std::function and function type deduced by "using" don't have same type

Question

Here is a small example to show the difference of two function types that are not the same:

#include <iostream>
#include <functional>
#include <type_traits>

template <typename T>
using BinaryOperator = T(const T&, const T&);

int main() {

    std::cout << std::boolalpha 
              << std::is_same<
                      std::function<int(const int&, const int&)>, 
                      BinaryOperator<int>
                 >::value 
              << std::endl;

    return 0;
}

This prints false which is confusing to me. Both types seems to be equivalent. How are they different?

Answer 1

Both types seems to be equivalent. How are they different?

Well... no: they are different types.

If you look at the std::function's page in cppreference.com, you can see that std::function is a class with partial specialization (only the specialization is defined) declared as follows

template <class>
class function; // undefined

template <class R, class... Args>
class function<R(Args...)>;

So your BynaryOperator<int> isn't equivalent to std::function<int(const int&, const int&)> but is equivalent to its template argument.

You can see that is true

std::is_same<std::function<int(const int&, const int&)>, 
             std::function<BinaryOperator<int>>
>::value //  ^^^^^^^^^^^^^^...................^