf(" %u %u", *&p , &*p);
getch();
}

(A) 5 Address
(B) Address Address
(C) Address 5
(D) Compilation error
(E) None of above

Explanation: b
Since * and & always cancel to each other.
i.e. *&a = a
so *&p = p which store address of integer i
&*p = &*(&i) //since p = &i
= &(*&i)
= &i
So second output is also address of i

18.

What will be output of following program?

#include<stdio.h>
void main(){
int i = 100;
printf("value of i : %d addresss of i : %u",i,&i);
i++;
printf("\nvalue of i : %d addresss of i : %u",i,&i);
getch();
}

(A)
value of i : 100 addresss of i : Address
value of i : 101 addresss of i : Address
(B)
value of i : 100 addresss of i : Address
value of i : 100 addresss of i : Address
(C)
value of i : 101 addresss of i : Address
value of i : 101 addresss of i : Address
(D) Compilation error
(E) None of above

Explanation: a
Within the scope of any variable, value of variable may change but its address will never change in any modification of variable.

19.

What will be output of following program?

#include<stdio.h>
void main(){
char far *p =(char far *)0x55550005;
char far *q =(char far *)0x53332225;
*p = 25;
(*p)++;
printf("%d",*q);
getch();
}

(A) 25
(B) Address
(C) Garbage
(D) Compilation error
(E)None of above

Explanation: e
Far address of p and q are representing same physical address. Physical address of
0x55550005 = 0x5555 * ox10 + ox0005 = 0x55555
Physical address of
0x53332225 = 0x5333 * 0x10 + ox2225 = 0x55555
*p = 25, means content at memory location 0x55555 is assigning value 25
(*p)++ means to increase the content by one at memory the location 0x5555 so now content of memory location at 0x55555 is 26
*q also means content at memory location 0x55555 which is 26

20.

What will be output of following program?

#include<stdio.h>
void main(){
int I = 3;
int *j;
int **k;
j = &i;
k = &j;
printf(“%u %u %u”,i,j,k);
}

(A) 3 Address 3
(B) 3 Address Address
(C) 3 3 3
(D) Compilation error
(E) None of above

Explanation: b
Here 6024, 8085, 9091 is any arbitrary address, it may be different.

C question set 6

2012-02-22T06:43:00.002-08:00

1=

main()
{
int i = 3;
for (;i++=0;) printf("%d",i);
}
Answer:

Compiler Error: Lvalue required.
Explanation:

As we know that increment operators return rvalues and hence it cannot appear on the left hand side of an
assignment operation.

2=

main()
{
int i=10,j=20;
j = i, j?(i,j)?i:j:j;
printf("%d %d",i,j);
}
Answer:

10 10
Explanation:

The Ternary operator ( ? : ) is equivalent for if-then-else statement. So the question can be written as:
if(i,j)
{
if(i,j)
j = i;
else
j = j;
}
else
j = j;

3=

const char *a;
char* const a;
char const *a;

-Differentiate the above declarations.
Answer:

'const' applies to char * rather than 'a' ( pointer to a constant char )
*a='F' : illegal
a="Hi" : legal
'const' applies to 'a' rather than to the value of a (constant pointer to char )
*a='F' : legal
a="Hi" : illegal
Same as 1.

4=

main()
{
int i=4,j=7;
j = j || i++ && printf("YOU CAN");
printf("%d %d", i, j);
}
Answer: 4 1

Explanation:

The boolean expression needs to be evaluated only till the truth value of the expression is not known.
j is not equal to zero itself means that the expression's truth value is 1. Because it is followed by || and
true || (anything) => true where (anything) will not be evaluated. So the remaining expression is not
evaluated and so the value of i remains the same. Similarly when && operator is involved in an expression,
when any of the operands become false, the whole expression's truth value becomes false and hence the
remaining expression will not be evaluated.
false && (anything) => false where (anything) will not be evaluated.

5=

main()
{
register int a=2;
printf("Address of a = %d",&a);
printf("Value of a = %d",a);
}
Answer:

Compier Error: '&' on register variable
Rule to Remember:
& (address of ) operator cannot be applied on register variables.

C question set 5

2012-02-22T06:35:00.000-08:00

1===========

Is the following statement a declaration/definition. Find what does it mean?
int (*x)[10];
Answer:

Definition.
x is a pointer to array of(size 10) integers.
Apply clock-wise rule to find the meaning of this definition.

