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From: "redi at gcc dot gnu.org" <gcc-bugzilla@gcc.gnu.org>
To: gcc-bugs@gcc.gnu.org
Subject: [Bug libstdc++/113007] `std::variant` converting constructor and
 `operator=` compile while the C++ Standard says they must not
Date: Thu, 14 Dec 2023 01:03:21 +0000
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https://gcc.gnu.org/bugzilla/show_bug.cgi?id=3D113007

--- Comment #2 from Jonathan Wakely <redi at gcc dot gnu.org> ---
(In reply to Pavel Novikov from comment #0)
> e.g. for `std::variant<std::monostate, bool, int64_t, double>` the overlo=
ad
> set would be
>=20
>     void F(std::monostate) {}
>     void F(bool) {}
>     void F(int64_t) {}
>     void F(double) {}

No, because it says:

"build an imaginary function FUN(Ti) for each alternative type Ti for which=
 Ti
x[] =3D {std=E2=80=8B::=E2=80=8Bforward<T>(t)}; is well-formed for some inv=
ented variable x."

That array initializer would be ill-formed for all except int64_t.
std::monostate just can't be initialized from an int, and for double and bo=
ol
it's a narrowing conversion (because of the braces).

I think libstdc++ is correct.=