On Tue, Apr 21, 2009 at 9:13 PM, Chris Rebert <c...@rebertia.com> wrote:
> On Tue, Apr 21, 2009 at 9:08 AM, Doron Tal <doron.tal.l...@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
(Chris - sorry for the multiple replies, I forgot to reply all)
I actually did not attempt to modify this dictionary (the one which locals()
returns) explicitly.
The simplest assignment command, such as 'a = 2' modifies this dictionary
implicitly, and it's working perfectly well.
I expected exec to work the same, but apparently I was wrong. Is there is a
way to exec a file "more" correctly? thus avoid the need to resort to
awkward solutions such as using the locals() dictionary?
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