Python throws UnboundLocalError when dealing with outer variables

Asked Oct 12 '21 at 09:07

Active Sep 09 '22 at 12:37

Viewed 27 times

player_name = ''


def main():
    if not player_name:
        player_name = input('Please input your name')
    else:
        input('Your name is {}'.format(player_name))


main()

and it throws:

UnboundLocalError: local variable 'player_name' referenced before assignment.

why can't main() function reference out variables? thanks!

edited Oct 12 '21 at 16:38

Python learner

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asked Oct 12 '21 at 09:07

Ymin Hu

you need `global player_name` because the parser sees the variable as local otherwise, because you're assigning to it – Jean-François Fabre Oct 12 '21 at 09:10
why do you need to define it before ? – AnonymousScientificUser Oct 12 '21 at 09:11
Does this answer your question? [Don't understand why UnboundLocalError occurs (closure)](https://stackoverflow.com/questions/9264763/dont-understand-why-unboundlocalerror-occurs-closure) – SamBob Oct 12 '21 at 09:11
@SamBob thanks, i think that answers my question. I was using this snippet in a bigger program which needs to manipulate global variables. – Ymin Hu Oct 12 '21 at 09:13

Python throws UnboundLocalError when dealing with outer variables

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