CODEWITHSUNDEEP
c++inheritance
Jakub M.
Jakub M.

Reputation: 33827

inheritance, calling overloaded method from base method

I have two classes:

class A {
    int get();
    string str();
}

string A::str() {
    return string_with( get() );
}

class B : public A{
    int get();
}

if I do:

A a
B b
b.str()

b will invoke A::str() (good) and it ill use A::get() method (bad!). I want, that when I invoke b.str(), the B::get() is used by str.

How to make it work?

Upvotes: 0

Views: 132

Answers (3)

XiaBing
XiaBing

Reputation: 11

You can use virtual keyword. and then use pointer

class A
{
public:
    virtual int get() { cout << "A::get()" << endl; return 0;}
    string str()
    {
        cout << "A::str()" << endl;
        get();
        return "";
    }
};

class B : public A
{
public:
    virtual int get() { cout << "B::get()" << endl; return 0; }
};

B* b = new B; b->str();

the output: A::str() B::get()

Upvotes: 1

akappa
akappa

Reputation: 10490

In the magical word of Object-Oriented programming, there are two ways of calling a method: static and dynamic dispatch.

In static dispatch, the code called when you do something like a.do_it() is determined statically, that is, it is determined upon the type of the variable a.

In dynamic dispatch, the code called is determined dynamically, i.e., it is determined upon the type of the object referenced by a.

C++, of course, supports both. How do you tell the compiler which type of dispatch do you want? Simple: by default you have static dispatch, unless you put the virtual in the method declaration.

Upvotes: 2

EboMike
EboMike

Reputation: 77752

Just make it virtual. That's exactly what virtual is for.

Write

virtual int get();

in A's definition. And, just to make the code more understandable, do the same in B's.

By the way, I'm assuming that you meant to write class B : public A.

Upvotes: 3

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