SFINAE: std::enable_if as function argument
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 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);