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 -- https://mail.python.org/mailman/listinfo/python-list
