polymorphism - virtual function

[al]

class Base
{
public:
virtual void method();
};

class Derive : public Base
{
public:
void method();
};


Base *b = new Base;
b->method();//Base::method() called

Base *d = new Derive;
d->method();//Derive::method( ) called

Why b->method() trigger Base::method() whereas d->method() Derive::method( )?

Thanks!

[Victor Bazarov]

Re: polymorphism - virtual function

"al" <allin@168.ne t> wrote...[color=blue]
> class Base
> {
> public:
> virtual void method();
> };
>
> class Derive : public Base
> {
> public:
> void method();
> };
>
>
> Base *b = new Base;
> b->method();//Base::method() called
>
> Base *d = new Derive;
> d->method();//Derive::method( ) called
>
> Why b->method() trigger Base::method() whereas d->method()[/color]
Derive::method( )?

Because 'b' is a pointer to a complete object of class Base, and
'd' is a pointer to a subobject of a complete object of class Derive.
What matters is the type of the object at its creation (the actual,
the complete, object).

V

[Jeff Schwab]

Re: polymorphism - virtual function

Victor Bazarov wrote:[color=blue]
> "al" <allin@168.ne t> wrote...
>[color=green]
>>class Base
>>{
>> public:
>> virtual void method();
>>};
>>
>>class Derive : public Base
>>{
>> public:
>> void method();
>>};
>>
>>
>>Base *b = new Base;
>>b->method();//Base::method() called
>>
>>Base *d = new Derive;
>>d->method();//Derive::method( ) called
>>
>>Why b->method() trigger Base::method() whereas d->method()[/color]
>
> Derive::method( )?
>
> Because 'b' is a pointer to a complete object of class Base, and
> 'd' is a pointer to a subobject of a complete object of class Derive.
> What matters is the type of the object at its creation (the actual,
> the complete, object).
>
> V[/color]

What Victor said is true for any "virtual" method, including the one
you've called here. If you want to change the behavior, just don't
include the word "virtual" in the method declaration.

[Billy O'Connor]

Re: polymorphism - virtual function

"al" <allin@168.ne t> writes:
[color=blue]
> class Base
> {
> public:
> virtual void method();
>
> };
>
> class Derive : public Base
> {
> public:
> void method();
> };
>
>
> Base *b = new Base;
> b->method();//Base::method() called
>
> Base *d = new Derive;
> d->method();//Derive::method( ) called
>
> Why b->method() trigger Base::method() whereas d->method() Derive::method( )?[/color]

Well, b is a Base, and d is a Derived. What behavior were you
expecting?

[al]

Re: polymorphism - virtual function

Victor Bazarov <v.Abazarov@com Acast.net> wrote in message
news:ugrIb.7064 51$Fm2.609354@a ttbi_s04...[color=blue]
> "al" <allin@168.ne t> wrote...[color=green]
> > class Base
> > {
> > public:
> > virtual void method();
> > };
> >
> > class Derive : public Base
> > {
> > public:
> > void method();
> > };
> >
> >
> > Base *b = new Base;
> > b->method();//Base::method() called
> >
> > Base *d = new Derive;
> > d->method();//Derive::method( ) called
> >
> > Why b->method() trigger Base::method() whereas d->method()[/color]
> Derive::method( )?
>
> Because 'b' is a pointer to a complete object of class Base, and
> 'd' is a pointer to a subobject of a complete object of class Derive.
> What matters is the type of the object at its creation (the actual,
> the complete, object).
>
> V
>
>[/color]
Thanks! This definitely helps me understand the topic.

If removing "virtual" from method() declaraton of class Base, then why
d->method() triggers Base::method() since 'd' is a pointer to the actual
object of class Derive?

