Subj : Re: Typecasting question
To   : borland.public.cpp.borlandcpp
From : Greg Chicares
Date : Mon Dec 13 2004 09:51 pm

On 2004-12-13 6:33 PM, Rick wrote:
> 
>     class X{};
>     class Y: public X
>     {
>         public:
>             int aValue;
>     };
> 
>     X    *objects[10];
> 
>     objects[0] = new Y();
> 
> when I want to call a property from the Y class, I don't know how to do 
> this. I tried typecasting but it didn't work:
>     someValue = (X)objects[0]->aValue;

Don't cast to X: you want a Y. And don't use an
explicit '(Y)' cast: it's not type checked.

Use a polymorphic base class as in the example below
(preferably putting common functionality into the
base interface) and then use dynamic_cast.

> aValue isn't a member of the X class, how to tell that objects[0] is 
> actually an instance of class Y?

dynamic_cast can tell. Here's a complete example:

#include <iostream>
#include <ostream>
#include <stdexcept>

class X
{
   public:
     virtual ~X() {}
};

class Y: public X
{
   public:
     Y() : aValue(3) {}
     int aValue;
};

int main()
{
     X* objects[10];
     objects[0] = new Y();

     Y* yp = dynamic_cast<Y*>(objects[0]);
     if(yp)
         std::cout << yp->aValue << std::endl;
     else
         throw std::runtime_error("Your error message here");
}

.