在C ++ 20中使用类型名需要/概念吗? [英] Using typename in C++20 requires / concept?

问题描述

请考虑以下C ++ 20程序:

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;
}

这是格式错误的吗?如果没有，输出应该是什么?

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

我期望输出为 1 0 ，但是我在clang 1 1 中观察到.谁是对的，为什么?

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

推荐答案

这里的概念只是命名类型 A< BadArg> ，它并没有做任何事情来触发它的实例化.这里没有任何东西会导致格式不正确的 A< BadArg> :: X 的实例化.

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.

如果 did 仍然正确，那么无论如何您都不会得到 false ，您会得到格式错误的程序.例如，您是否做过:

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>{}; };

然后 WorksWithA< BadArg> 将触发 A< BadArg> 的实例化，该实例将尝试查找 BadArg :: X ，即现在是替代替代环境之外的失败.不是 false ，编译错误.

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.

如果想要的结果为 false ，则必须将 A 模板限制为现有类型:

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;
};

现在两种形式(您的原始和我的替代品)都将为 WorksWithA< BadArg> 生成 false .

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

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