Mel a écrit : (snip) > (What Diez said.) From what I've seen, f.bar creates a bound method > object by taking the unbound method Foo.bar and binding its first > parameter with f.
Nope. it's Foo.__dict__['bar'] (that is, the function bar defined in the namespace of class Foo) that creates a bound instancemethod object when looked up on a Foo instance - or an unbound instancemethod object when looked up on Foo. FWIW, types.UnboundMethodType and types.MethodType are both aliases to instancemethod type. -- http://mail.python.org/mailman/listinfo/python-list
