Should the expression x && (~x) return 1 or 0? And does it depend on the compiler?

Question

In C, given "short x = 0xFFF0" what is the output of "x && (~x)". This question was proposed to me on a quiz and the answer for it was 0. Although when compiled 1(true) is returned. Why is that?

int main()
{
    short x = 0xFFF0;
    printf("%X\n", x && (~x));
    return 0;
}

What I know is that "x = 0xFFF0" and "~x = 0x000F". when we do "logical and" its 1 && 1 which returns 1(true). However, the professor thinks otherwise by stating that the "bitwise and" should be done first 0xFFF0 & 0x000F = 0x0000 which is 0 (false). Is this a problem with an outdated compiler or something unexplainable?

Answer 1

Let's look at x && (~x) part by part.

• x by itself is not zero and will therefore count as true in boolean contexts as when using the logical && (AND).
• ~x by itself is not zero either, so true.
• true && true is true:

#include <stdbool.h>
#include <stdio.h>

int main() {
    short x = 0xFFF0;
    
    printf("%s\n", x != 0 ? "true" : "false");         // true
    printf("%s\n", ~x != 0 ? "true" : "false");        // true
    printf("%s\n", true && true ? "true" : "false");   // true
}