xmlrpc filebug API raises TypeError exception

If you don't specify any arguments to the xmlrpc API, you get some strange behaviour:

>>> from xmlrpclib import ServerProxy
>>> s = ServerProxy("https://xmlrpc.launchpad.net/malone/")
>>> s.filebug()
Traceback (most recent call last):
  File "<stdin>", line 1, in ?
  File "/System/Library/Frameworks/Python.framework/Versions/2.3/lib/python2.3/xmlrpclib.py", line 1032, in __call__
  File "/System/Library/Frameworks/Python.framework/Versions/2.3/lib/python2.3/xmlrpclib.py", line 1319, in __request
  File "/System/Library/Frameworks/Python.framework/Versions/2.3/lib/python2.3/xmlrpclib.py", line 1083, in request
  File "/System/Library/Frameworks/Python.framework/Versions/2.3/lib/python2.3/xmlrpclib.py", line 1222, in _parse_response
  File "/System/Library/Frameworks/Python.framework/Versions/2.3/lib/python2.3/xmlrpclib.py", line 745, in close
xmlrpclib.Fault: <Fault -1: 'Unexpected Zope exception: TypeError: Missing argument to filebug(): params'>

This should instead first raise an Unauthorized exception, because the user didn't even log in! BjornT thinks this may be a Zope 3 issue.