https://gcc.gnu.org/bugzilla/show_bug.cgi?id=91397

--- Comment #2 from Marc Glisse <glisse at gcc dot gnu.org> ---
        if (g == 0) return (char *)malloc(0);
        for (;;)        
                ;

so the only way this can return is if g is 0. This means that in j, k is -1,
and you are calling memcpy with a huge argument. So at least in the reduced
testcase, the warning makes some sense.

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