Why this program output 11?

Question

#include <stdio.h>
#define A 1 + 2
#define B 3 + 4
int main()
{
int var = A*B;
printf("%d\n", var);
}

Answer 1

var is evaluated from the two expressions as:
var = 1 + 2 * 3 + 4
And because multiplication takes precedence:
var = 1 + (2 * 3) + 4 = 1 + 6 + 4 = 11

Answer 2

Thank youu!