Short circuit behavior of logical expressions in C in this example

Question

PROGRAM

#include <stdio.h>

int main(void)
{
    int i, j, k;

    i = 1; j = 1; k = 1;

    printf("%d ", ++i || ++j && ++k);
    printf("%d %d %d", i, j, k);

  return 0;
}

OUTCOME

1 2 1 1

I was expecting 1 1 2 2. Why? Because the && has precedence over ||. So I followed these steps: 1) j added 1, so j now values 2... 2) k added 1, so k now values 2... 3) 2 && 2, evaluates to 1... 4) No need of further evaluation as the right operand of || is true, so the whole expression must be true because of short circuit behavior of logical expressions...

Why am I wrong?

Answer 1

Precedence affects only the grouping. && has a higher precedence than || means:

++i || ++j && ++k

is equivalent to:

++i || (++j && ++k)

But that doesn't mean ++j && ++k is evaluated first. It's still evaluated left to right, and according to the short circuit rule of ||, ++i is true, so ++j && ++k is never evaluated.