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From: "amonakov at gcc dot gnu.org" <gcc-bugzilla@gcc.gnu.org>
To: gcc-bugs@gcc.gnu.org
Subject: [Bug target/111655] [11/12/13/14 Regression] wrong code generated
 for __builtin_signbit and 0./0. on x86-64 -O2
Date: Mon, 02 Oct 2023 11:08:12 +0000
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https://gcc.gnu.org/bugzilla/show_bug.cgi?id=3D111655

Alexander Monakov <amonakov at gcc dot gnu.org> changed:

           What    |Removed                     |Added
----------------------------------------------------------------------------
                 CC|                            |amonakov at gcc dot gnu.org
     Ever confirmed|0                           |1
            Summary|wrong code generated for    |[11/12/13/14 Regression]
                   |__builtin_signbit and 0./0. |wrong code generated for
                   |on x86-64 -O2               |__builtin_signbit and 0./0.
                   |                            |on x86-64 -O2
         Resolution|DUPLICATE                   |---
             Status|RESOLVED                    |NEW
   Last reconfirmed|                            |2023-10-02

--- Comment #9 from Alexander Monakov <amonakov at gcc dot gnu.org> ---
It's true that the sign of 0./0 is unpredictable, but we can fold it only w=
hen
the division is being eliminated by the folding.=20

It's fine to fold

  t =3D 0./0;
  s =3D __builtin_signbit(t);

to

  s =3D 0

with t eliminated from IR, but it's not OK to fold

  t =3D 0./0
  s =3D __builtin_signbit(t);

to

  t =3D 0./0
  s =3D 0

because the resulting program runs as if 0./0 was evaluated twice, first to=
 a
positive NaN (which was used for signbit), then to a negative NaN (which fed
the following computations). This is not allowed.

This bug was incorrectly classified as a dup. The fix is either to not fold
this, or fold only when we know that the division will be eliminated (e.g. =
the
only use was in the signbit). Reopening.=