execfile (exec?) create non consistent locals() state

Chris Rebert clp2 at rebertia.com
Tue Apr 21 20:13:32 CEST 2009


On Tue, Apr 21, 2009 at 9:08 AM, Doron Tal <doron.tal.list at gmail.com> wrote:
> Hi,
>
> Recently I tried to execute a python file using execfile (exec performed
> just the same for that reason).
> I encountered the behavior below:
>
> """
> $ cat execme.py
> a = 2
> $ python
> Python 2.4.3 (#1, May 24 2008, 13:57:05)
> [GCC 4.1.2 20070626 (Red Hat 4.1.2-14)] on linux2
> Type "help", "copyright", "credits" or "license" for more information.
>>>> def execfile_func():
> ...     execfile('execme.py')
> ...     print 'locals() = %s' % str(locals())
> ...     print a
> ...
>>>> execfile_func()
> locals() = {'a': 2}
> Traceback (most recent call last):
>   File "<stdin>", line 1, in ?
>   File "<stdin>", line 4, in execfile_func
> NameError: global name 'a' is not defined
>>>>
> """
>
> After execfile, the a variable can be found in locals(), however any direct
> reference (e.g., print a) fails.
> Is it expected?

Yes. See http://docs.python.org/library/functions.html#locals (emphasis mine):

locals()
    [...]
    Warning: The contents of this dictionary should not be modified;
***changes may not affect the values of local variables used by the
interpreter***.
    [...]

Cheers,
Chris
-- 
I have a blog:
http://blog.rebertia.com



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