https://gcc.gnu.org/bugzilla/show_bug.cgi?id=89536
--- Comment #16 from Eric Botcazou <ebotcazou at gcc dot gnu.org> --- > So, for BIT_AND_EXPR we only handle the case where the result of > BIT_AND_EXPR is known to be non-zero. That means both operands have to be > non-zero (and have at least one common bit). Now, if say the operation is > done on int and we know that rhs1 (or rhs2 or both) have [0, 1] range, that > means we can safely assume that BIT_AND_EXPR is non-zero implies that rhs1 > (or rhs2 or both) are 1. But why would the [0, 1] range hold in the BIT_AND_EXPR case and not in the BIT_NOT_EXPR case? Either we take the range as granted in both cases or in none of them. If the former, then we can apply the exclusion game in the BIT_NOT_EXPR case too.
