SFINAE on Error in Dependent Type 导致意外的硬错误

Question

我有可以简化为如下代码的代码：

#include <type_traits>

template <typename T>
struct dependent
{
    using type = typename T::type;
};

template <typename T>
typename dependent<T>::type
foo(const T& x);

bool foo(bool x) { return x; }

int main()
{
    foo(true);
}

使用 g++ 9.3 编译失败 --std=c++17 错误：

test.cpp: In instantiation of 'struct dependent<bool>':
test.cpp:11:1:   required by substitution of 'template<class T> typename dependent<T>::type foo(const T&) [with T = bool]'
test.cpp:17:13:   required from here
test.cpp:6:11: error: 'bool' is not a class, struct, or union type
    6 |     using type = typename T::type;
      |           ^~~~

这不是我所期望的。我希望尝试用 bool 替换 template <typename T> typename dependent<T>::type foo(const T& x) 中的 T 会失败，这不是错误。似乎 SFINAE 不适合我，但我不知道为什么。

来自 SFINAE 上的非官方参考中的示例：

Substitution proceeds in lexical order and stops when a failure is encountered.

template <typename A>
struct B { using type = typename A::type; };

template <
  class T,
  class   = typename T::type,      // SFINAE failure if T has no member type
  class U = typename B<T>::type    // hard error if T has no member type
                                   // (guaranteed to not occur as of C++14)
> void foo (int);

我在 class U = typename B<T>::type 上遇到这种情况，但是 "guaranteed to not occur as of C++14" 位似乎表明从 C++14 开始这不应该发生。给出了什么？

Answer 1

问题是 dependent<T> 有 type，但那个可能格式不正确导致硬故障。

你可能会让 dependent SFINAE 变得友好：

template <typename T, typename Enabler = void>
struct dependent
{
};

template <typename T>
struct dependent<T, std::void_t<typename T::type>>
{
    using type = typename T::type;
};

Demo

SFINAE on Error in Dependent Type 导致意外的硬错误

SFINAE on Error in Dependent Type causes unexpected hard error

c++

sfinae

c++14