Re: Polymorphism and vectors

[Anakreon <anakreonmejdi at yahoo dot gr>]

> This mailing list is for the development OF libstdc++, not WITH, so you 
> shouldn't really be mailing here unless you think you've found a bug in 
> the library. gcc-help (or a newsgroup, such as comp.lang.c++) would be 
> more approriate. However, I'll answer your question anyway :)
I didn't know about the comp.lang.c++ list. I knew that the question was not
apropriate for this list but had to ask. Sorry about this. I'll post to the 
right list
next time.
> >  vector<base*> vect;
> > ...
> >  for (vector<base*>::iterator it = vect.begin();
> >        it != vect.end(); it++) {
> >     cout << typeid(*it).name() << endl;
> >  }
> >
> >  
> Here you are asking for the type of a pointer to base, and so thats what 
> you get!
> If you want the type of the thing the pointer points to, you have to do 
> typeid(**it).name(), which then gets you what you would expect. This is 
> broadly  the right way to do what you are trying to do (although  
> getting one of the various reference-counted pointers will make your 
> life much easier in terms of not having to manually delete things as and 
> when you eject them from the vector).
Ah, I see. This works perfectly. Thanks.
> > ...
> >  vector<base> v2;
> >  
>  >...
>
> >  for (vector<base>::iterator it = v2.begin();
> >        it != v2.end(); it++) {
> >     cout << typeid(*it).name() << endl;
> >  }
> >  
>
> Here you have a problem with something known as slicing. Basically the 
> vector will convert all the objects to bases when they are inserted, by 
> ignoring the non-base part. This is bad! :)
Actually I knew about this. I placed the code in order to see if any difference 
would
exist in the output generated by the two loops.

Thanks Chris.