Interview Questions

C Back

main()
{
extern int i;
i=20;
printf("%d",sizeof(i));
}
Ans: 

Compiler error: undefined symbol out in function main.
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..

#include main()
{
struct xx {
int x;
struct yy {
char s;
struct xx *p;
};
struct yy *q;
};
}
Ans: 

Compiler Error
In the end of nested structure yy a member have to be declared..

#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);
}
Ans: 

Compiler error: Cannot modify a constant value.
Initialization should not be done for structure members inside the structure declaration.

#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);
}
Ans: 

mmmm aaaa nnnn
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");.

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

1
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)..

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

100.

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

1
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. .

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

Compiler Error: Constant expression required in function main.
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. .

void main()
{
int i=5;
printf("%d",i+++++i);
}
Ans: 

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

void main()
{
int i=5;
printf("%d",i++ + ++i);
}
Ans: 

Output Cannot be predicted exactly.
Side effects are involved in the evaluation of i.