Can't deduce template parameter with std::enable_if as function parameter

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.

Piotr Skotnicki · Accepted Answer · 2020-06-07 15:20:02Z

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);

I see. How would that work in a constructor which doesn't have a return type though?
What is the significance of using int as the second argument of std::enable_if? Is it arbitrary? Or is it because int = 0 is safe?

Alan Stokes · Accepted Answer · 2015-12-19 09:25:26Z

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

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