Wrong result when comparing xor and int

[Valentin Crone]

Hello,
I'm on Ubuntu 14.04 LTS (64bits)
With gcc:
gcc --version
gcc (Ubuntu 4.8.4-2ubuntu1~14.04) 4.8.4
Copyright (C) 2013 Free Software Foundation, Inc.
This is free software; see the source for copying conditions. There is NO
warranty; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.

lsb_release -rd
Description: Ubuntu 14.04.3 LTS
Release: 14.04

apt-cache policy gcc
gcc:
  Installé : 4:4.8.2-1ubuntu6
  Candidat : 4:4.8.2-1ubuntu6
 Table de version :
 *** 4:4.8.2-1ubuntu6 0
        500 http://fr.archive.ubuntu.com/ubuntu/ trusty/main amd64 Packages
        100 /var/lib/dpkg/status

So:
3 ^ 1 = 2 ok?

Try this code:
#include <stdio.h>
int main()
{
 printf("%d\n", 3 ^ 1);
 if(3 ^ 1 == 1)//Bug in this test
 {
  printf("SURPRISE !\n");
 }
 else
 {
  printf("FAIL !\n");
 }
 return 0;
}

Just compile this with defaut options:
gcc -o youprogram yourfile.c

Then, run, and you will see:
2
SURPRISE !

So, normally it will show "FAIL !", because 3 ^ 1 = 2, and 3 ^ 1 == 1 must return 0, but this test return 1 !

If you do this:

int v = 3 ^ 1;
if(v == 1)//This test work normally
{
}

And:
if((int) (3 ^ 1) == 1)//This test work normally
{
}

Globally, if you have:
int x, y, z;//Choose random values for this

if(x ^ y == z)//Return always true
{
}

The invert is bugged too:

if(z == y ^ x)//Same thing
{
}

I don't know if this bug is fixed on more recent releases, but a patch is welcome for this version :)

Thank you :)

[Colin Watson (cjwatson)]

This is a standard gotcha in C operator precedence: == has higher precedence than ^, and so your expression is parsed as 3 ^ (1 == 1), which is indeed non-zero. To fix it, use explicit parentheses as follows: (3 ^ 1) == 1.

It would perhaps be nice if the C language had been defined differently (Wikipedia mentions this, for instance: https://en.wikipedia.org/wiki/Operators_in_C_and_C%2B%2B#Criticism_of_bitwise_and_equality_operators_precedence), but we're stuck with it now, and GCC certainly cannot arbitrarily decide to change the language's operator precedence.

[Valentin Crone (va-crone)]

Ok, with parentheses it's ok.
I don't know that about the C, but it's a strange definition because:
(1 + 2) == 3 is same as 1 + 2 == 3
and
(3 ^ 1) == 2 is not the same as 3 ^ 1 == 2

[Colin Watson (cjwatson)]

Yes, it is strange, and that exact same analogy is mentioned in the Wikipedia link I gave. However, it's part of the language specification and we're stuck with it now.