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On 02/12/2013 10:50 AM, Terry Reedy wrote:
> On 2/12/2013 8:27 AM, Dave Angel wrote:
>> On 02/12/2013 06:46 AM, Helmut Jarausch wrote:
>
>>> <snip>
>>>
>>> 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.

The OP presumably wanted to restore the original values of the original
variables. The above "assignment" won't help a bit with that.


>
> 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"].
>

My claim was based on the assumption that the earlier assignment had
been fixed by some kind of copy. If not, there's nothing to restore.

I also retracted my use of that trick anyway, since being corrected by
MRAB. I only tested it in top-level code, not inside a function.


--
DaveA