单独定义的 enable_if 类成员函数

Question

I'm using enable_if class member functions to iterate over a variational template argument. Here is a minimal example (without the actual variadic)

#include <iostream>

template<int size> class Test {
    public:
        template<int i = 0> typename std::enable_if<i == size, void>::type test() {}

        template<int i = 0> typename std::enable_if<i < size, void>::type test() {
            std::cout << "cycle: " << i << '\n';
            test<i + 1>();
        }
};

int main(int, char**) {
    Test<10> a;
    a.test<>();
}

It works just fine but now I'm having problems with dependencies and decided to separate the declarations and definitions. I tried this:

#include <iostream>

template<int size> class Test {
    public:
        template<int i = 0> void test();
};

template<int size>
template<int i> typename std::enable_if<i == size, void>::type Test<size>::test() {}

template<int size>
template<int i> typename std::enable_if<(i < size), void>::type Test<size>::test() {
    std::cout << "cycle: " << i << '\n';
    test<i + 1>();
}

int main(int, char**) {
    Test<10> a;
    a.test<>();
}

but GCC says that error: out-of-line definition of 'test' does not match any declaration in 'Test<size>' . I managed to make it work by including definitions for both cases of test . My question is: why doesn't this work? Shouldn't the compiler only find one of the declarations for any i ? Thank you in advance for any help!

Answer 1

template<int i = 0> 
typename std::enable_if<i == size, void>::type test() { }

template<int i = 0> 
typename std::enable_if<i < size, void>::type test() { /* ... */ }

The two member functions above are completely different, they just happen to have the same name test . They have different signatures and must be declared separately. It's similar to writing:

template<int i = 0> 
int test() { }

template<int i = 0> 
float test() { /* ... */ }

Would you expect to be able to have a single declaration for both of those in your class definition?

Answer 2

You need to add the declaration inside the class with the matching signature

#include <iostream>

template<int size> class Test {
    public:
        template<int i = 0> typename std::enable_if<i == size, void>::type test();
    template<int i = 0> typename std::enable_if<i < size, void>::type test();
};

template<int size>
template<int i> typename std::enable_if<i == size, void>::type Test<size>::test() {}

template<int size>
template<int i> typename std::enable_if<(i < size), void>::type Test<size>::test() {
    std::cout << "cycle: " << i << '\n';
    test<i + 1>();
}

int main(int, char**) {
    Test<10> a;
    a.test<>();
}