On Mar 20, 5:24 pm, Duncan Booth <duncan.bo...@invalid.invalid> wrote: > Joaquin Abian <gatoyga...@gmail.com> wrote: > > "User-defined method objects may be created when getting an attribute > > of a class (perhaps via an instance of that class), if that attribute > > is a user-defined function object, an unbound user-defined method > > object, or a class method object. When the attribute is a user-defined > > method object, a new method object is only created if the class from > > which it is being retrieved is the same as, or a derived class of, the > > class stored in the original method object; otherwise, the original > > method object is used as it is." > > > It is a bit of a tongue-twister for me. What the last sentence means? > > Please, I beg for a simple example of the different objects (user > > defined function, user defined method, class method) refered. > >>> class A(object): > > def foo(self): pass > > >>> class B(object): > > def bar(self): pass > > >>> B.baz = B.bar > >>> B.foo = A.foo > >>> instance = B() > >>> B.foo > > <unbound method A.foo>>>> B.bar > > <unbound method B.bar>>>> B.baz > > <unbound method B.bar>>>> instance.foo > > <unbound method A.foo>>>> instance.bar > > <bound method B.bar of <__main__.B object at 0x00000000036B7780>>>>> > instance.baz > > <bound method B.bar of <__main__.B object at 0x00000000036B7780>>>>> > B.__dict__['bar'] > > <function bar at 0x00000000036C5048>>>> B.__dict__['baz'] > > <unbound method B.bar>>>> B.__dict__['foo'] > > <unbound method A.foo> > > So, we have a function 'bar' stored in B's dict. When you access the > function as the attribute B.bar Python 2.x will create an unbound method > object. When you access the function through instance.bar Python creates a > bound method. Note that every time you access instance.bar it creates > another new method object: > > >>> instance.bar is instance.bar > > False > > B.baz is an unbound method stored directly in B's dict. When you access > instance.baz you get a new bound method object (it's at the same memory > location as the previous one but that's only because the lifetimes don't > overlap). Somewhat suprisingly the same also happens if you access B.baz: > it creates a new unbound method from the existing unbound method: > > >>> B.bar is B.__dict__['bar'] > > False > > B.foo is an unbound method stored directly in B's dict, but it is a method > of an A and B doesn't subclass A, so when you try to access B.foo you just > get the stored unbound method it isn't converted into a new object.
Thanks Duncan, I think I got the idea. Your explanation together with the comment from Terry about the absence of unbound method objects in python 3 cleared some dust from my neuron(s) Thanks! JA -- http://mail.python.org/mailman/listinfo/python-list
