id(a) == id(b) and a is not b --> bug?

Gerrit Holl gerrit at nl.linux.org
Fri Jun 6 21:43:36 CEST 2003


Erik Max Francis schreef op vrijdag  6 juni om 02:22:35 +0000:
> >>> class C:
> ...  def m(self): pass
> ... 
> >>> c = C()
> >>> c.m is c.m
> 0
> >>> id(c.m) == id(c.m)
> 1
> 
> It's happening because accessing a method dynamically creates an unique
> object and goes away when you're finished with it.

There's probably a perfectly good reason for that ;)

> So when an
> expression needs both at the same time (c.m is c.m), two unique objects
> are created, but when an expression doesn't need both at the same time
> (id(c.m) == id(c.m)), an object gets created with an ID, is destroyed,
> and then a second object is created with the same ID (since the last one
> just got reclaimed) and their IDs compare equal.  Weird, huh?  (It
> surprised me when I first saw it, too.)

Ah, thanks for this explanation!

BTW, is there any way I could have found out about this myself, expect
reading the source (which I can't because of lack of knowledge)?

yours,
Gerrit.

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