无法使SFINAE工作

Cannot get SFINAE to work

本文关键字：工作 SFINAE      更新时间：2023-10-16

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？

您需要几个typename才能工作：

// 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;
}
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