Accessing Base class function using a pointer to a derived class

**John Harrison** · Jul 19 '05, 02:46 PM

Re: Accessing Base class function using a pointer to a derived class

"Abhijit Deshpande" <abhijit.deshpa nde@lycos.com> wrote in message
news:8d1b01ba.0 306280456.12ff7 ba1@posting.goo gle.com...[color=blue]
> Is there any elegant way to acheive following:
>
> class Base {
> public:
> Base() {}
> virtual ~Base() {}
> virtual void Method() { cout << "Base::Meth od called"; return;
> }
> };
>
> class Derived : public Base {
> public:
> Derived() {}
> ~Derived()
> void Method() { cout << "Derived::Metho d called"; return; }
> };
>
> int main() {
> Derived deriveObj;
> Base * basePtr = 0;
>
> basePtr = <some kind of cast????> &deriveObj;
>
> basePtr->Method();
> }
>
> In the above code, the call "basePtr->Method" should print
> "Base::Meth od called" and not "Derived::Metho d called".
>
> Is there any way to assign address of "deriveObj" to "basePtr" so that
> the virtual mechanism is bypassed and call to member function "Method"
> actually calls the member function from the "class Base" and not from
> "class Derived".
>
> Thanks in advance,
> Regards,
> Abhijit.[/color]

You can do this

Derived deriveObj;
Base * basePtr = 0;

basePtr = &deriveObj;

basePtr->Base::Method() ; // calls Base::Method

but there is no way to bypass the virtual machanism when you assign a
pointer.

john

**Rolf Magnus** · Jul 19 '05, 02:46 PM

Re: Accessing Base class function using a pointer to a derived class

Abhijit Deshpande wrote:
[color=blue]
> Is there any elegant way to acheive following:
>
> class Base {
> public:
> Base() {}
> virtual ~Base() {}
> virtual void Method() { cout << "Base::Meth od called"; return;
> }
> };
>
> class Derived : public Base {
> public:
> Derived() {}
> ~Derived()
> void Method() { cout << "Derived::Metho d called"; return; }
> };
>
> int main() {
> Derived deriveObj;
> Base * basePtr = 0;
>
> basePtr = <some kind of cast????> &deriveObj;
>
> basePtr->Method();
> }
>
> In the above code, the call "basePtr->Method" should print
> "Base::Meth od called" and not "Derived::Metho d called".
>
> Is there any way to assign address of "deriveObj" to "basePtr" so that
> the virtual mechanism is bypassed and call to member function "Method"
> actually calls the member function from the "class Base" and not from
> "class Derived".[/color]

basePtr->Base::Method() ;

But please think thrice before doing that. Looks like bad design to me.

**John Harrison** · Jul 19 '05, 02:46 PM

Re: Accessing Base class function using a pointer to a derived class

"Craig Thomson" <craig@craigtho mson.me.uk> wrote in message
news:bdklpg$4r6 $1@news.ukfsn.o rg...[color=blue]
> Hi,
>[color=green]
> > Is there any way to assign address of "deriveObj" to "basePtr" so that
> > the virtual mechanism is bypassed and call to member function "Method"
> > actually calls the member function from the "class Base" and not from
> > "class Derived".[/color]
>
> I'm a little hazy on the whole casting business, but as I understand it[/color]
what[color=blue]
> you want is:
>
> main() {
> Derived deriveObj;
> Base * basePtr = 0;
>
> basePtr = dynamic_cast<Ba se*>(&DeriveObj );
>
> basePtr->Method();
> }
>
> or
>
> main() {
> Derived deriveObj;
> Base * basePtr = 0;
>
> basePtr = static_cast<Bas e*>(&DeriveObj) ;
>
> basePtr->Method();
> }
>
> Dynamic cast is safer because it does run time type checking but as long[/color]
as[color=blue]
> your going from derived to base class static cast should work fine too.
> Google for dynamic_cast or static_cast and you should get all the
> information you need.
>[/color]

Neither type of cast is necessary when converting from a derived class
pointer to a base class pointer, and neither cast achieves what the OP wants
which is to override the virtual function calling mechanism.

You've got this mixed up with converting from a base class pointer to a
derived class pointer, when some sort of cast is necessary.

john

**Abhijit Deshpande** · Jul 19 '05, 02:49 PM

Re: Accessing Base class function using a pointer to a derived class

Thanks John and Ralf for the solution.

But I was wondering, should following piece of code work?
In addition to "class Base" and "class Derived", we define one more
class,

class DummyBase() {

public:

DummyBase() {
}

~DummyBase() {
}

void Method() {
Base::Method();

return;
}
};

int main() {
Derived deriveObj;
DummyBase * dummyBasePtr = reinterpret_cas t<DummyBase
*>(&deriveObj) ;

dummyBasePtr->Method();

return 0;
}

This should print "Base::Meth od called". Is there anything
conceptually wrong in above piece of code?? Because, I tried it using
GNU g++ on RedHat linux 7.2, and it still prints "Derived::Metho d
called".

Regards,
Abhijit.

