C LANGUAGE QUESTIONS WITH SOLUTION

C language objective types questions and answers with solution or explanation

1.

#include<stdio.h>

int main(){

int a=0;

#if (a==0)

printf("Equal");

#else if

printf("Not equal");

#endif

return 0;

}

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

#include<stdio.h>

int main(){

for(;NULL;)

printf("cquestionbank");

return 0;

}

(A)	c
(B)	bank
(C)	cquestionbank
(D)	Infinite loop
(E)	Compilation error

Explanation:

Here NULL is micro constantan. Value of this symbolic constant is 0 or 0L as defined stdio.h:

#ifndef NULL

# if defined(__TINY__) || defined(__SMALL__) || defined(__MEDIUM__)

# define NULL 0

# else

# define NULL 0L

# endif

#endif

So corresponding intermediate file of above code will be:

int main(){

for(;0;)

printf("cquestionbank");

return 0;

}

As you know in c :

0: Means false

Non- zero: True

So for loop should not execute any time because intitial condtion is false. But it is bug of turbo c compiler. Loop will execute one time and it will print : cquestionbank

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

#include<stdio.h>

int main(){

int x=25;

if(!!x)

printf("%d",!x);

else

printf("%d",x);

return 0;

}

Explanation:

! is negation operator.

!x = 0 if x is non-zero number.

!x = 1 if x is equal to zero.

So,

!!x

=! (! x)

=! (! 25)

=! (0)

As we know in c:

Zero: It represents false.

Non-zero: It represents false.

if (1) means condition is true hence if part will execute.

!x =! 1 = 0

Hence printf statement will print zero in the console window.

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

#include<stdio.h>

int main(){

float a=0.5, b=0.9;

if(a&&b>0.9)

printf("Sachin");

else

printf("Rahul");

return 0;

}

(A)	Sachin
(B)	Rahul
(C)	null
(D)	Run time error
(E)	Compilation error

Explanation:

Consider on the expression:

a && b > 0.9

As we know > operator enjoy higher precedence than && operator.

So expression will be

a && (b > 0.9)

=0.5 && (0.9 > 0.9)

Properties of > (Greater than operator)

a > b returns 1 if a is greater than b.

a > b return 0 if a is less than or equal to b.

So our expression became:

= 0.5 && 0

In c zero represents false and non-zero represents true.

= true && false

= false

= 0

So, if (0) means condition is false. Hence else part will execute.

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

#include<stdio.h>

int main(){

int x=5, y=10;

if(!(!x) && x)

printf("%d",x);

else

printf("%d",y);

return 0 ;

}

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

#include<stdio.h>

int main(){

char ch=321;

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

return 0 ;

}

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

#include<stdio.h>

int main(){

int a,b;

a = -3- -3;

b = -3 - - (-3 );

printf("%d %d",a,b);

return 0 ;

}

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

#include<stdio.h>

int main(){

int x;

x= -2 + 11 - 7 * 9 % 6 / 12;

printf("%d",x);

return 0 ;

}

(A)	6
(B)	7
(C)	8
(D)	9
(E)	Compilation error

Explanation:

Consider on the expression

x = -2 + 11 - 7 * 9 % 6 / 12

There are seven different type of operator in the above expression:

Operator	Precedence	Associative
-	1	Right to left
* , % , /	2	Left to right
- , +	3	Left to right
=	4	Right to left

First of all unary minus will perform the operation

x = (-2) + 11 - 7 * 9 % 6 / 12

Now multiplication operator will perform the operation

x = (-2) + 11 - (7 * 9) % 6 / 12

x = (-2) + 11 - 63 % 6 / 12

After this modular division operator will perform operation:

x = (-2) + 11 - (63 % 6) / 12

x = (-2) + 11 - 3 / 12

After this division will perform the operation:

x = (-2) + 11 - (3 / 12)

x = (-2) + 11 - 0

Now binary plus operator will perform the operation:

x = ((-2) + 11) - 0

x = 9 - 0

Now binary minus operator will perform the operation:

x = 9

At the end assignment operator will perform the operation and it will assign 9 to variable x.

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

#include<stdio.h>

int main(){

int x=3, y=4, z=4;

printf("%d", (z>=y>=x?100:200));

return 0 ;

}

10.

