16 is the output of following code. Explain

Question

#include<stdio.h>
#define square (a) a*a
void main()
{
int i=16/square(4) ;
prints("%d", i) ;
}

Answer 1

coffeeunderrun suggested a good and necessary correction to the macro. More parentheses would be needed if you should call it with an expression like square(i+3). That would incorrectly expand to (i+3*i+3). It should be ((i+3)*(i+3)).

Here is a better definition:
#define square(a) ((a)*(a))