On Wednesday 12 September 2007 21:27, Paul McNett wrote:
> [EMAIL PROTECTED] wrote:
> > Nate,
> > That's exactly what I came up with. We were just wondering if that was
> > the only way to do it.
> >
> > What's cool about this is I can refer to the object by doing the
> > following...
> >
> > for i in range(1, 10):
> > exec("o=self.biz%s = dabo.biz.dBizObj()" % i)
> > o.CurrentSql=....
> > o.do_something_else
> >
> > Thanks
> > Larry Long
> >
> > p.s If I executed the code above, does
> > o=None
> > Release the bizobj? I don't think it does. if not, then how can I
> > release it?
>
> Objects in Python are released by the garbage collector when there are
> no more references to them from anywhere. So whether you use the exec()
> or setattr() method (but you should use setattr) the bizobjs you create
> in that loop will stay alive as long as o is in scope (just the single
> iteration in the for loop) and as long as the object referred to as
> 'self' is alive, if you never delete the reference from self using e.g.
> delattr().
>
> So your short answer is: no, calling o=None will not release the bizobj
> (because self has a reference to it).
>
> The longer answer requires asking more questions:
>
> + If you don't want to keep the bizobj around, why bother attaching it
> to self? IOW:
>
> for i in range(10):
> o = dabo.biz.dBizobj()
> o.UserSQL = ...
>
> Now the bizobj goes out of scope after the iteration.
>
> + You probably do want to keep the bizobj around, so that makes me
> wonder why you asked about setting o=None to release it, when o is only
> active inside the for loop.
>
> But, here's how to release the bizobj based on what I know about what
> you are doing:
>
> for i in range(1, 10):
> name_of_biz_to_delete = "biz%s" % i
> delattr(self, name_of_biz_to_delete)
>
> This will work if you haven't saved a reference to that bizobj in any
> other object.
I believe the issue revolves around a recursive method Larry is attempting to
write. The method adds/determines the cost for parent and children records
and he is concerned that the objects he is recreating will use to much
memory. Bottom line after obtaining the data (the cost in this case) the
object is no longer needed.
As I understand it the object will not go out of scope because of the
recursion.
def start(self):
...do something that includes creating an object
if something:
call end()
def end(self):
do something
call start()
It would be nice to release the object after obtaining the cost.
--
John Fabiani
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