https://gcc.gnu.org/bugzilla/show_bug.cgi?id=64309
--- Comment #6 from Marc Glisse <glisse at gcc dot gnu.org> --- (In reply to Marek Polacek from comment #5) > I don't think so. I tried to come up with a more general transformation > that would simplify ((CST << n) & CST) != 0, but I haven't found anything If both CST are (possibly different) powers of 2 it works. ((c<<n)&c)==c also works. ((pow2<<p)&(pow2<<n))!=0. > yet. So maybe just this? > ((1 << n) & 1) != 0 -> n == 0 That looks like what richi posted. For scalars, != 0 is useless and this could just be: (1<<n)&1 -> n==0 > ((1 << n) & 1) == 0 -> n != 0
