Question about name scope

[Ethan Furman]

Ethan Furman wrote:
> Ethan Furman wrote:
>> Ian Kelly wrote:
>>> I am not a dev, but I believe it works because assigning to locals()
>>> and assigning via exec are not the same thing.  The problem with
>>> assigning to locals() is that you're fundamentally just setting a
>>> value in a dictionary, and even though it happens to be the locals
>>> dict for the stack frame, Python can't figure out that it should go
>>> and update the value of the optimized local to match.  exec, on the
>>> other hand, compiles and executes an actual STORE_NAME operation.  Of
>>> course, if the particular local variable hasn't been optimized by the
>>> compiler, then updating locals() works just fine (although you
>>> probably should not rely on this):
>>>
>>>>>> def f(x, y):
>>> ...     locals()[x] = y
>>> ...     print locals()[x]
>>> ...     exec 'print ' + x
>>> ...
>>>>>> f('a', 42)
>>> 42
>>> 42
>>
>> Definitely should rely on it, because in CPython 3 exec does not 
>> un-optimize the function and assigning to locals() will not actually 
>> change the functions variables.
> 
> 
> Ouch, that should have been *not* rely on it; not because it doesn't 
> work (exec uses locals() if one is not specified), but because it is 
> easy for the names in the function to get out of sync with the names in 
> the functions locals() (or __dict__).

I should stop answering now  :(  Ignore the __dict__ comment, it is 
incorrect.

~Ethan~