On Tue, 26 Dec 2023 at 22:39, Egor via Std-Proposals <
std-proposals_at_[hidden]> wrote:

>
> 25.12.2023 05:19, Edward Catmur пишет:
>
> This won't always work. If the conversion operator is templated, it will
>> receive the base class type, not the derived type, and the behavior for the
>> two can be different (one of them can be rejected by SFINAE but not the
>> other, etc).
>>
>
> It seems to work; am I missing something? https://godbolt.org/z/3c5fjnqo4
>
> `std::is_convertible_v<const U&, secret_base<T>>` and `std::is_convertible_v<const
> U&, optional<T>>` are not equivalent. You could have arbitrary
> constraints on `A::operator T` that permit conversions to one and not the
> other.
>
That's the trick; A can't know about `secret_base<T>` (since it's secret),
so it can't condition conversions on that type, and any conversion that
reaches `optional<T>` can always be extended with a derived-to-base
conversion.

If `A::operator T` constrains out `std::optional<U>` (and
`std::in_place_t`), then the converting constructor of `std::optional<U>`
won't accept `A`. But that is almost certainly what the user intended!
https://godbolt.org/z/dzv68bWhG