enable_if method specialization

Question

template
struct A
{
    A operator%( const T& x);
};

template
A A::operator%( const T& x ) {          


        

Answer 1

With C++20

You can achieve that simply by adding requires to restrict the relevant template function:

template // the generic case, no restriction
A operator% ( const Q& right ) const {
    return A(std::fmod(x, right));
}

template requires std::is_integral_v && std::is_integral_v
A operator% ( const Q& right ) const {
    return A(x % right);
}

The requires clause gets a constant expression that evaluates to true or false deciding thus whether to consider this method in the overload resolution, if the requires clause is true the method is preferred over another one that has no requires clause, as it is more specialized.

Code: https://godbolt.org/z/SkuvR9