Is a derived class considered a member of the base class?

From this [emphasis added]:

If some member function vf is declared as virtual in a class Base, and some class Derived, which is derived, directly or indirectly, from Base, has a declaration for member function with the same

  • name
  • parameter type list (but not the return type)
  • cv-qualifiers
  • ref-qualifiers

Then this function in the class Derived is also virtual (whether or not the keyword virtual is used in its declaration) and overrides Base::vf (whether or not the word override is used in its declaration).

Base::vf does not need to be accessible or visible to be overridden. (Base::vf can be declared private, or Base can be inherited using private inheritance. Any members with the same name in a base class of Derived which inherits Base do not matter for override determination, even if they would hide Base::vf during name lookup.)

class B {
    virtual void do_f(); // private member
    void f() { do_f(); } // public interface
struct D : public B {
    void do_f() override; // overrides B::do_f

int main()
    D d;
    B* bp = &d;
    bp->f(); // internally calls D::do_f();

Overriding the virtual function does not count as accessing. However if you need to call the base implementation, that would be impossible to do from Derived.

class Base {
    virtual ~Base() = default;

    void do_something() {

    virtual void do_something_impl() { /*Default implementation*/ }

class Derived : public Base {
    void do_something_impl() override {
        //Base::do_something_impl() is not accessible from here:
        std::cout << "Hello, world!\n";