https://gcc.gnu.org/bugzilla/show_bug.cgi?id=113777
--- Comment #3 from Eric Botcazou <ebotcazou at gcc dot gnu.org> --- > This works with __attribute__((may_alias)) though, so what's special with > __attribute__((__hardbool__)) ? Replying myself: this creates an enumeration type under the hood, so this is a duplicate of PR debug/113519.