On Tue, 03 Jun 2008 23:08:46 +0200, Christian Heimes wrote:
> [EMAIL PROTECTED] 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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