Re: Does object have function?

From:
James Kanze <james.kanze@gmail.com>
Newsgroups:
comp.lang.c++
Date:
Mon, 1 Nov 2010 11:11:15 -0700 (PDT)
Message-ID:
<3fd625f2-d6b0-44c3-8d5e-a15b74e98f42@c20g2000yqj.googlegroups.com>
On Oct 30, 12:04 am, "Daniel T." <danie...@earthlink.net> wrote:

Joshua Maurice <joshuamaur...@gmail.com> wrote:

On Oct 29, 4:46 am, "Daniel T." <danie...@earthlink.net> wrote:

 Joshua Maurice <joshuamaur...@gmail.com> wrote:

With this multiple inheritance design, I would guess
that you probably want to virtually inherit from Fooer
as well (not done in the above code).


Virtual inheritance would only be necessary if Fooer had
member-variables. Inheriting interfaces (classes with only
pure virtual functions and no member-variables,) does not
require virtual inheritance.


I'm not so sure about this. Let me think about it.

At the very least, if you inherit from such an "interface"
class twice, then you will have two distinct base class
sub-objects, and they will have distinct addresses. (IIRC,
the intent of the standard is that two objects (complete or
sub-objects) of the same type should be distinct objects iff
they have distinct addresses. However, I recall that the
wording might have been changed to remove this requirement.
Not sure what the situation is.) I would think that that is
counter-intuitive. I think that a programmer might simply
assume that if he has two distinct Fooer (sub)objects
(distinct according to distinct addresses), then he has two
distinct complete objects. However, such inference would be
incorrect without virtual inheritance.


The empty base class optimization means that the above is not
the case. A base class that contains no data members need not
have a distinct address.


First, the empty base class optimization is not required by
a compiler. And a class with virtual functions is never really
empty. Whether the base class has data members or not is
irrelevant when considering whether inheritance should be
virtual or not.

In any case, virtual inheritance only comes into play if there is a
diamond inheritance pattern, which there is not in this example.


That's the critical distinction.

--
James Kanze

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