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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