srini4vasan@gmail.com wrote:
> 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.

"Phill W." <p-.-a-.-w-a-r-d-@-o-p-e-n-.-a-c-.-u-k> wrote in message
news:f8cmbu$lui$1@south.jnrs.ja.net...
> srini4vasan@gmail.com wrote:
>> 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.
>
> 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.