Accessing base class method using derived class object

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srini4vasan@gmail.com - 26 Jul 2007 12:54 GMT

#include<iostream>
using namespace std;

class A
{
public:
    void display()
    {
        cout<<"A display"<<endl;
    }

};

class B : public A
{
public:
    void display()
    {
        cout<<" B display "<<endl;
    }
};

int main()
{
    B b1;
    b1.display();

    return 0;
}

In this program can you please tell me how can i access base class
method ie.,display() using derived class object.ie.,b1.

A single line has to be included in this main funtion.

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Phill W. - 27 Jul 2007 12:53 GMT

> class A
> {
[quoted text clipped - 27 lines]
>
> A single line has to be included in this main funtion.

Sounds decidedly home-work-ish to me, but what the hey...

/If/ B.display() shadows A.display(), i.e. there's no overriding
involved - my C++ is a /little/ rusty ;-) - then you can do this
directly.
Simply tell ("up-cast") 'B' to behave as if it were an 'A'.

((A)b1).display();

However, if B.display() /overrides/ A.display() - don't think this is
the case, but just for completeness - then you're stuck.
Unless you expose an alternative method on 'B' that calls the base
implementation in 'A', then you can't access the base implementation in
A; and that's as it should be.

HTH,
Phill W.

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Ben Voigt [C++ MVP] - 02 Aug 2007 18:36 GMT

>> class A
>> {
[quoted text clipped - 39 lines]
> However, if B.display() /overrides/ A.display() - don't think this is the
> case, but just for completeness - then you're stuck.

There's no "virtual", so no dynamic dispatch. Calling through a reference
to A will call the base implementation.

> Unless you expose an alternative method on 'B' that calls the base
> implementation in 'A', then you can't access the base implementation in A;
> and that's as it should be.
>
> HTH,
> Phill W.

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Accessing base class method using derived class object

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