SFINAE: std::enable_if as function argument - c++

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, e.g., 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 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 <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