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
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