Gary Wessle wrote: > Ryan Forsythe <[EMAIL PROTECTED]> writes: > > >>Gary Wessle wrote: >> >>>the example was an in-accuretlly representation of a the problem I am >>>having. my apologies. >>> (snip) > I finally was able to duplicate the error with a through away code > as follows, > > **************************************************************** > acc = [1,2,3] > > def a(): > b = [4, 5, 6] > acc = acc + b > print len(acc) > > a() > > **************** error **************** > Traceback (most recent call last): > File "a.py", line 12, in ? > a() > File "a.py", line 9, in a > acc = acc + b > UnboundLocalError: local variable 'acc' referenced before assignment
This is a FAQ: http://www.python.org/doc/faq/programming/#what-are-the-rules-for-local-and-global-variables-in-python For short: if a name is 'assigned' (in Python, the correct term is 'bound') in the local scope, it'll be considered a local name. If it's *only* accessed, it'll be looked up in the enclosing namespace - here the so-called 'global' (which means: 'module') namespace. The dirty solution is to declare 'acc' as global in the function: def a(): b = [4, 5, 6] global acc acc = acc + b print len(acc) but this is really BadCode(tm). As a general rule, functions should not silently modify or rebind global variables - this leads to maintenance nightmares. In this case, you should manage to either 1/ pass acc as a param to a(), or 2/ have a() return the sequence to be added to acc: # 1 acc = [1,2,3] def a(alist): alist.extend([4, 5, 6]) return alist acc = a(acc) print acc, len(acc) # 2 acc = [1,2,3] def a(): return [4, 5, 6] acc.extend(a()) print acc, len(acc) The Right Thing(tm) to do of course depends on the real code, so it may be yet another solution, but it's impossible to decide with dummy code... -- bruno desthuilliers python -c "print '@'.join(['.'.join([w[::-1] for w in p.split('.')]) for p in '[EMAIL PROTECTED]'.split('@')])" -- http://mail.python.org/mailman/listinfo/python-list