SFINAE: std::enable_if 作为函数参数

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:

template<typename T>
void fun(const typename std::enable_if_t<std::is_integral<T>::value, T>& val) {
    std::cout << "fun<int>";
}

template<typename T>
void fun(const typename std::enable_if_t<std::is_floating_point<T>::value, T>& val) {
    std::cout << "fun<float>";
}

int main()
{
    fun(4);
    fun(4.4);
}

This way I would have to write:

fun<int>(4);
fun<double>(4.4);

How would I avoid that?

Compiler complains that it can't deduce the parameter T .

Answer 1

The examples are wrong, since T is in a non-deduced context . Unless you call the function like fun<int>(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, eg, in a return type syntax:

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

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

DEMO

Also, the typename s 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 <typename T,
              typename std::enable_if<std::is_integral<T>::value, int>::type = 0>
    A(const T& val)
    {
        std::cout << "A<int>";
    }

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

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);

Answer 2

To allow deduction you need a function parameter that is straightforwardly based on T . You then need to figure out where to put your enable_if (which indeed does not allow T to be deduced). Common options are on the return type or on an extra default parameter that you ignore.

Some good examples here: http://en.cppreference.com/w/cpp/types/enable_if