在 C++20 中使用 typename 需要 / 概念？

Question

Please consider the following C++20 program:

#include <iostream>

template<typename T>
struct A {
    using X = typename T::X;
};

template<typename T>
constexpr bool WorksWithA = requires { typename A<T>; };

struct GoodArg {
    using X = int;
};

struct BadArg {
};

int main() {
    std::cout << WorksWithA<GoodArg> << std::endl;
    std::cout << WorksWithA<BadArg> << std::endl;
}

Is this ill-formed? And if not, what should the output be?

I was expecting the output to be 1 0 but I observe in clang 1 1 . Who's right and why?

$ clang++ --version
clang version 10.0.0-4ubuntu1 
$ clang++ test.cc -std=c++20
$ ./a.out 
1
1

Answer 1

The concept here is just naming the type A<BadArg> , it doesn't do anything to trigger instantiation of it. Nothing here leads to the instantiation of A<BadArg>::X which would be ill-formed.

If it did though, then you wouldn't get false anyway, you'd get an ill-formed program. For instance, had you done:

template<typename T>
constexpr bool WorksWithA = requires { A<T>{}; };

Then WorksWithA<BadArg> would trigger instantiation of A<BadArg> which would try to look up BadArg::X , which is now a failure outside of the immediate context of substitution. Not false , compile error.

If you want a result that's false , you'll have to constrain the A template on the type existing:

template <typename T>
    requires requires { typename T::X; }
struct A {
    using X = typename T::X;
};

And now both formulations (your original and my substitute) would yield false for WorksWithA<BadArg> .