Select class constructor using enable_if

Question

Consider following code:

#include 
#include 

template 
struct A {
    int val = 0;

    template 

        

Answer 1

Usually this is done using an anonymous defaulted argument :

A(int n, typename std::enable_if::type* = 0) : val(n) {};

You can not use template parameters from the class to SFINAE out methods. SO one way is to add a dummy type replacing int :

see: http://ideone.com/2Gnyzj

#include 
#include 

template 
struct A {
    int val = 0;

    template::type
            >
    A(Integer n) : val(n) {};

    A(...) {}
    /* ... */
};

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

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

This works because you use a member template parameter to SFINAE out the constructor but the test is always true so it doesn't pollute your checks