On Mon, 27 Oct 2025 at 13:07, Jonathan Wakely <cxx_at_[hidden]> wrote:

>
>
> On Mon, 27 Oct 2025 at 13:02, Frederick Virchanza Gotham via Std-Proposals
> <std-proposals_at_[hidden]> wrote:
>
>> On Mon, Oct 27, 2025 at 12:13 PM Jonathan Wakely wrote:
>> >
>> > This is problematic for arbitrary types, which might not be default
>> constructible. In generic code, you don't know how to construct a value to
>> return.
>> >
>> > But you can return std::type_identity<int>{} or
>> std::type_identity<double>{} and then use decltype(select_type<1>())::type.
>>
>> #include <type_traits>
>> #include <semaphore>
>>
>> template<int N> consteval auto select_type_detail(void)
>> {
>> /**/ if constexpr ( 0 == N ) return (int
>> *)nullptr;
>> else if constexpr ( 1 == N ) return (double
>> *)nullptr;
>> else if constexpr ( 2 == N ) return
>> (std::binary_semaphore*)nullptr;
>>
>
> You can't use this to return reference types (nor abominable functions
> types, although that's less of a problem in practice).
>

std::type_identity<T> works for *every* type T.

Returning a value of type T doesn't work for references, non-default
constructible types, abstract classes, function types.

Returning a T* doesn't work for non-referenceable types.



>
>
>> }
>>
>> template<int N>
>> using select_type = std::remove_pointer_t< decltype(
>> select_type_detail<N>() ) >;
>>
>> #include <iostream>
>> #include <typeinfo>
>>
>> int main(void)
>> {
>> select_type<2> var(0);
>> std::cout << typeid(var).name() << std::endl;
>> }
>> --
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>> Std-Proposals_at_[hidden]
>> https://lists.isocpp.org/mailman/listinfo.cgi/std-proposals
>>
>