Steven D'Aprano <steve+pyt...@pearwood.info> added the comment:
The behaviour and error message is correct, and your interpretation is incorrect. You are not assigning to a closure variable on line 4; you are printing an unbound local variable on line 3, precisely as the error message says. That may not match your *intention*, but Python doesn't do what we want it to do, only what we tell it to do :-) Like global variables, nonlocals (closures) are NOT defined by "does this name match a name in the surrounding scope?". Rather, *local* variables are defined by assignment: any assignment in the body of the function defines a local, unless otherwise declared as nonlocal or global. > Instead, the error should point to line 4 and report an illegal assignment to > a read-only closure variable. That can't happen without a huge and backwards-incompatible change to Python semantics. (Possibly in Python 5000?) The only reason I'm not closing this as "Not a bug" is in case someone would like to suggest an improvement to error message. Perhaps: UnboundLocalError: local variable 'a' referenced before assignment -- did you forget a nonlocal or global declaration? ---------- nosy: +steven.daprano _______________________________________ Python tracker <rep...@bugs.python.org> <https://bugs.python.org/issue37568> _______________________________________ _______________________________________________ Python-bugs-list mailing list Unsubscribe: https://mail.python.org/mailman/options/python-bugs-list/archive%40mail-archive.com
