Can a using declaration refer only to a specific overload based on const qualification?

Question

If the base class has both const and non-const version of a function, can I refer to only one or the other in the derived class by the using keyword?

struct Base
{
protected:
    int x = 1;

    const int&  getX() const     {return x;}
    int&        getX()           {return x;}
};

struct Derived : public Base
{
public:
    using Base::getX; // Can we refer to the const or the non-const only?
};

If no, whats the simplest way to make const getX() public in Derived without repeating the function body?

Answer 1

using always brings the entire overload set for the given name with it. You cannot invoke using for a particular function, only for its name, which includes all uses of that name.

You will have to write a derived-class version of the function, with your preferred signature, that calls the base class. The simplest way would be with decltype(auto) getX() const { return Base::getX(); }.