]On 6 Jan 2022, at 22:50, Taylor Blau <me@xxxxxxxxxxxx> wrote: > > On Wed, Jan 05, 2022 at 01:23:10PM +0000, Jessica Clarke wrote: >> [...] In most cases this is clear, as normally at least one operand is >> provably a plain integer, but if both operands are uintptr_t and have >> no indication they're just plain integers then it is ambiguous, and >> the current implementation will arbitrarily, but deterministically, >> pick the left-hand side, due to empirical evidence that it is more >> likely to be correct. > > Wouldn't a simpler, less invasive fix be to instead write the expression > so that the left-hand operand is a pointer? IOW, shouldn't the following > work (with no other changes): > > ent->util = (void *)((uintptr_t)what | ent->util); > > ? > > (I dropped the explicit cast on the right-hand side, since ent->util is > already a uintptr_t, and the left-hand side has an explicit cast, so > there isn't any type promotion going on here). That would still warn as it’s still ambiguous. It just happens that the arbitrary choice picks the right side. Better to clean up the code to remove the ambiguity and clarify it to the programmer. Jess
