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

Steven Bethard steven.bethard at gmail.com
Wed Dec 8 21:13:12 CET 2004

Caleb Hattingh wrote:
> 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.

Yes, that was my intent.  In the same way that my "x" was not 
initialized, your "x" is not updated.  locals() is readable but not 
writable in any case where locals() is not globals(), I believe.

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

Unfortunately, I don't think there is a Rule(tm) because the behavior of 
locals() (and globals()?) are implementation details.  I remember 
hearing an explanation of why locals() is not writable that had to do 
with something about efficiency and the call stack (or something like 


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