Select class constructor using enable_if

Question

Consider following code:

#include <iostream>
#include <type_traits>

template <typename T>
struct A {
    int val = 0;

    template <class = typename std::enable_if<T::value>::type>
    A(int n) : val(n) {};
    A(...) { }

    /* ... */
};

struct YES { constexpr static bool value = true; };
struct NO { constexpr static bool value = false; };

int main() {
    A<YES> y(10);
    A<NO> n;
    std::cout << "YES: " << y.val << std::endl
              << "NO:  " << n.val << std::endl;
}

I want to selectively define constructor A::A(int) only for some types using enable_if. For all other types there is default constructor A::A(...) which should be the default case for compiler when substitution fails. However this makes sense for me compiler (gcc version 4.9.0 20130714) is still complaining

sfinae.cpp: In instantiation of 'struct A': sfinae.cpp:19:11:
required from here sfinae.cpp:9:5: error: no type named 'type' in
'struct std::enable_if'
A(int n) : val(n) {};

Is something like this possible for constructor? Is this possible with another constructor(s) (copy-constructor and move-constructor)?

Answer 1

With C++20

You can achieve that simply by adding requires to the constructor:

A(int n) requires T::value : val(n) { }

The requires clause gets a constant expression that evaluates to true or false deciding thus whether to consider this method in the overload resolution, if the requires clause is true, or ignore it otherwise.

Code: https://godbolt.org/z/948z41zKK