Re: compare two objects, if they have the same vtable

From:

=?UTF-8?B?RXJpayBXaWtzdHLDtm0=?= <Erik-wikstrom@telia.com>

Newsgroups:

comp.lang.c++.moderated

Date:

Thu, 2 Oct 2008 19:29:03 CST

Message-ID:

<Jk8Fk.2966$U5.12478@newsb.telia.net>

On 2008-10-01 20:38, Albert Zeyer wrote:

Mathias Gaunard schrieb:

On 29 sep, 18:29, Albert Zeyer <albert.z...@rwth-aachen.de> wrote:

I have an easy base class:

class Base {
virtual void func();

}

And I want to do an easy type-equality check based on the vtable-entry
of func.

typeid(obj1) == typeid(obj2).

I don't want to add any RTTI to the project as it is
not really needed here (because in my case, it's enough to just compare
the vtable).

RTTI just compares the pointers to the vtable.

I thought RTTI adds runtime type information to each class and typeid is
based on these additional information.

No only for classes for which the type can not be statically determined
(polymorphic classes) and a good compiler can probably avoid adding the
type information for classes for which it will never be used.

I don't need the overhead of runtime type information just to compare
the vtable which is there anyway.

The overhead will be some additional memory in the data-segment of your
application, which will probably be swapped out until you need it.

The same thing could be written in a non-portable, ugly way, as
*(void**)&obj1 == *(void**)&obj2.

Hm, just to get sure that I understand that:

You assume here that there is a pointer to the vtable at the beginning
of the object.

Yes.

--
Erik Wikstr??m

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