Access to an overridden member of a base class from a derived class object

Question

I have two classes:

class A { public: int i; }; class B : public A { public: int i; }; 

Suppose I created an object for class B

 B b; 

Is it possible to access A::i using B ?

Answer 1

Is it possible to access A :: i using b?

Yes! What about bA::i ? ;)

Answer 2

Yes:

 int main() { B b; bi = 3; bA::i = 5; A *pA = &b; B *pB = &b; std::cout << pA->i << std::endl; std::cout << pB->i << std::endl; } 

Answer 3

Yes, you can. Read this to find out.

The override method provides a new implementation of an element inherited from the base class. A method overridden by an override declaration is known as an overridden base method. An overridden base method must have the same signature as the override method.

Inside a derived class that has an override method, you can still access the overridden base method with the same name using the base keyword. For example, if you have a virtual method MyMethod () and a method to override a derived class, you can access the virtual method from the derived class by calling:

base.MyMethod ()

Answer 4

Perhaps, as others answered. But in the example you posted, the base and derived members are the same, the data type was not overrated.

This would be relevant for derived classes that define a new data type for base members, as shown in this post: C ++ Inheritance. Changing Object Data Types

Answer 5

Two ways:

 struct A{ A():i(1){} int i; }; struct B : A{ B():i(0), A(){} int i; }; int main(){ B b; cout << bA::i; cout << (static_cast<A&>(b)).i; }