[Bug libstdc++/113007] `std::variant` converting constructor and `operator=` compile while the C++ Standard says they must not

[redi at gcc dot gnu.org]

https://gcc.gnu.org/bugzilla/show_bug.cgi?id=113007

--- 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 overload
> set would be
> 
>     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[] = {std​::​forward<T>(t)}; is well-formed for some invented 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 bool
it's a narrowing conversion (because of the braces).

I think libstdc++ is correct.