2===========

What is the output for the program given below
typedef enum errorType{warning, error, exception,}error;
main()
{
error g1;
g1=1;
printf("%d",g1);
}
Answer:

Compiler error: Multiple declaration for error
Explanation:

The name error is used in the two meanings. One means that it is a enumerator constant with value 1.
The another use is that it is a type name (due to typedef) for enum errorType. Given a situation the
compiler cannot distinguish the meaning of error to know in what sense the error is used:
error g1;
g1=error;
// which error it refers in each case? When the compiler can distinguish between usages then it will not
issue error (in pure technical terms, names can only be overloaded in different namespaces).
Note:

the extra comma in the declaration,
enum errorType{warning, error, exception,}
is not an error. An extra comma is valid and is provided just for programmer's convenience.

3============

#ifdef something
int some=0;
#endif

main()
{
int thing = 0;
printf("%d %d\n", some ,thing);
}
Answer:

Compiler error : undefined symbol some
Explanation:

This is a very simple example for conditional compilation. The name something is not already known to the
compiler making the declaration int some = 0; effectively removed from the source code.

4============

#if something == 0
int some=0;
#endif

main()
{
int thing = 0;
printf("%d %d\n", some ,thing);
}
Answer:

0 0
Explanation:

This code is to show that preprocessor expressions are not the same as the ordinary expressions. If a name
is not known the preprocessor treats it to be equal to zero.

5============

main()
{
char *p = "ayqm";
printf("%c",++*(p++));
}

Answer: z

6============

main()
{
char *p = "ayqm";
char c;
c = ++*p++;
printf("%c",c);
}
Answer:

b
Explanation:

There is no difference between the expression ++*(p++) and ++*p++. Parenthesis just works as a visual
clue for the reader to see which expression is first evaluated.

7============

#include<stdio.h>

main()
{
char *p = "ayqm";
printf("%c...%c....%c",*p,++*(p++),*p);
}

Answer:y...z....a

8============

int aaa() {printf("Hi");}
int bbb(){printf("hello");}
iny ccc(){printf("bye");}

main()
{
int ( * ptr[3]) ();
ptr[0] = aaa;
ptr[1] = bbb;
ptr[2] =ccc;
ptr[2]();
}
Answer:

bye
Explanation:

int (* ptr[3])() says that ptr is an array of pointers to functions that takes no arguments and returns the
type int. By the assignment ptr[0] = aaa; it means that the first function pointer in the array is initialized
with the address of the function aaa. Similarly, the other two array elements also get initialized with the
addresses of the functions bbb and ccc. Since ptr[2] contains the address of the function ccc, the call to
the function ptr[2]() is same as calling ccc(). So it results in printing "bye".

9============
main()
{
while (strcmp("some","some\0"))
printf("Strings are not equal\n");
}
Answer:

No output
Explanation:

Ending the string constant with \0 explicitly makes no difference. So "some" and "some\0" are equivalent.
So, strcmp returns 0 (false) hence breaking out of the while loop.

10===========

main()
{
char str1[] = {'s','o','m','e'};
char str2[] = {'s','o','m','e','\0'};
while (strcmp(str1,str2))
printf("Strings are not equal\n");
}
Answer:

"Strings are not equal"
"Strings are not equal" ...
Explanation:

If a string constant is initialized explicitly with characters, '\0' is not appended automatically to the string.
Since str1 doesn't have null termination, it treats whatever the values that are in the following positions as
part of the string until it randomly reaches a '\0'. So str1 and str2 are not the same, hence the result.

C question set 4

2012-02-22T03:23:00.006-08:00

1-----------------------------------------------------------------------------------------

What are the following notations of defining functions known as?
int abc(int a,float b)
{
/* some code */
}
int abc(a,b)
int a; float b;
{
/* some code*/
}
Answer:

i. ANSI C notation
ii. Kernighan & Ritche notation (K&R Notation)

2-----------------------------------------------------------------------------------------

main()
{
char *p;
p="%d\n";
p++;
p++;
printf(p-2,300);
}
Answer:

300
Explanation:
The pointer points to % since it is incremented twice and again decremented by 2, it points to '%d\n' and 300 is
printed.

3-----------------------------------------------------------------------------------------

main()
{
int i;
i = abc();
printf("%d",i);
}
abc()
{
_AX = 1000;
}
Answer: 1000
Explanation:< blockquote>Normally the return value from the function is through the information from the
accumulator. Here _AH is the pseudo global variable denoting the accumulator. Hence, the value of the
accumulator is set 1000 so the function returns value 1000.

4-----------------------------------------------------------------------------------------

int i;
main(){
int t;
for ( t=4;scanf("%d",&i)-t;printf("%d\n",i))
printf("%d--",t--);
}
// If the inputs are 0,1,2,3 find the o/p
Answer:

4--0
3--1
2--2
Explanation:

Let us assume some x= scanf("%d",&i)-t the values during execution will be,
t i x
4 0 -4
3 1 -2
2 2 0

5--------------------------------------------------------------------------------------------

main(){
int a= 0;int b = 20;char x =1;char y =10;
if(a,b,x,y)
printf("hello");
}
Answer:

hello
Explanation:

The comma operator has associativity from left to right. Only the rightmost value is returned and the
other values are evaluated and ignored. Thus the value of last variable y is returned to check in if.
Since it is a non zero value if becomes true so, "hello" will be printed.

