Easy questions from a python beginner

[Rhodri James]

On Mon, 12 Jul 2010 13:56:38 +0100, bart.c <bartc at freeuk.com> wrote:

> "Steven D'Aprano" <steve at REMOVE-THIS-cybersource.com.au> wrote in  
> message news:4c3aedd5$0$28647$c3e8da3 at news.astraweb.com...
>> On Mon, 12 Jul 2010 09:48:04 +0100, bart.c wrote:
>>
>>> That's interesting. So in Python, you can't tell what local variables a
>>> function has just by looking at it's code:
>
>>> def foo(day):
>>>  if day=="Tuesday":
>>>   x=0
>>>  print ("Locals:",locals())
>>>
>>> #foo("Monday")
>>>
>>> Does foo() have 1 or 2 locals?
>>
>> That's easy for CPython: it prepares two slots for variables, but only
>> creates one:
>>
>>>>> foo("Monday")
>> ('Locals:', {'day': 'Monday'})
>>>>> foo.func_code.co_varnames
>> ('day', 'x')
>>>>> foo.func_code.co_nlocals
>> 2
>>
>> So, the question is, is x a local variable or not? It's not in locals,
>> but the function clearly knows that it could be.
>
> So Alf P.S. could be  right; x exists, but Python pretends it doesn't  
> until it's assigned to.

CPython, not Python.  And as Steven said, x *doesn't* exist.  Allowance is  
made by that specific implementation of the interpreter because x *might*  
exist, but in this particular case it doesn't and a more dynamic  
implementation might choose not to reserve a slot just in case.  x is  
created until it's actually used.

-- 
Rhodri James *-* Wildebeeste Herder to the Masses