[Victor Bazarov]

Re: polymorphism - virtual function

"al" <allin@168.ne t> wrote...[color=blue]
> Victor Bazarov <v.Abazarov@com Acast.net> wrote in message
> news:ugrIb.7064 51$Fm2.609354@a ttbi_s04...[color=green]
> > "al" <allin@168.ne t> wrote...[color=darkred]
> > > class Base
> > > {
> > > public:
> > > virtual void method();
> > > };
> > >
> > > class Derive : public Base
> > > {
> > > public:
> > > void method();
> > > };
> > >
> > >
> > > Base *b = new Base;
> > > b->method();//Base::method() called
> > >
> > > Base *d = new Derive;
> > > d->method();//Derive::method( ) called
> > >
> > > Why b->method() trigger Base::method() whereas d->method()[/color]
> > Derive::method( )?
> >
> > Because 'b' is a pointer to a complete object of class Base, and
> > 'd' is a pointer to a subobject of a complete object of class Derive.
> > What matters is the type of the object at its creation (the actual,
> > the complete, object).
> >
> > V
> >
> >[/color]
> Thanks! This definitely helps me understand the topic.
>
> If removing "virtual" from method() declaraton of class Base, then why
> d->method() triggers Base::method() since 'd' is a pointer to the actual
> object of class Derive?[/color]

No, 'd' is not a pointer to the actual object of class Derive. It is,
as I already said, a pointer to a subobject. If you don't use 'virtual'
in declaring the member function[s], the binding is static and does not
respect the fact that the subobject of type Base is really part of the
object created as 'Derive'.

The difference, hence, is static binding versus dynamic binding.

Victor

[Cy Edmunds]

Re: polymorphism - virtual function

"al" <allin@168.ne t> wrote in message
news:dmsIb.5668 66$0v4.22876237 @bgtnsc04-news.ops.worldn et.att.net...[color=blue]
> Victor Bazarov <v.Abazarov@com Acast.net> wrote in message
> news:ugrIb.7064 51$Fm2.609354@a ttbi_s04...[color=green]
> > "al" <allin@168.ne t> wrote...[color=darkred]
> > > class Base
> > > {
> > > public:
> > > virtual void method();
> > > };
> > >
> > > class Derive : public Base
> > > {
> > > public:
> > > void method();
> > > };
> > >
> > >
> > > Base *b = new Base;
> > > b->method();//Base::method() called
> > >
> > > Base *d = new Derive;
> > > d->method();//Derive::method( ) called
> > >
> > > Why b->method() trigger Base::method() whereas d->method()[/color]
> > Derive::method( )?
> >
> > Because 'b' is a pointer to a complete object of class Base, and
> > 'd' is a pointer to a subobject of a complete object of class Derive.
> > What matters is the type of the object at its creation (the actual,
> > the complete, object).
> >
> > V
> >
> >[/color]
> Thanks! This definitely helps me understand the topic.
>
> If removing "virtual" from method() declaraton of class Base, then why
> d->method() triggers Base::method() since 'd' is a pointer to the actual
> object of class Derive?
>
>[/color]

Well, that's the distinction between virtual and not virtual! If not virtual
the compiler only takes note of the declared type of the pointer and
effectively bit slices down to a Base type (which is OK because a Derive IS
a Base). If virtual it looks up the actual type and proceeds accordingly.

By the way, Derive::method is virtual also. Once a method is declared
virtual you can't make it non-virtual later.

--
Cy

http://home.rochester.rr.com/cyhome/

[P.V.S.S.VIKAS]

Re: polymorphism - virtual function

Billy O'Connor <billyoc@gnuyor k.org> wrote in message news:<87u13hwtp r.fsf@dps11.gnu york.org>...[color=blue]
> "al" <allin@168.ne t> writes:
>[color=green]
> > class Base
> > {
> > public:
> > virtual void method();
> >
> > };
> >
> > class Derive : public Base
> > {
> > public:
> > void method();
> > };
> >
> >
> > Base *b = new Base;
> > b->method();//Base::method() called
> >
> > Base *d = new Derive;
> > d->method();//Derive::method( ) called
> >
> > Why b->method() trigger Base::method() whereas d->method() Derive::method( )?[/color]
>
> Well, b is a Base, and d is a Derived. What behavior were you
> expecting?[/color]

these lines have no sense in
i'am new to this world.learning to mve around kindly coperate.