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To: python-list@python.org
From: Terry Reedy <tjreedy@udel.edu>
Subject: Re: Python3 exec locals - this must be a FAQ
Date: Tue, 12 Feb 2013 10:50:11 -0500
References: <anuoc6F8kvaU1@mid.dfncis.de> <511A434D.4030309@davea.name>
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On 2/12/2013 8:27 AM, Dave Angel wrote:
> On 02/12/2013 06:46 AM, Helmut Jarausch wrote:

>> I've tried but didn't find an answer on the net.

You should start with the fine manual, which is on the net as well as 
included with at least the Windows distribution. It has a nice index 
that includes an entry for locals (built-in function).

>> The exec function in Python modifies a copy of locals() only.
>> How can I transfer changes in that local copy to the locals of my
>> function
>> ** without **  knowing the names of these variables.

You cannot. Just accept that.

>> E.g.  I have a lot of local names.
>>
>> Doing
>>
>> _locals= locals()

This merely gives you a handle of the dict returned by locals() for when 
you want to do more than just pass it to (for example) exec). This is 
unusual because it is not very useful.

> This doesn't copy everything.

I have no idea what you mean. The locals() dict contains all local and 
nonlocal names, excluding global names, which are in the globals() dict.

>> expr=compile(input('statements assigning to some local variables '),
>>                                                   'user input','exec')
>> exec(expr,globals(),_locals)
>>
>> How can I "copy" the new values within _locals to my current locals.
>>
>> If I knew that  Var1  has changed I could say
>> Var1 = _locals['Var1'] but what to do in general?

If you want to put a value back into the function local namespace, this 
is the only thing you can do. In CPython, at least, all function local 
names must be explicit and known when the function statement is executed 
and the function object is created. Read the Library manual entry for 
locals(), including the highlighted note.

> locals()["Var1"] = _locals["Var1"]  will set the same Var1 local.

Huh??? The dict returned by this locals call is the same dict returned 
by the previous locals call and bound to _locas. So the above does 
nothing. It is the same thing as _locals["Var1"] = _locals["Var1"].

-- 
Terry Jan Reedy