Re: Confused .. What is happenning here

From:

"mliptak" <Mehturt@gmail.com>

Newsgroups:

comp.lang.c++

Date:

28 Mar 2007 08:13:20 -0700

Message-ID:

<1175094800.678662.200290@b75g2000hsg.googlegroups.com>

On Mar 28, 4:50 pm, "Mathematician" <mathemtician1234567...@yahoo.com>
wrote:

On Mar 28, 6:17 am, "mliptak" <Meht...@gmail.com> wrote:

On Mar 28, 3:34 pm, "James Kanze" <james.ka...@gmail.com> wrote:

On Mar 28, 2:08 pm, "mliptak" <Meht...@gmail.com> wrote:

On Mar 28, 12:44 pm, "Erik Wikstr=F6m" <eri...@student.chalmers.se>
wrote:

On 28 Mar, 12:25, vb.h...@gmail.com wrote:

I am new to C++ and was just reading about polymorphism. I trie=

d to

write a very simple program. Then a curious thought came into m=

y mind.

And instead of using pointer in polymorphism, i used a referenc=

e. And

both of them printed the same thing.
I want to know what is going on under the hood.

Polymorphism. Polymorphism is independent of pointers or referenc=

(but you must use them to make it work). And as you see it works =

just

No, you don't need pointers nor references in order to invoke member
functions polymorphically.

I'm not sure what you're trying to say. Dynamic polymorphism
only occurs when the dynamic type of an object can differ from
the static type, and in C++, that pretty much means pointers or
references.

What I meant is that virtual op can be invoked from non-virtual in the
base class, e.g.:

class Base
{
public:
void f()
{ g(); }
virtual void g() {}

};

class Derived : public Base
{
public:
void g() {}

};

int main()
{
Derived d;
d.f();
return 0;

}

--
James Kanze (GABI Software) mailto:james.ka...@gmail.com
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virtual key is meaningless, whether you use it or not, the output's
still unchanged, you just handle a simple scope problem that you call
g() via f() on purpose

of course it's not..
if Base::g() is not virtual, then Base::g() would be invoked
however in my example, Derived::g() is invoked