On Tue, Jun 26, 2018 at 8:04 PM, Antoon Pardon <antoon.par...@vub.be> 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.
Would you expect the same behaviour from this function? def test(): a = 1 b = 2 result = locals() return result ChrisA -- https://mail.python.org/mailman/listinfo/python-list