C Programming and Technical

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

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C Other Question

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 Q.
main()
{i
nt i=0;
for(;i++;printf("%d",i)) ;
printf("%d",i);
}

A. 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).