Why this code outputs "9 7", but not "9 16"???

Question

#include <stdio.h>

#define sqr(i) i*i
int main(){
printf("%d %d", sqr(3), sqr(3+1));
return 0;
}

Answer 1

Operator precedence. You see, because sqr is a macro, not a function, this is what the compiler actually sees: (3+1*3+1)

Use : #define sqr(i) ((i)*(i))

Answer 2

Because sqr(i) will be replaced with i*i.

sqr(3+1) = 3+1*3+1 = 7

Change the definition to

#define sqr(i) (i)*(i)

sqr(3+1) = (3+1)*(3+1) = 16