Why this code gives the output as 28? (Code mentioned below)

Question

#include<stdio.h>
#define square(x) x*x
void main()
{
int i;
i=(28/square(4));
printf("%d",i);
}

Answer 1

Each presence of `square( <number> )` will be replaced with `<number> * <number>` before compilation phase begins. So ...

i = ( 28 / square( 4 ) );

Will be replaced into something like this

i = ( 28 / 4 * 4 );

If you want 28 to be divided by 16 (4 * 4), then wrap the macro arguments by parentheses inside the macro body as follows

#define square( x ) ( ( x ) * ( x ) )

This way, the line will be replaced into something like this

i = ( 28 / ( 4 * 4 ) );

(Edited)

Answer 2

Taking this one step further: always wrap your arguments in parentheses

#define square(x) ((x)*(x))

Why? Imagine a call "square(x+y)" 😉