Two override examples in C ++

/*

 * main.cc

 *

 *  Created on: 2008-8-5

 *      Author: Administrator

 */

#include <iostream>

class Base

{

public:

    virtual Base *clone(){ std::cout << "Base::clone() \n"; return new Base;}

};

class Derived : public Base

{

public:

    Derived *clone(){ std::cout << "Derived::clone() \n"; return new Derived;}

};

int main(int arg, char **args)

{

    Derived *pd = new Derived;

    pd->clone();

    Base *pb = pd;

    pb->clone();

    return 0;

}

Result:

Derived: Clone ()
Derived: Clone ()

We can see that although the return types are not exactly the same, the return value in the parent class is a pointer or reference. When the subclass override, we can return the derivation of this pointer (or reference). In this case, override succeeded.

Let's look at the following example:

#include <iostream>

class Base

{

public:

    virtual Base *clone(){ std::cout << "Base::clone() \n"; return new Base;}

    virtual void copy(Base&){ std::cout << "Base::copy() \n"; }

};

class Derived : public Base

{

public:

    Derived *clone(){ std::cout << "Derived::clone() \n"; return new Derived;}

    void copy(Derived&) { std::cout << "Derived::copy() \n"; }

};

int main(int arg, char **args)

{

    Derived d;

    Base b;

    Derived *pd = new Derived;

    pd->copy(d);

    Base *pb = pd;

    pb->copy(b);

    return 0;

}

Result:

Derived: Copy ()
Base: Copy ()

In this case, override does not occur, and hide does.

Set

void copy(Derived&) { std::cout << "Derived::copy() \n"; }

Changed:

void copy(Base&) { std::cout << "Derived::copy() \n"; }

Override is successful!