python 2.3's lambda behaves old fashioned

[Fredrik Lundh]

Uwe Schmitt wrote:

> I just tried (Python 2.3)
>
>     li = [ lambda x: x*a for a in range(10) ]
>
> which results in
>
>     li[0](1) = 9
>     ...
>     li[9](1) = 9
>
> In order to achieve the intended result I had to fall back on the
> following trick:
>
>     li = [ lambda x,a=a: x*a for a in range(10)]
>
> which leads to the expected result.
>
> Any explanations ???

repeat after me:

free variables bind to names, default arguments bind to objects.
free variables bind to names, default arguments bind to objects.
free variables bind to names, default arguments bind to objects.
(etc)

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