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Organization: Axon Sistemas
To: Lukas Smith <lsmith@php.net>
Date: Tue, 10 May 2005 12:16:14 -0300
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Subject: Re: [PHP] Re: [PHP-DEV] make new return a reference to existent object
From: lpedretti@suserver.com (Leonardo Pedretti)

How do i obtain the object again having just the key when i want to return =
it=20
in another call, if i don't have a reference somewhere?
The point would be: object asked id #1, it is not already loaded, load it a=
nd=20
return it, end. Then, in another call: object asked id #1, it is already=20
loaded, return the same instance already in memory, end. And outside of thi=
s=20
function, if all references of that object are deleted, then the object is=
=20
unloaded from memory and so destructed.
The behavior i want to reach is that if an instance of some specific class =
has=20
the same $id that another instance, then they are the same instance.
It's for a database persistent object framework.
Everyone that tries to program such a framework gets to the same problem.

Regards
Leo

On Tuesday 10 May 2005 11:57, you wrote:
> Leonardo Pedretti wrote:
> > Yes, the question was pointed in another way, the destructors are not
> > called until the last reference to an object has been removed, but if y=
ou
> > store a reference in an array so you lately return a reference to the
> > already loaded instance instead of creating an aliasing of that instanc=
e,
> > then you get the undesirable behavior that the object is never destroyed
> > until the script ends, because there is always a reference in that array
> > and you can not check when references are removed in any way.
> > And leaving the array growing while each object that exists in a huge
> > database is slowly loaded and never unloaded, is not an option.
>
> then dont store a reference to the object but instead just store the key
> in the array.
>
> regards,
> Lukas

=2D-=20

Leonardo Pedretti
Axon Sistemas
L=EDder de Equipo Proyecto Basalto