On Mon, 14 Apr 2025 15:46:54 GMT, Jaikiran Pai <j...@openjdk.org> wrote:
> It then means that the other field will never move out of its initial value > and thus will never be considered "stable". For all non-zero hashcodes, it matters that `hash` would be constant-foldable, which means `if (h == 0 && !hashIsZero)` would be optimized to `if (false && !hashIsZero)` -> `if (false)` -> poof. So it would not matter if `hashIsZero` is actually foldable. Same goes the other way around for the hashcode that _is_ zero: now `!hashIsZero` would be folded to `false` -- with a wrinkle that `int h = hash;` read would likely still be there... You can massage the code a bit to return `0` if `hashIsZero`, but I don't think it is worth it. ------------- PR Comment: https://git.openjdk.org/jdk/pull/24625#issuecomment-2802411963
