template<typename T, typename U>
constexpr std::remove_reference_t<T>&& move_as(U& u) noexcept
{
     return static_cast<T&&>(static_cast<U&&>(u));
}

Would work as a possible implementation (example)

On 04.06.2025 11:13, Tymi wrote:
>
> "moving" meaning "using std::move" meaning casting to an rvalue
> reference, let's not point such things out when we know what we are
> talking about :/
>
>
> > I don't think your std:: expected example works, it didn't have a
> conversion operator. What you want is just std:: move(*exp) or
> *std::move(exp).
>
> My std::expected example does not currently work, but std::move_as
> opens a possibility for std::expected to implement such a conversion.
> std::move(*myExpected) works, but std::move_as<Value>(myExpected)
> seems cleaner to me.
>
> > And your string example works with just static_cast<std::string&&>(obj).
>
> And no, static_cast<std::string&&>(obj) does not work.
>
> On 04.06.2025 11:08, Jonathan Wakely wrote:
>>
>>
>> On Wed, 4 Jun 2025, 09:03 Tymi via Std-Proposals,
>> <std-proposals_at_[hidden]> wrote:
>>
>> I find the move_as example more intuitive and generally "safer".
>> Also my first thought was that std::string(std::move(c)) copied
>> c._value and created a new object out of it, instead of just
>> moving c._value.
>>
>>
>> You seem to be saying "moving" when you mean "casting to rvalue
>> reference", which is not moving.
>>
>>
>> I don't think your std:: expected example works, it didn't have a
>> conversion operator. What you want is just std:: move(*exp) or
>> *std::move(exp).
>>
>> And your string example works with just static_cast<std::string&&>(obj).
>>
>> So I think the existing solutions are fine.
>>
>>