why is self not passed to id()?

[Maric Michaud]

Le Thursday 04 September 2008 22:26:53 Ruediger, vous avez écrit :
> Hello!
>

Hello,

> Executing following little program gives me an TypeError.
>
> What makes me wonder is that foo does get an argument passed while bar
> doesn't. Can anyone explain why??????
>

Because id is a builtin written in the core language and doesn't subscribe to 
the descritpor protocol python functions has.

>
> class foo(list):
>     __hash__ = lambda x: id(x)
>

Wow ! You are really going on trouble with this, believe me there is a real 
good reason for list not to be hashable. A dictionnary or set containing some 
of your foo is virtually inconsistent, read carefully the manual about 
prerequesites for dict keys, they *need* to be immutable.


> class bar(list):
>     __hash__ = id
>
> _s_ = set()
> _s_.add(foo())
> _s_.add(bar())
>
> rue at linux:~> python test01.py
> Traceback (most recent call last):
>   File "test01.py", line 9, in <module>
>     _s_.add(bar())
> TypeError: id() takes exactly one argument (0 given)
>
> --
> http://mail.python.org/mailman/listinfo/python-list



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Maric Michaud
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