CODEWITHSUNDEEP
c++oopinheritance
user3533014
user3533014

Reputation:

Calling base method however only second level base method exists

Imagine a huge code base with many many inheritance. At some point a line comes in your way in the class you have to work with.

inherited::Load();

This is do what it says. It loads the base object into memory. inherited is a typedef for the base class.
However the actual parent class does not have a Load method.

As a quick example:

class Base {
public:
  Base() {};
  void Load() {
    cout << "Base ok" << endl;
  }
};

class Derived : public Base {
public:
  Derived() {};
};

class MostDerived : public Derived {
public:
  MostDerived(){};
  void Load() {
    Derived::Load();
    cout << "Child ok"<< endl;
  }
};

Here, if we call MostDerived.Load() it will call its parent's Load method, but the Derived class doesn't have its own Load, only Base's Load method.

What is actually going on here? Why there is no compilation issue? Is the Derived class copies all the method of the base, so it will have a Load method what is prints "Base ok"? Or does Derived simply calls forward the Base method?

I found a related question Do ALL virtual functions need to be implemented in derived classes? and the answer says

It inherits the bar implementation from its ancestor class.

For me it still miss leading that explicitly calling the Derived::Load() method works. However, there is only a Base::Load() method exists.

Upvotes: 0

Views: 45

Answers (1)

463035818_is_not_an_ai
463035818_is_not_an_ai

Reputation: 122516

What is actually going on here? Why there is no compilation issue? Is the Derived class copies all the method of the base, so it will have a Load method what is prints "Base ok"?

Yes. A derived class inherits all methods from the base class. (yes really all of them, whether it can access them depends on the access specified for the methods).

For me it still miss leading that explicitly calling the Derived::Load() method works. However, there is only a Base::Load() method exists.

Sloppy speaking the Derived:: only tells the compiler to look for names accesible from Derived. As Derived does inherit from Base, it does have a Load method.

Maybe your confusion can be cleared up a bit by noting that Derived::Load indeed does refer to Base::Load:

#include <type_traits>
#include <iostream>
struct Base {
    void Load() {}
};
struct Derived : Base {};

int main() {
    std::cout << std::is_same_v< decltype(&Derived::Load), void (Derived::*) ()>;
    std::cout << std::is_same_v< decltype(&Derived::Load), void (Base::*) ()>;
}

This prints 01, because Derived::Load is indeed a method of Base.

I hope the example is not adding confusion ;). It uses std::is_same to see if two types are the same. &Derived::Load gives me a member function pointer to the Load method in Derived whose type I infer via decltype. I compare that once to pointer to method of Derived and then to pointer to method of Base. It turns out that Derived::Load is a method of Base.

Upvotes: 2

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