无法让SFINAE工作

这是我在SFINAE的第一次尝试:

#include 
#include 

struct C1 {
    using T = int;
};

struct C2 {
    using T = void;
};

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

// For classes that declare T = void
template 
void f(C &c,
       std::enable_if::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::value), int>::type’ is not a type 

我究竟做错了什么？

你需要几个typenames才能工作:

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

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

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