C++ inheritance and name hiding [duplicate]

Answer 1

This means that it is just the member function name to hide the same name member function in the base class, and the signature is not taken into consideration

Correct. That is the way the language was designed. I don't really like that language design feature either. But it is well documented and you don't have a choice.

Answer 2

I think you already understood. Its called Name Hiding and this is how the language is made. But i don't see it as an inconvenience as you get pretty comfortable with these convention eventually when you spend time with C++. When you define a non virtual method with the same name as Base method it hides the Base class method in Derived class so you are getting the error for

obj.methodA();

To avoid hiding of Base class methods in Derived class, you can do :

obj.Base::methodA();

For more info: http://www.programmerinterview.com/index.php/c-cplusplus/c-name-hiding/

Answer 3

That is the way the language is designed - names in inner scopes hide names in outer scopes. Here the derived class acts as an inner scope and the base class as an outer scope.

The fact that the functions would be considered overloads if they were in the same scope, doesn't matter. The hiding works on the names, not on individual functions.

You can override the default, by explicitly adding the name to the derived class:

class Derived : public Base {
public:
    using Base::methodA;   // Now it is visible!

    void methodA(int i) { cout << "Derived.methodA(int i)" << endl ;}
};