我可以用std :: enable_if摆脱模板专业化吗？

Question

I read about std::enable_if for:

function overloading based on arbitrary properties of type

So I was trying to overload class ctors via enable_if (like below) but i get error saying that enable_if cannot be used to disable declaration and this in both lines when I used std::enable_if :

#include <iostream>
#include <type_traits>
#include <typeinfo>


template <typename T>
class cls
{
public:

    cls (T a, typename std::enable_if< std::is_same<T, int>::value >::type  * Dummy = 0)
    {
        std::cout << "Ctor for int\n";
    }

    cls (T a, typename std::enable_if< std::is_same<T, char>::value >::type  * Dummy = 0)
    {
        std::cout << "Ctor for char\n";
    }
};

int main()
{

    cls a(10);
    cls b('x');

    return 0;
}

So is it possible to overload ctors using enbale_if .

Answer 1

The problem is, eg when you instantiate a cls<int> , you'll always get a failed requirement from the 2nd constructor overloaing, ie std::enable_if< std::is_same<T, char>::value >::type with T is int . And it's same for cls<char> .

You can make the constructors templates to make SFINAE taking effect; which will get rid of template specializations from the overload set and won't cause compile error. eg

template <typename X = T>
cls (X a, typename std::enable_if< std::is_same<X, int>::value >::type  * Dummy = 0)
{
    std::cout << "Ctor for int\n";
}

template <typename X = T>
cls (X a, typename std::enable_if< std::is_same<X, char>::value >::type  * Dummy = 0)
{
    std::cout << "Ctor for char\n";
}

LIVE

Answer 2

You don't seem to need enable_if here, just a regular constructor:

cls(T a) { std::cout << "Ctor for T\n"; }

To disable a template class specialization do not provide its definition:

template<> class cls<void>; // disabled specialization.