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Frequently Asked c interview questions and answers for freshers Page 38

(#M40029266) C QUESTION C Keep an EYE

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

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(#M40029262) C QUESTION C Keep an EYE

Q.
#include
main()
{i
nt i=1,j=2;
switch(i)
{
case 1: printf("GOOD");
break;
case j: printf("BAD");
break;
}
}

A. 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

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(#M40029259) C QUESTION C Keep an EYE

output?
Q.
main()
{i
nt i;
printf("%d",scanf("%d",&i)); // value 10 is given as input here
}

A. 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.

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(#M40029258) C QUESTION C Keep an EYE

output??
Q.
void main()
{i
nt i=5;
printf("%d",i+++++i);
}

A. Compiler Error

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

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(#M40029255) C QUESTION C Keep an EYE

output?
main()
{i
nt i=1;
while (i2)
goto here;
i++;
}} fun()
{
here:
printf("PP");
}

A. 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

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(#M40029253) C QUESTION C Keep an EYE

output??
main()
{
char *p;
p="Hello";
printf("%cn",*&*p);
}

A. H

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.

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(#M40029252) C QUESTION C Keep an EYE

output??
main()
{i
nt i=400,j=300;
printf("%d..%d");
}

A. 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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(#M40029250) C QUESTION C Keep an EYE

output??
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.

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(#M40029249) C QUESTION C Keep an EYE

output?
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).

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(#M40029243) C QUESTION C Keep an EYE

output?
main()
{
char *p="hai friends",*p1;
p1=p;
while(*p!='') ++*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 ‘’ and p1
points to p thus p1doesnot print anything

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