Re: Chaining of dispose and finalize

[John Cupitt <jcupitt gmail com>]

On Sat, 04 Sep 2004 00:39:27 +1000, Russell Shaw <rjshaw netspace net au> wrote:
> I understand that, and is what i do. What i mean is that most
> objects do in class_init something like:
>
>    gobject_class->dispose=myobject_dispose;
>
> But this makes the original dispose code in gobject
> inaccessible. Same for gobject_class->dispose=myobject_finalize.

That assignment isn't overwriting gobject's dispose, it's setting the
gobject dispose for this class. If you have a class called myclass
derived from gobject, then that class has two ->dispose() function
pointers: the standard gobject one, and the one you set. That's
because every time you make a class, you get a fresh copy of every
class above this one in the inheritance hierarchy.

Classes and class instances don't behave the same. When you make a
class instance, you are building a single big, overlapping struct. You
can override things in your parent classes.

When you derive a new class, you build a lot of structs: one for each
class in the hierarchy. When you chain up at the end of dispose,
you're jumping up one struct in this pile of structs.

(hope I've understood you here, sorry if I've missed the point)