Bitwise Operations on signed chars

Question

Does the following code rely on undefined behavior if the platform's char type is signed and some of the parameters are in the negative range (for example, char_bitmatch('\\xf0', '\\xc0', '\\x20') )?

static constexpr bool char_bitmatch(char c, char pos, char neg)
{
    return (c & pos) == pos
        && !(c & neg);
}

---

Context

The reason I am asking this is because in GCC 8.1.0 with -O3 , I am seeing a behavior which can only be caused by char_bitmatch('\\xf0', '\\xc0', '\\x20') erroneously returning true . This code behaves as expected:

static constexpr bool char_bitmatch(char c_in, char pos_in, char neg_in)
{
    auto c   = static_cast<unsigned char>(c_in);
    auto pos = static_cast<unsigned char>(pos_in);
    auto neg = static_cast<unsigned char>(neg_in);

    return (c & pos) == pos
        && !(c & neg);
}

From my understanding, this should not have fixed the issue -- & should work the same between signed char and unsigned char .

This leads me to a few conclusions (but I don't know which is correct):

1. Use of unsigned char fixes an undefined behavior.
2. I am still relying on undefined behavior -- the "fix" is luck and the actual bug lies elsewhere in my code.
3. There is a bug in GCC 8.1.0 optimization -- the "fix" is a voodoo incantation that makes GCC do the right thing.

Answer 1

Interesting. I think your assumption that char_bitmatch is returning true is, erm, false.

When I run this code:

#include "stdio.h"

static constexpr bool char_bitmatch(char c, char pos, char neg)
{
    return (c & pos) == pos
        && !(c & neg);
}

int main (void)
{
    constexpr bool b = char_bitmatch ('\xf0', '\xc0', '\x20');
    printf ("%d\n", b);
}

I get:

0

So I think the problem lies elsewhere in your code.

I used the same compiler as you - run it at Wandbox (choice of compilers available).

Also, == pos is redundant, no?