On 26 June 2018 at 11:09, Chris Angelico <ros...@gmail.com> wrote: > 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
Regardless of the answer to that question, the message here is basically "people don't have good intuition of how locals() works". Or to put it another way, "code that uses locals() is already something you should probably be checking the docs for if you care about the details of what it does". Which agrees with my immediate reaction when I saw the original question :-) Paul -- https://mail.python.org/mailman/listinfo/python-list
