How does the compiler determine between a function using SFINAE and a standard function if both are viable?

Question

Consider the following code:

#include <iostream>
#include <type_traits>

template <typename T>
class A
{
public:
    // Allow func to be called if T is the const version of T2
    // e.g., T is 'int const' and T2 is 'int'
    template <typename T2,
              typename = typename std::enable_if<
                                          std::is_same<T, T2 const>::value>::type>
    void func(A<T2> const &)
    {
        std::cout << "Conversion" << std::endl;
    }

    // Allow func to be called for same version of T
    void func(A const &)
    {
        std::cout << "No conversion" << std::endl;
    }
};

int main()
{
    A<int const> a;

    a.func(A<int const>{});

    return 0;
}

This code, when compiled with GCC-8.3 compiles and produces the output No conversion - it selected the version of func that does not use std::enable_if. However, if I comment out the second version of func, it will still compile and now produce the output Conversion. In other words, both versions of func within A are usable for this method. Given that both overloads are viable, what specific rule is the compiler using to select func(A const &) over the other version (func(A<T2> const &))?

Answer 1

The rule is that if a non function template and a function template specialization have the same signature, then the non function template is chosen over the template specialization. This can be found in [over.match.best]/2

Given these definitions, a viable function F1 is defined to be a better function than another viable function F2 if for all arguments i, ICSi(F1) is not a worse conversion sequence than ICSi(F2), and then

[...]

• F1 is not a function template specialization and F2 is a function template specialization [...]