are free variables part of the local dict ?

[Cesar Douady]

In article <3C4372A4.77AD176F at gol.ge>, "Giorgi Lekishvili" <gleki at gol.ge>
wrote:

> Hi!
> What you'd say about this?
>>>> def foo(i):
>  global x
>  x=1
>  def bar(i):
>   x=x+i
>   print locals()
>  bar(i)
> 
> Or, have I undestood you in a wrong manner?
> 
> Grtz,
> Giorgi
> 
> 
>>>> foo(2)
> Traceback (innermost last):
>   File "<pyshell#22>", line 1, in ?
>     foo(2)
>   File "<pyshell#21>", line 7, in foo
>     bar(i)
>   File "<pyshell#21>", line 5, in bar
>     x=x+i
> UnboundLocalError: Local variable 'x' referenced before assignment

Where is your problem ? because x is assigned in bar, x is a local and
you reference it before assigning to it.
This, by the way, implies that there is no way (that I know of, I'd be
glad to be contradicted) to update a free variable.

>>>> def foo(i):
>  global x
>  x=1
>  def bar(i):
>   x
>   x=x+i
>   print locals()
>  bar(i)
> 
> 
>>>> foo(2)
> Traceback (innermost last):
>   File "<pyshell#30>", line 1, in ?
>     foo(2)
>   File "<pyshell#29>", line 8, in foo
>     bar(i)
>   File "<pyshell#29>", line 5, in bar
>     x
> UnboundLocalError: Local variable 'x' referenced before assignment
>>>>

idem.

>>>>
>