updating locals() and globals() (WAS: How do I do this? (eval() on the left hand side))

[Caleb Hattingh]

Steve,

I don't think I understand.  Here is what I just tried:

'>>> def f():
	x = 3
	d = locals()
	print x
	print d['x']
	d['x'] = 5
	print x

	
'>>> f()
3
3
3
'>>>

In your example, x had not yet been initialised, maybe.  What I am seeing  
is that "x" does not seem to update when being assigned to - I guess this  
is what you are referring to by being unreliable.

But "unreliable" sounds kinda vague and intermittent, and I assume that is  
not the case here - What is the Rule(tm)?

Thanks
Caleb



On Wed, 08 Dec 2004 16:59:23 GMT, Steven Bethard  
<steven.bethard at gmail.com> wrote:

> Peter Hansen wrote:
>> Nick Coghlan wrote:
>>
>>> Generally, altering the contents of the dicts returned by locals() and  
>>> globals() is unreliable at best.
>>   Nick, could you please comment on why you say this about globals()?
>> I've never heard of any possibility of "unreliability" in updating
>> globals() and, as far as I know, a large body of code exists which
>> does in fact rely on this -- much of mine included. ;-)
>
> Updating locals() is unreliable.  Updating globals() is fine, AFAIK.
>
> http://docs.python.org/lib/built-in-funcs.html
>
> I believe that the only time that locals() is updatable is when locals()  
> is globals():
>
>  >>> locals() is globals()
> True
>  >>> x
> Traceback (most recent call last):
>    File "<interactive input>", line 1, in ?
> NameError: name 'x' is not defined
>  >>> locals()['x'] = 3
>  >>> x
> 3
>
>
>  >>> def f():
> ...     print locals() is globals()
> ...     locals()['x'] = 3
> ...     print x
> ...
>  >>> f()
> False
> Traceback (most recent call last):
>    File "<interactive input>", line 1, in ?
>    File "<interactive input>", line 4, in f
> NameError: global name 'x' is not defined
>
>
> Steve