Q about object identity

Marc 'BlackJack' Rintsch bj_666 at gmx.net
Wed Jun 4 06:51:34 CEST 2008


On Tue, 03 Jun 2008 23:08:46 +0200, Christian Heimes wrote:

> vronskij at gmail.com schrieb:
>> Hello,
>> 
>> I am testing object identity.
>> 
>> If I do it from the interpreter, I get strange results.
>> 
>>>>> print [] is []
>> False
>> 
>>>>> print id([]), id([])
>> 3083942700 3083942700
>> 
>> 
>> 
>> Why is that? Isn't this an error?
> 
> No, it's not an error. You are getting this result because the list
> implementation keeps a bunch of unused list objects in a free list. It's
> an optimization trick. Python has to create two different list objects
> for "[] is []" while it can reuse the same list object for id([]) == id([]).

I don't think you need optimization tricks for the explanation.  In ``[]
is []`` there need to be two lists that exist at the same time to be
compared by the ``is`` operator.  With ``id([]) == id([])`` just the
id numbers have to exist at the same time, so the execution may look like
this:

• create empty list
• call `id()` with it
• remember first id number
• when `id()` returns, the list is not referenced anymore and can be
  garbage collected
• create empty list, and here the list object might get allocated at the
  same memory location as the first empty list → same id.
• call `id()` with it
• remember second id number
• garbage collect second empty list
• compare both numbers

Ciao,
	Marc 'BlackJack' Rintsch



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