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

I think this can't work with a single defaulted template parameter, because its value needs to be resolved when the class template is instantiated.

We need to defer the substitution to the point of constructor template instantiation. One way is to default the template parameter to T and add an extra dummy parameter to the constructor:

template<typename U = T>
A(int n, typename std::enable_if<U::value>::type* = 0) : val(n) { }