C Programming and Technical

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.

Read Solution (Total 0)

C Other Question

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. 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