C++ real method overloading?

Question

I have the classes A and B. B derives from A and overloads the method WhoAreYou(), when I now create a variable of type A and set the value to a B object and then call WhoAreYou(), the method of A is called. Look at this:

class A{
public:
    virtual void WhoAreYou(){
        cout << "I am A!"<<endl;
    }
};
class B: public A{
public:
    void WhoAreYou(){
        cout << "I am B!" << endl;
    }
};


int main(int argc, char ** argv){
    A a = B();
    a.WhoAreYou(); //Output: I am A!
}

Is there a way to overload the method so, that in this case the WhoAreYou() method of B would be called? When I must cast the object first, a method overloading doesn´t really make sense in my opinion...

Thanky for your help!

B overrides WhoAreYou. Overloading would mean it also exists with a different set of arguments( e.g. WhoAreYou(int) ) — xtofl
– xtofl, Commented Feb 21, 2011 at 12:32

Erik · Accepted Answer · 2011-02-21 12:30:35Z

5

Your "a" variable is an A, not a B. A a = B(); creates a new temporary B, then creates an A by copying the A part of B.

Try this:

B b;
A * a = &b;
a->WhoAreYou();

answered Feb 21, 2011 at 12:30

Erik

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4 Comments

Van Coding Over a year ago

This works! Didn't know that there is a defference between pointers and normal objects in this case. Thanks!

xtofl Over a year ago

Or use a reference: A& a = b;.

Erik Over a year ago

@FlashFan: Your variable "a" is an A. Nothing you do will change that, you declared it to be an A. Using references or pointers, you can refer to or point to something else, which is at least an A, as my sample shows.

Van Coding Over a year ago

Hey xtofl, I saw your answer and wanted to mark it as correct. But you deleted it...

Community · Accepted Answer · 2017-05-23 12:04:24Z

1

The problem has to do with slicing. I asked the exact same question with this topic.

edited May 23, 2017 at 12:04

CommunityBot

11 silver badge

answered Feb 21, 2011 at 12:32

Marlon

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