Martin v. Löwis <mar...@v.loewis.de> added the comment: As David says, this is not a bug. del l indicates that there is a local variable to be deleled, but when the del statement is executed, there is no local variable. The error message is confusing in this case: there actually is no later assignment to l (in the function at all). Typically, when you have an unbound local, it is because of a later assignment, such as
def foo(): a = l + 1 l = 2 In this specific example, there is no later assignment - yet it is still an unbound local. So that you get the exception is not a bug. I was going to suggest that the error message could be better, but I can't think of any other error message that is better and still correct, hence closing it as won't fix. ---------- nosy: +loewis resolution: -> wont fix status: open -> closed _______________________________________ Python tracker <rep...@bugs.python.org> <http://bugs.python.org/issue5092> _______________________________________ _______________________________________________ Python-bugs-list mailing list Unsubscribe: http://mail.python.org/mailman/options/python-bugs-list/archive%40mail-archive.com