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

#include<stdio.h>

int main(){

int a=30, b=40, x;

x=(a!=10) && (b=50);

printf("%d",x);

return 0 ;

}

11.

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

#include<stdio.h>

int main(){

float x=12.25, y=13.65;

if(x=y)

printf("x and y are equal");

else

printf("x and y are not equal");

return 0 ;

}

12.

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

#include<stdio.h>

int main(){

int i=1, j=1;

for(;j;printf("%d%d\t",i,j))

j=i++ <= 5;

return 0 ;

}

(A)	1 2 3 4 5
(B)	11 12 13 14
(C)	21 31 41 51 61 70
(D)	Infinite loop
(E)	Compilation error

Explanation:

Syntax of for loop is:

for (exp1;exp2;exp3){

}

Note: all Exp1, Exp2, Exp3 and <statements> are optional.

Order of evaluation is:

In first cycle

1. Exp1

2. Exp2

3. <Statement>

4. Exp2

In rest cycles

1. Exp2

2. <Statements>

3. Exp3

Now we come to our question.

In first cycle

1. It has not exp1

2. Exp2 is j which is equal to 1 i.e. condition is true

3. <Statements> is j=i++ <= 5

j = 1 <= 5

j = 1

i = 2 (Due to postfix increment operator)

4. Exp3 is printf statement which will print i as 2 and j as 1

In second cycle

1. Exp2 is j which is equal to 1 i.e. condition is true

2. <Statements> is j=i++ <= 5

j = 2 <= 5

j = 1

i = 3 (Due to postfix increment operator)

3. Exp3 is printf statement which will print i as 3 and j as 1

In third cycle

1. Exp2 is j which is equal to 1 i.e. condition is true

2. <Statements> is j=i++ <= 5

j = 3 <= 5

j = 1

i = 4 (Due to postfix increment operator)

3. Exp3 is printf statement which will print i as 4 and j as 1

In fourth cycle

1. Exp2 is j which is equal to 1 i.e. condition is true

2. <Statements> is j=i++ <= 5

j = 4 <= 5

j = 1

i = 5 (Due to postfix increment operator)

3. Exp3 is printf statement which will print i as 4 and j as 1

In fifth cycle

1. Exp2 is j which is equal to 1 i.e. condition is true

2. <Statements> is j=i++ <= 5

j = 5 <= 5

j = 1

i = 6 (Due to postfix increment operator)

3. Exp3 is printf statement which will print i as 6 and j as 1

In sixth cycle

1. Exp2 is j which is equal to 1 i.e. condition is true

2. <Statements> is j=i++ <= 5

j = 6 <= 5

j = 0

i = 7 (Due to postfix increment operator)

3. Exp3 is printf statement which will print i as 7 and j as 0

In seventh cycle

1. Exp2 is j which is equal to 0 i.e. condition is false. So loop will terminate.

13.

What function is used to release the allocated memory space?

14.

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

#include<stdio.h>

auto int a=5;

int main(){

int x;

x=~a+a&a+a<<a;

printf("%d",x);

return 0;

}

15.

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

#include<stdio.h>

void main(){

int a=5;

{

int b=10;

++b;

++a;

{

int a=20;

++a;

a=++b;

}

++a;

++b;

printf("%d %d",a,b);

}

printf(" %d",a);

}

16.

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

#include<stdio.h>

#include<conio.h>

void main(){

int a[]={0,1,2,3,4,5,6,7,8,9,10};

int i=0,num;

num=a[++i+a[++i]]+a[++i];

printf("%d",num);

}

17.

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

#include<stdio.h>

#include<conio.h>

void main(){

int i=3,val;

val=sizeof f(i)+ +f(i=1)+ +f(i-1);

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

}

int f(int num){

return num*5;

}

18.

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

#include<stdio.h>

#include<conio.h>

void main(){

int i;

(i=8)+=1;

printf("%d",i);

}

19.

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

#include<stdio.h>

#include<conio.h>

void main(){

char c=-'a';

printf("%d",c);

}

20.

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

#include<stdio.h>

int main(){

int num,a=5;

num=-a--;

printf("%d %d",num,a);

}

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C LANGUAGE QUESTIONS WITH SOLUTION

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(A)	x and y are equal
(B)	x and y are not equal
(C)	It will print nothing
(D)	Run time error
(E)	Compilation error