c++ - 使用 CRTP 和其他模板重载继承

Question

我在重载某些运算符时遇到了问题。

具体来说，我有一个模板，interface_mixin<T> ，它具有传统的 CRTP 设置，即 Derived : public interface_mixin<Derived> .

现在需要重载运算符哪一边是对象，另一边是模板，即:

template<typename T, typename Derived> ... operator..
(T t, const interface_mixin<Derived>& d) {
    ... 
}
template<typename T, typename Derived> ... operator..
(const interface_mixin<T>& t, Derived d) {
    ... 
}
template<typename T, typename Derived> ... operator..
(const interface_mixin<T>& t, const interface_mixin<Derived>& d) {
    ... 
}

但是，我的编译器 (VS2010) 不会接受这个，调用模糊重载。我怎样才能说服它接受这些过载？

现在，我正在尝试使用 SFINAE 来尝试清除其他重载。但是即使逻辑看起来不错，编译器还是选择了错误的重载。

template<typename T, typename Derived> 
typename std::enable_if<
    !std::is_base_of<
        interface_mixin<T>, 
        T
    >::value, 
    and<
        equality_rule<T>, 
        Derived
    >
>::type operator>>(T t, const interface_mixin<Derived>& d) {
    return and<equality_rule<T>, Derived>(equality_rule<T>(std::move(t)), d.crtp_cast());
}
template<typename T, typename Derived> 
typename std::enable_if<
    !std::is_base_of<
        interface_mixin<Derived>, 
        Derived
    >::value, 
    and<
        T, 
        equality_rule<Derived>
    >
>::type operator>>(const interface_mixin<T>& t, Derived d) {
    return and<T, equality_rule<Derived>>(t.crtp_cast(), equality_rule<Derived>(std::move(d)));
}
template<typename T, typename Derived> and<T, Derived> operator>>(const interface_mixin<T>& t, const interface_mixin<Derived>& d) {
    return and<T, Derived>(t.crtp_cast(), d.crtp_cast());
}

However, VS is picking the wrong overload, and the logic won't make sense when the wrong overload is picked.

Answer 1

你的问题的第一部分与第二部分不同，你没有在那里指定函数体的返回类型。
如果返回类型是 void - 此代码适用于我的 MSVC2010 编译器:

#include <iostream>

template<class T> struct interface_mixin {};

template<typename T, typename Derived>
typename std::enable_if<!std::is_base_of<interface_mixin<T>, T>::value, void>::type
operator>>(T t, const interface_mixin<Derived>& d) { std::cout << "1\n"; }

template<typename Derived, typename T>
typename std::enable_if<!std::is_base_of<interface_mixin<T>, T>::value, void>::type
operator>>(const interface_mixin<Derived>& d, T t) { std::cout << "2\n"; }

template<typename DerivedL, typename DerivedR>
void
operator>>(const interface_mixin<DerivedL>& t, const interface_mixin<DerivedR>& d)
{ std::cout << "3\n"; }

struct Foo : interface_mixin<Foo> {};
struct Bar : interface_mixin<Bar> {};

int main()
{
    Foo f;
    Bar b;

    1 >> f;
    f >> 1;
    f >> b;
}

我不知道问题是与返回类型有关还是其他原因，但是 enable_if 应该可以解决这个问题。