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From: "ppalka at gcc dot gnu.org" <gcc-bugzilla@gcc.gnu.org>
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Subject: [Bug c++/98712] Incorrect defaulted operator== at compile time and
 runtime when declared constexpr and inheriting (c++20)
Date: Tue, 19 Jan 2021 01:49:23 +0000
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https://gcc.gnu.org/bugzilla/show_bug.cgi?id=3D98712

Patrick Palka <ppalka at gcc dot gnu.org> changed:

           What    |Removed                     |Added
----------------------------------------------------------------------------
   Last reconfirmed|                            |2021-01-19
             Status|UNCONFIRMED                 |NEW
     Ever confirmed|0                           |1
                 CC|                            |ppalka at gcc dot gnu.org

--- Comment #2 from Patrick Palka <ppalka at gcc dot gnu.org> ---
Confirmed.

It seems that when operator=3D=3D is declared constexpr, we synthesize it e=
agerly
from check_bases_and_members before doing layout_class_type, which means th=
at
at the point of synthesis the type's TYPE_FIELDS lacks the FIELD_DECLs that
correspond to its bases, so build_comparison_op ends up completely ignoring=
 the
type's bases.

When operator=3D=3D is not declared constexpr, we synthesize the method laz=
ily at
the point of first use, and thus well after layout_class_type has been done
from finish_struct_1.=