Why result is 8 but not 2?

Question

#include<stdio.h>

#define sq(x) x*x

int main()

{

    int i = 8/sq(2);

    printf("%d", i);

}

Answer 1

Macro call is not a function call but a textual replacement before compilation. If you replace sq(2) in the expression, you get
int i = 8/2*2
Since multiplication and division have the same precedence, it is evaluated left to right. That is, (8/2)*2, and that is 8.

Answer 2

Jay Matthews better yet, this would produce the expected results:

#define sq(x) ((x)*(x))

Answer 3

It is because same precedence of * & / , and left to right evaluation.