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From: Ian Kelly <ian.g.kelly@gmail.com>
Date: Wed, 1 Feb 2012 16:00:02 -0700
Subject: Re: Question about name scope
To: Ethan Furman <ethan@stoneleaf.us>
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Xref: x330-a1.tempe.blueboxinc.net comp.lang.python:19748

On Wed, Feb 1, 2012 at 3:53 PM, Ethan Furman <ethan@stoneleaf.us> wrote:
> --> def f(x, y):
>
> ... =A0 =A0 locals()[x] =3D y
> ... =A0 =A0 print(vars())
> ... =A0 =A0 exec('print (' + x + ')')
> ... =A0 =A0 print(x)
> ...
> --> f('a', 42)
>
> {'y': 42, 'x': 'a', 'a': 42}
> 42
> a
>
> Indeed -- the point to keep in mind is that locals() can become out of sy=
nc
> with the functions actual variables. =A0Definitely falls in the camp of "=
if
> you don't know *exactly* what you are doing, do not play this way!"

Sure, but that's not actually out of sync.  The argument of your exec
evaluates to 'print (a)'.  You get two different results because
you're actually printing two different variables.  You can get the
dict temporarily out of sync:

>>> def f(x, y):
...     frob =3D None
...     loc =3D locals()
...     loc[x] =3D y
...     print(loc)
...     print(locals())
...     print(loc)
...
>>> f('frob', 42)
{'y': 42, 'x': 'frob', 'frob': 42, 'loc': {...}}
{'y': 42, 'x': 'frob', 'frob': None, 'loc': {...}}
{'y': 42, 'x': 'frob', 'frob': None, 'loc': {...}}

In this case, 'frob' is updated to 42 in the dict, but the optimized
local is not updated.  Calling locals() again refreshes the dict.