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?

CodePudding user response：

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(); }.