C++ 友元函数模板重载和 SFINAE 在 clang++、g++、vc++ 中的不同行为（C++14 模式）

因此，以下代码在 clang++ (3.8.0) 下构建并成功运行，但在 g++ (6.3.0) 和 vc++ (19.10.24903.0) 下均失败。 g++ 和 vc++ 都抱怨运算符&& 的重新定义。

有谁知道这里哪个编译器有问题。对于无法编译代码的编译器，编译错误的解决方法是什么？

#include <functional>
#include <iostream>

template <typename T>
struct awaitable
{
    friend awaitable<void> operator&&(awaitable a1, awaitable a2)
    {
        std::cout << "operator&&(awaitable a1, awaitable a2) - T: " << typeid(T).name() << std::endl;
        return awaitable<void>{};
    }

    template <typename U = T, typename std::enable_if<!std::is_same<U, void>::value>::type* = nullptr>
    friend awaitable<void> operator&&(awaitable<void> a1, awaitable<U> a2)
    {
        std::cout << "operator&&(awaitable<void> a1, awaitable<U> a2) - U: " << typeid(T).name() << std::endl;
        return awaitable<void>{};
    }

    template <typename U = T, typename std::enable_if<!std::is_same<U, void>::value>::type* = nullptr>
    friend awaitable<void> operator&&(awaitable<U> a1, awaitable<void> a2)
    {
        std::cout << "operator&&(awaitable<U> a1, awaitable<void> a2) - U: " << typeid(T).name() << std::endl;
        return awaitable<void>{};
    }
};

int main(int argc, const char * argv[])
{
    awaitable<int> a1, a2, a3, a4;
    auto ar = a1 && (a1 && a2) && (a2 && a3) && a4;
}

铿锵++：http://coliru.stacked-crooked.com/a/cb01926bbcacdfb0 http://coliru.stacked-crooked.com/a/cb01926bbcacdfb0

g++: http://coliru.stacked-crooked.com/a/73d17a5ae26f22eb http://coliru.stacked-crooked.com/a/73d17a5ae26f22eb

vc++: http://webcompiler.cloudapp.net/ http://webcompiler.cloudapp.net/

SFINAE 在模板实例化级别工作，即struct awaitable<T>，不在模板的个人成员级别。awaitable<void>是一个有效的实例化，因此它实例化了该类的所有 3 个成员的声明，并复制了后 2 个成员。

这并不是两个定义相互冲突 - 而是每个定义都与自身冲突（example http://rextester.com/XRLN87339) (更多细节 https://stackoverflow.com/questions/6972368/stdenable-if-to-conditionally-compile-a-member-function).

解决方法1

在类外定义辅助运算符（当然，与您所拥有的不完全相同 - 这些将是朋友any模板实例）

#include <functional>
#include <iostream>

template <typename T>
struct awaitable
{
    friend awaitable<void> operator&&(awaitable a1, awaitable a2)
    {
        std::cout << "operator&&(awaitable a1, awaitable a2) - T: " << typeid(T).name() << std::endl;
        return {};
    }

    template<typename U>
    friend std::enable_if_t<!std::is_void<U>::value, awaitable<void>> operator&&(awaitable<void> a1, awaitable<U> a2);

    template<typename U>
    friend std::enable_if_t<!std::is_void<U>::value, awaitable<void>> operator&&(awaitable<U> a1, awaitable<void> a2);

};

template<typename U>
std::enable_if_t<!std::is_void<U>::value, awaitable<void>> operator&&(awaitable<void> a1, awaitable<U> a2)
{
    std::cout << "operator&&(awaitable<void> a1, awaitable<U> a2) - U: " << typeid(U).name() << std::endl;
    return {};
}

template<typename U>
std::enable_if_t<!std::is_void<U>::value, awaitable<void>> operator&&(awaitable<U> a1, awaitable<void> a2)
{
    std::cout << "operator&&(awaitable<U> a1, awaitable<void> a2) - U: " << typeid(U).name() << std::endl;
    return {};
}

int main(int argc, const char * argv[])
{
    awaitable<int> a1, a2, a3, a4;
    auto ar = a1 && (a1 && a2) && (a2 && a3) && a4;
}

解决方法2

根本不使用 SFINAE，而是使用awaitable。请注意，专业化是相反的 - 基本实现是一个特殊情况awaitable<void>专业化适用于其他一切。

#include <functional>
#include <iostream>

template <typename T, bool isvoid = std::is_void<T>::value>
struct awaitable
{
    friend awaitable<void> operator&&(awaitable a1, awaitable a2)
    {
        std::cout << "operator&&(awaitable a1, awaitable a2) - void" << std::endl;
        return {};
    }
};

template <typename T>
struct awaitable<T, false>
{
    friend awaitable<void> operator&&(awaitable a1, awaitable a2)
    {
        std::cout << "operator&&(awaitable a1, awaitable a2) - T: " << typeid(T).name() << std::endl;
        return {};
    }

    friend awaitable<void> operator&&(awaitable<void> a1, awaitable<T> a2)
    {
        std::cout << "operator&&(awaitable<void> a1, awaitable<T> a2) - U: " << typeid(T).name() << std::endl;
        return {};
    }

    friend awaitable<void> operator&&(awaitable<T> a1, awaitable<void> a2)
    {
        std::cout << "operator&&(awaitable<T> a1, awaitable<void> a2) - void" << std::endl;
        return {};
    }
};

int main(int argc, const char * argv[])
{
    awaitable<int> a1, a2, a3, a4;
    auto ar = a1 && (a1 && a2) && (a2 && a3) && a4;
}