Re: Is it possible to call virtual method in derived class from base class without declaring it abstract?

From:
Jaco Naude <naude.jaco@gmail.com>
Newsgroups:
comp.lang.c++
Date:
Tue, 28 Sep 2010 05:05:57 -0700 (PDT)
Message-ID:
<30a9cb47-a055-4a64-9d52-77c000b1d11c@h4g2000yqp.googlegroups.com>
On Sep 28, 12:01 pm, Jaco Naude <naude.j...@gmail.com> wrote:

On Sep 28, 9:49 am, cpp4ever <n2xssvv.g02gfr12...@ntlworld.com> wrote:

On 09/28/2010 10:07 AM, Jaco Naude wrote:

Hi

Not sure if this is a stupid question, but here it goes:

Is it possible to call a virtual method in the derived class from a
base class? I know that the method can be made abstract but then the
whole class also becomes and abstract base class.

class BaseClass {
    virtual bool method(bool call_derived) {
        if (call_derived)
            // I want to call method() in DerivedClass wi=

thout knowing

anything about it.
        else
            return false;
    }
}

class DerivedClass : public BaseClass {
    bool method() { return true; }
}

Thus when I do this in my main function:
BaseClass base;
base.method(true); // Should return true;
base.method(false); // Should return false;

Is this even possible at all? Or is the only way to do this by
declaring method() = 0 in the base class (like a normal interface
class).

Thanks,
Jaco


If you wish to ensure that the base class member function is called you
need to call the member function using a fully qualified name

Hence in DerivedClass you would need to call as follows

BaseClass::method(); // call BaseClass function
method(); // calls DerivedClass function

This will work even if it is called via a BaseClass pointer to a
DerivedClass object.

HTH

cpp4ever


Thanks for the answers. The problem is not a piece of code that does
not work, I'm trying to figure out if how something works in C++. I
created a proper small example which answers my question:

#include <QtCore/QCoreApplication>
#include <stdio.h>

class Base
{
protected:
    virtual bool virtualBlah() const {
        return false;
    }
public:
    bool blah() const {
        return virtualBlah();
    }

};

class Derived: public Base
{
protected:
    virtual bool virtualBlah() const {
        return true;
    }

};

int main(int argc, char *argv[])
{
    Base* base = new Base();
    Derived* derived = new Derived();

    if (base->blah())
        printf("Call on base class instance: True\n");
    else
        printf("Call on base class instance: False\n");

    Base* derived_base_ptr = reinterpret_cast<Base*> (derived);
    if (derived_base_ptr->blah())
        printf("Call on derived class instance: True\n");
    else
        printf("Call on derived class instance: True\n");

    return a.exec();

}

The output of the program is:
Call on base class instance: False
Call on derived class instance: True

I always thought that if you call blah() on derived_base_ptr it will
call the base class implementation. To get to the derived class
implementation I thought you need a Derived*, unless you make the base
class an abstract base class.

Thank you for your help, it is appreciated.
Jaco


The was a typo in my previous post,

    Base* derived_base_ptr = reinterpret_cast<Base*> (derived);
    if (derived_base_ptr->blah())
        printf("Call on derived class instance: True\n");
    else
        printf("Call on derived class instance: True\n");

should have been:

    Base* derived_base_ptr = reinterpret_cast<Base*> (derived);
    if (derived_base_ptr->blah())
        printf("Call on derived class instance: True\n");
    else
        printf("Call on derived class instance: False\n");

I only saw all the other responses now. Thanks!
I understand it now.
Cheers
Jaco

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