6----------------------------------------------------------------------------------------------

main(){
unsigned int i;
for(i=1;i>-2;i--)
printf("c aptitude");
}
Explanation:

i is an unsigned integer. It is compared with a signed value. Since the both types doesn't match, signed
is promoted to unsigned value. The unsigned equivalent of -2 is a huge value so condition becomes false and
control comes out of the loop.

7------------------------------------------------------------------------------------------------

func(a,b)
int a,b;
{
return( a= (a==b) );
}
main()
{
int process(),func();
printf("The value of process is %d !\n ",process(func,3,6));
}
process(pf,val1,val2)
int (*pf) ();
int val1,val2;
{
return((*pf) (val1,val2));
}

Answer: The value if process is 0 !
Explanation:
The function 'process' has 3 parameters - 1, a pointer to another function 2 and 3, integers. When this
function is invoked from main, the following substitutions for formal parameters take place: func for pf, 3 for
val1 and 6 for val2. This function returns the result of the operation performed by the function 'func'.
The function func has two integer parameters. The formal parameters are substituted as 3 for a and 6 for b.
since 3 is not equal to 6, a==b returns 0. therefore the function returns 0 which in turn is returned by the
function 'process'.

8-----------------------------------------------------------------------------------------------

void main()
{
void *v;
int integer=2;
int *i=&integer;
v=i;
printf("%d",(int*)*v);
}
Answer:

Compiler Error. We cannot apply indirection on type void*.
Explanation:

Void pointer is a generic pointer type. No pointer arithmetic can be done on it. Void pointers are normally used for,
Passing generic pointers to functions and returning such pointers.
As a intermediate pointer type.
Used when the exact pointer type will be known at a later point of time.

9--------------------------------------------------------------------------------------------------

void main()
{
int i=i++,j=j++,k=k++;
printf("%d%d%d",i,j,k);
}
Answer:

Garbage values.
Explanation:

An identifier is available to use in program code from the point of its declaration. So expressions such
as i = i++ are valid statements. The i, j and k are automatic variables and so they contain some garbage
value. Garbage in is garbage out (GIGO).

10---------------------------------------------------------------------------------------------------

void main()
{
static int i=i++, j=j++, k=k++;
printf("i = %d j = %d k = %d", i, j, k);
}
Answer:

i = 1 j = 1 k = 1
Explanation:

Since static variables are initialized to zero by default.

C question set 3

2012-02-22T03:18:00.001-08:00

(Q-1)

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.

(Q-2)

main()
{
extern out;
printf("%d", out);
}
int out=100;
Answer:

100
Explanation:

This is the correct way of writing the previous program.

(Q-3)

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.
2 4 7 8 3 4 2 2 2 3 3 4
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.

(Q-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
a
0 1 2 3 4
100 102 104 106 108
p
100 102 104 106 108
1000 1002 1004 1006 1008
ptr
1000
2000
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.

(Q-5)

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'.

(Q-6)

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:

ck
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'.

(Q-7)

main()
{
int i, n;
char *x = "girl";
n = strlen(x);
*x = x[n];
for(i=0; i‹n; ++i)
{
printf("%s\n",x);
x++;
}
}

Answer:

(blank space)
irl
rl
l
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.

(Q-8)

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).

(Q-9)

What are the files which are automatically opened when a C file is executed?
Answer: stdin, stdout, stderr (standard input,standard output,standard error).

(Q-10)

main(){
int * j;
void fun(int **);
fun(&j);
}
void fun(int **k) {
int a =0;
/* add a stmt here*/
}
Answer: *k = &a
Explanation:

The argument of the function is a pointer to a pointer.

C question set 2

2012-02-22T03:03:00.000-08:00

(Q-6)

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.

(Q-7)

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

(Q-8)

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

Answer:1

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

(Q-9)

#define f(g,g2) g##g2
main()
{
int var12=100;
printf("%d",f(var,12));
}
Answer: 100

(Q-10)

main()
{
int i=0;

for(;i++;printf("%d",i)) ;
printf("%d",i);
}
Answer: 1
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).

c question set 1

2012-02-22T02:48:00.001-08:00

(Q-1)
#include<stdio.h>
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 â€œibj!gsjfoetâ€ and since p reaches '\0' and p1 points to p thus
p1doesnot print anything.

(Q-2)

#include‹stdio.h›

#define a 10

main()

{

#define a 50

printf("%d",a);

}

Answer: 50

Explanation:

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

(Q-3)

#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

(Q-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.

(Q-5)

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.