Question

这是我第一次尝试SFINAE：

#include <type_traits>
#include <iostream>

struct C1 {
    using T = int;
};

struct C2 {
    using T = void;
};

// For classes that declare T = int
template <class C>
void f(C &c,
       std::enable_if<!std::is_same<typename C::T, void>::value, int>::type = 0) {
    std::cout << "With T" << std::endl;
}

// For classes that declare T = void
template <class C>
void f(C &c,
       std::enable_if<std::is_same<typename C::T, void>::value, int>::type = 0) {
    std::cout << "Without T" << std::endl;
}

int main() {
    C1 c1;
    f(c1); // With T
    C2 c2;
    f(c2); // Without T
    return 0;
}

编译器（gcc 4.8.2）抱怨：

‘std::enable_if<!(std::is_same<typename C::T, void>::value), int>::type’ is not a type

我做错了什么？

Answer 1

您需要几个typename才能实现此目的：

// For classes that declare T = int
template <class C>
void f(C &c,
       typename std::enable_if<!std::is_same<typename C::T, void>::value, int>::type = 0) {
    std::cout << "With T" << std::endl;
}

// For classes that declare T = void
template <class C>
void f(C &c,
       typename std::enable_if<std::is_same<typename C::T, void>::value, int>::type = 0) {
    std::cout << "Without T" << std::endl;
}

或者如果您的编译器支持C ++ 14，您可以使用std::enable_if_t：

// For classes that declare T = int
template <class C>
void f(C &c,
       std::enable_if_t<!std::is_same<typename C::T, void>::value, int> = 0) {
    std::cout << "With T" << std::endl;
}

// For classes that declare T = void
template <class C>
void f(C &c,
       std::enable_if_t<std::is_same<typename C::T, void>::value, int> = 0) {
    std::cout << "Without T" << std::endl;
}

无法让SFINAE工作

1 个答案: