Guido van Rossum writes:

...

...

I think you could subclass the metaclass, override __new__, and delete the bogus __getstate__ from the type's __dict__. Then you'll get the default pickling behavior which ignores slots; that should work just fine in your case. :-)

Ooh, that's sneaky! But I can't quite see how it works. The error message I quoted at the top about __getstate__ happens when you try to pickle an instance of the class. If I delete __getstate__ during __new__, it won't be there for pickle to find when I try to do the pickling. What will keep it from inducing the same error?

Just try it. There are many ways to customize pickling, and if __getstate__ doesn't exist, pickling is done differently.

Since this doesn't work: >>> d = type('d', (object,), { '__slots__' : ['foo'] } ) >>> pickle.dumps(d()) I'm still baffled as to why this works: >>> class mc(type): ... def __new__(self, *args): ... x = type.__new__(self, *args) ... del args[2]['__getstate__'] ... return x ... >>> c = mc('c', (object,), { '__slots__' : ['foo'], '__getstate__' : lambda self: tuple() } ) >>> pickle.dumps(c()) 'ccopy_reg\n_reconstructor\np0\n(c__main__\nc\np1\nc__builtin__\nobject\np2\nNtp3\nRp4\n.' especially since: >>> dir(d) == dir(c) 1 I don't see the logic in the source for object.__reduce__(), so where is it? OK, I see it in typeobject.c. But now: >>> c.__getstate__ OK, this seems to indicate that my attempt to remove __getstate__ from the class __dict__ was a failure. That explains why pickling c works, but not why you suggested that I remove __getstate__ inside of __new__. Did you mean for me to do something different? I note that c's __slots__ aren't pickled at all, which I guess was the point of the __getstate__ requirement: >>> x = c() >>> x.foo = 1 >>> pickle.dumps(x) == pickle.dumps(c()) 1 Fortunately, in our case the __slots__ are empty so it doesn't matter. -- Dave Abrahams Boost Consulting www.boost-consulting.com