On 26-06-18 12:39, Steven D'Aprano wrote: > 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?
Sure, I would now expect any |permutation of either a) [1, 2] or b) [1, 2, ||UnboundLocalError|] -- Antoon Pardon. |||| -- https://mail.python.org/mailman/listinfo/python-list
