On Sun, Oct 14, 2012 at 7:08 PM, Cameron Simpson <c...@zip.com.au> wrote: > On 14Oct2012 18:32, Ian Kelly <ian.g.ke...@gmail.com> wrote: > | 'attr_name' is not in locals because while it's a local variable, it > | has not been assigned to yet. It has no value and an attempt to > | reference it at that point would result in an UnboundLocalError. > > Can you elaborate a bit on that? The only place in my code that > unset_object is set is as a default parameter in make_file_property > (snippet): > > def make_file_property(attr_name=None, unset_object=None, poll_rate=1): > print >>sys.stderr, "make_file_property(attr_name=%r, unset_object=%r, > poll_rate=%r): locals()=%r" % (attr_name, unset_object, poll_rate,locals()) > def made_file_property(func): > print >>sys.stderr, "made_file_property(func=%r): locals()=%r" % (func, > locals()) > if attr_name is None: > attr_name = '_' + func.__name__ > > and attr_name is set there also. > > Is attr_name omitted from locals() in made_file_property _because_ I > have an assignment statement?
Yes. Syntactically, a variable is treated as local to a function if it is assigned to somewhere in that function and there is no explicit global or nonlocal declaration. > If that's the case, should I be doing this (using distinct names for the > closure variable and the function local variable): > > def make_file_property(attr_name=None, unset_object=None, poll_rate=1): > print >>sys.stderr, "make_file_property(attr_name=%r, unset_object=%r, > poll_rate=%r): locals()=%r" % (attr_name, unset_object, poll_rate,locals()) > def made_file_property(func): > print >>sys.stderr, "made_file_property(func=%r): locals()=%r" % (func, > locals()) > if attr_name is None: > my_attr_name = '_' + func.__name__ > else: > my_attr_name = attr_name > lock_name = my_attr_name + '_lock' > def getprop(self): > with getattr(self, lock_name): > pass > return getattr(self, my_attr_name, unset_object) > > i.e. deliberately _not_ assigning to attr_name as as to _avoid_ masking > the outer attr_name from the inner locals()? > > BTW, doing that works. Is that The True Path for this situation? That's a perfectly good way to do it as long as you don't want to actually change the value of the outer attr_name. If you did, then you would either declare the variable as nonlocal (Python 3.x only) or use a container (e.g. a 1-element list), which would allow you to modify the contents of the container without actually assigning to the variable. > If so, I think I now understand what's going on: Python has inspected > the inner function and not placed the outer 'attr_name' into locals() > _because_ the inner function seems to have its own local attr_name > in use, which should not be pre-tromped. Exactly right. -- http://mail.python.org/mailman/listinfo/python-list
