Why does std::swap<std::array<int,3>> not compile?

Question

I would like to swap two arrays of ints of some fixed size.

Counterintuitively, the following does not compile because no matching instance of swap has been found.

#include <array>
#include <utility>

int main()
{
    std::array<int,3> a, b;
    std::swap< std::array<int,3> >( a, b ); 
    return 0;
}

I find this surprising. However, replacing the swap with std::swap(a,b) compiles and (according to VSCode) has signature

inline void std::swap<int, 3UL>(std::array<int, 3UL> &__one, std::array<int, 3UL> &__two)

which I cannot make sense of either.

Q: What is going on here?

Answer 1

The overload you are looking for is (from cppreference):

template< class T, std::size_t N >    
constexpr void swap( std::array<T, N>& lhs,
                     std::array<T, N>& rhs )
                         noexcept(/* see below */);

As the error reports, the compiler can't find a viable overload of std::swap() that matches std::swap<std::array<int,3>>.

This would be the "right" way to supply template arguments explictly:

#include <array>
#include <utility>

int main()
{
    std::array<int,3> a, b;
    std::swap<int,3>( a, b ); 
    return 0;
}

I doubt there is a situation where you actually want to do that though.

PS: You can also see that in the signature you get from VSCOde: std::swap<int, 3UL> is not std::swap<std::array<int,3UL>>. However, looking at implementation can be misleading sometimes and I rather suggest to consult documentation first.