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.

Reply via email to