Quick survey: locals in comprehensions (Python 3 only)
Steven D'Aprano
steve+comp.lang.python at pearwood.info
Tue Jun 26 06:39:15 EDT 2018
On Tue, 26 Jun 2018 12:04:16 +0200, Antoon Pardon wrote:
> On 26-06-18 11:22, Steven D'Aprano wrote:
>> On Tue, 26 Jun 2018 10:20:38 +0200, Antoon Pardon wrote:
>>
>>>> def test():
>>>> a = 1
>>>> b = 2
>>>> result = [value for key, value in locals().items()] return result
>> [...]
>>
>>> I would expect an UnboundLocalError: local variable 'result'
>>> referenced before assignment.
>> Well, I did say that there's no right or wrong answers, but that
>> surprises me. Which line do you expect to fail, and why do you think
>> "result" is unbound?
>
> I would expect the third statement to fail because IMO we call the
> locals function before result is bound. But result is a local variable
> so the locals function will try to reference it, hence the
> UnboundLocalError.
Ah, I see. Thanks for the explanation.
Given that locals() is capable of dealing with uninitialised local
variables without raising, would you like to revise your expectation?
--
Steven D'Aprano
"Ever since I learned about confirmation bias, I've been seeing
it everywhere." -- Jon Ronson
More information about the Python-list
mailing list