Hi Ken! Thank you for your long answer! It cleared up a lot of things for me. I come from coding in Actionscript (Flash) and there things are apparently quite different.
Is it true that all instances I create live in memory on the same level? That they are all somehow equal? And that all relationships between them are only by pointers? Because in Actionscript you can have instances inside other object's instances and once you dealloc the containing instance all the other object's instances it contained die with it. Think Matryoshka puppets or something like that. On 10/28/08, Ken Thomases <[EMAIL PROTECTED]> wrote: > Part of the problem is that object's don't have names, they have addresses. I see. So I can only access an instance from within some other instance if I always remember to pass along a pointer to it? > But the pointer isn't the object and the object isn't the pointer. This is what I got wrong I guess. > So, when you say "I create a subobject called 'mySubObject' inside my > main object 'myMainObject'" I translate that to mean, that you have a > pointer named myMainObject to an instance of some custom class. In > the implementation of that class, there's a method, and in that method > you create another object of some other(?) class and store its address > into a pointer named mySubObject. mySubObject might be a local > variable, an instance variable, or whatever. > > Now, what does "within 'mySubObject'" mean? I assume you mean within > an instance method of the class of which mySubObject is an instance. Thank you. You translated correctly and explained it to me at the same time! :-) Your following explanation did the trick for me. Thank you again! Cheers, John. _______________________________________________ Cocoa-dev mailing list (Cocoa-dev@lists.apple.com) Please do not post admin requests or moderator comments to the list. Contact the moderators at cocoa-dev-admins(at)lists.apple.com Help/Unsubscribe/Update your Subscription: http://lists.apple.com/mailman/options/cocoa-dev/archive%40mail-archive.com This email sent to [EMAIL PROTECTED]
