doubt in c code

Question

#include <stdio.h>

#define sqr(x) x*x

int main()
{
int x = 16/sqr(4);
printf("%d", x); //x=16
int y=sqr(4); // y=16

return 0;
}

But here still the result is 16.
Anyone can usually evaluate it to be as 16/16=1.

What process actually going on here.

Is there any difference between #define sqr(x) x*x and #define sqr(x) (x*x)

Answer 1

the first sqr() macro has no parentheses around the expression x * x
so division happens before multiplication.

16 / sqr(4) = (16 / 4) * 4 = 4 * 4 = 16

parentheses force the multiplication to happen first. with sqr(x) (x * x) you get 16 / (4 * 4) where result is 1 as you expected. Parentheses should always be used in macros because they guarantee right evaluation order.

there is also sqr((x) * (x)) which guarantees evaluation is correct when the argument itself is an expression like sqr(5 - 3).

Answer 2

Compiler first replaces sqr(x) with x*x and then you get 16/x*x at runtime which is calculated like this:
16/4*4=
4*4=
16

Answer 3

I think
sqr(4) is the same as 4*4
that is why you got the same result for both x and y