"John Harrison" <john_andronicu s@hotmail.com> wrote in message news:<bdkjor$u8 2sv$1@ID-196037.news.dfn cis.de>...[color=blue]
> "Craig Thomson" <craig@craigtho mson.me.uk> wrote in message
> news:bdklpg$4r6 $1@news.ukfsn.o rg...[color=green]
> > Hi,
> >[color=darkred]
> > > Is there any way to assign address of "deriveObj" to "basePtr" so that
> > > the virtual mechanism is bypassed and call to member function "Method"
> > > actually calls the member function from the "class Base" and not from
> > > "class Derived".[/color]
> >
> > I'm a little hazy on the whole casting business, but as I understand it[/color]
> what[color=green]
> > you want is:
> >
> > main() {
> > Derived deriveObj;
> > Base * basePtr = 0;
> >
> > basePtr = dynamic_cast<Ba se*>(&DeriveObj );
> >
> > basePtr->Method();
> > }
> >
> > or
> >
> > main() {
> > Derived deriveObj;
> > Base * basePtr = 0;
> >
> > basePtr = static_cast<Bas e*>(&DeriveObj) ;
> >
> > basePtr->Method();
> > }
> >
> > Dynamic cast is safer because it does run time type checking but as long[/color]
> as[color=green]
> > your going from derived to base class static cast should work fine too.
> > Google for dynamic_cast or static_cast and you should get all the
> > information you need.
> >[/color]
>
> Neither type of cast is necessary when converting from a derived class
> pointer to a base class pointer, and neither cast achieves what the OP wants
> which is to override the virtual function calling mechanism.
>
> You've got this mixed up with converting from a base class pointer to a
> derived class pointer, when some sort of cast is necessary.
>
> john[/color]

**Karl Heinz Buchegger** · Jul 19 '05, 02:49 PM

Re: Accessing Base class function using a pointer to a derived class

Abhijit Deshpande wrote:[color=blue]
>
> Thanks John and Ralf for the solution.
>
> But I was wondering, should following piece of code work?
> In addition to "class Base" and "class Derived", we define one more
> class,
>
> class DummyBase() {[/color]
[color=blue]
>
> public:
>
> DummyBase() {
> }
>
> ~DummyBase() {
> }
>
> void Method() {
> Base::Method();
>
> return;
> }
> };
>
> int main() {
> Derived deriveObj;
> DummyBase * dummyBasePtr = reinterpret_cas t<DummyBase
> *>(&deriveObj) ;
>[/color]

You are casting way to much!
What (if any) is the relationship of Derived and DummyBase.
Please show it with code and not with english descriptions.

[color=blue]
> dummyBasePtr->Method();
>
> return 0;
> }
>
> This should print "Base::Meth od called". Is there anything
> conceptually wrong in above piece of code??[/color]

Impossible to say without seeing the actual, complete code you used to test it.

--
Karl Heinz Buchegger
kbuchegg@gascad .at

**Abhijit Deshpande** · Jul 19 '05, 02:51 PM

Re: Accessing Base class function using a pointer to a derived class

Well, here is the complete piece of code, I tried..

class Base {

public:

Base() {
};

virtual ~Base() {
};

virtual void Method() {
printf("Base::M ethod called.\n");

return;
}
};

class Derived : public Base {

public:

Derived() {
};

~Derived() {
};

void Method() {
printf("Derived ::Method called.\n");

return;
}
};

class DummyBase : public Base {

public:

DummyBase() {
}

~DummyBase() {
}

void Method() {
Base::Method();

return;
}
};

int main() {

Derived derivedObj;
DummyBase *dummyBasePtr;

dummyBasePtr = reinterpret_cas t<DummyBase *>(&derivedObj) ;

dummyBasePtr->Method();
}

And the expected output is "Base::Meth od called.", whereas the actual
output is "Derived::Metho d called."

Regards,
Abhijit.

Karl Heinz Buchegger <kbuchegg@gasca d.at> wrote in message news:<3F0025CC. DD54EAA2@gascad .at>...[color=blue]
> Abhijit Deshpande wrote:[color=green]
> >
> > Thanks John and Ralf for the solution.
> >
> > But I was wondering, should following piece of code work?
> > In addition to "class Base" and "class Derived", we define one more
> > class,
> >
> > class DummyBase() {[/color]
>[color=green]
> >
> > public:
> >
> > DummyBase() {
> > }
> >
> > ~DummyBase() {
> > }
> >
> > void Method() {
> > Base::Method();
> >
> > return;
> > }
> > };
> >
> > int main() {
> > Derived deriveObj;
> > DummyBase * dummyBasePtr = reinterpret_cas t<DummyBase
> > *>(&deriveObj) ;
> >[/color]
>
> You are casting way to much!
> What (if any) is the relationship of Derived and DummyBase.
> Please show it with code and not with english descriptions.
>
>[color=green]
> > dummyBasePtr->Method();
> >
> > return 0;
> > }
> >
> > This should print "Base::Meth od called". Is there anything
> > conceptually wrong in above piece of code??[/color]
>
> Impossible to say without seeing the actual, complete code you used to test it.[/color]

Accessing Base class function using a pointer to a derived class

Accessing Base class function using a pointer to a derived class

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