SFINAE: std::enable_if as function argument

Question

So, I\'m following the example set by the code somewhere on this web page: http://eli.thegreenplace.net/2014/sfinae-and-enable_if/

Here\'s what I have:

t         


        

Answer 1

The examples are wrong, since T is in a non-deduced context. Unless you call the function like fun(4);, the code won't compile, but this is probably not what the author intended to show.

The correct usage would be to allow T to be deduced by the compiler, and to place a SFINAE condition elsewhere, e.g., in a return type syntax:

template 
auto fun(const T& val)
    -> typename std::enable_if::value>::type
{
    std::cout << "fun";
}

template 
auto fun(const T& val)
    -> typename std::enable_if::value>::type
{
    std::cout << "fun";
}

DEMO

Also, the typenames in your code contradict your usage of std::enable_if_t.

Use either c++11:

typename std::enable_if<...>::type

or c++14:

std::enable_if_t<...>

---

How would that work in a constructor which doesn't have a return type though?

In case of constructors, the SFINAE condition can be hidden in a template parameter list:

struct A
{    
    template ::value, int>::type = 0>
    A(const T& val)
    {
        std::cout << "A";
    }

    template ::value, int>::type = 0>
    A(const T& val)
    {
        std::cout << "A";
    }
};

DEMO 2

Alternatively, in c++20, you can use concepts for that:

A(const std::integral auto& val);

A(const std::floating_point auto& val);