Re: Call derived class function from base class reference

From:
"Victor Bazarov" <v.Abazarov@comAcast.net>
Newsgroups:
comp.lang.c++
Date:
Fri, 8 Jun 2007 14:06:53 -0400
Message-ID:
<f4c5ru$kv4$1@news.datemas.de>
Javier wrote:

Hello, is this possible?
I have a pure virtual function in the base class (to force the
programmers of the derived classes to have this function implemented)
but I want to call the derived class function from a base class
reference:

class A
{
public:
    virtual void function() = 0;
};

class B
:
public A
{
public:
    void function() { cout << "Hello" << endl; };
}

    ;

You've missed a semicolon.

int main()
{
    B derived();


'derived' here is a function. Drop the parentheses.

    A &base = dynamic_cast<A &>(B); // don't now if this is the way


Of course this is not correct. First of all, if you have dropped
the parentheses, you _could_ do

    A &base = dynamic_cast<A&>(derived);

but you don't need to, the conversion is _implicit_:

   A &base = derived;

    base.function(); // I need this type of call because I dont't
know the derived class type at runtime.


That should work OK, after you fix the declarations.

}

Of course I could have a function2 (private) and make it pure virtual,
and then have function() in base class calling the virtual function2:

class A
{
public:
    void function() { this->function_impl() };

protected:
    virtual void function_impl() = 0;
};

class B
:
public A
{
protected:
    void function_impl() { cout << "Hello" << endl; };
}

    ;

int main()
{
    B derived();
    A &base = dynamic_cast<A &>(B);

    A.function();
}

Is this good?


That's just as OK, given that you fix the declarations and get
rid of the dynamic_cast.

But... can I do it the first way?


Should be OK.

V
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