are free variables part of the local dict ?

[Giorgi Lekishvili]

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
>>> 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
>>>


Cesar Douady wrote:

> in :
>
> def foo():
>     x=1
>     def bar():
>         x
>         print locals()
>     bar()
> foo()
>
> the result is "{}", indicating that x is not in locals(). But in :
>
> def foo():
>     x=1
>     def bar():
>         x
>         y=1
>         print locals()
>     bar()
> foo()
>
> the result is "{'y':1, 'x':1}", indicating that the presence of y has made
> x part of locals().
>
> Note also that in :
>
> def foo():
>     x=1
>     class bar:
>         x
>         y=1
>     print dir(bar)
> foo()
>
> the result is "['__doc__', '__module__', 'y']", showing that x was not
> included.
>
> My comprehension is that free variables should never be part of locals()
> and that the second output is a bug. However, from looking at the
> interpreter code, it seems that this is done on purpose (i.e. free vars
> are part of the local dict only for functions, the difference between the
> first 2 codes still looks like a bug).
>
> Before sending a bug report on such a tricky subject, I would appreciate
> getting various opinions.
>
> Cesar.