Python __class__()

Question

In Python's documentation __class__ is described as an attribute. In the object type (the metaclass), __class__ appears to be a method.

If we do:

>>> class Foo:
        pass

>>> a = Foo()
>>> a.__class__ == type.__class__(a)
True

So, my questions are:

1. When we call a.__class__, are we really calling the method type.__class__(a)?
2. Is this the reason why __class__ is not a member of the __dict__ attribute of a?

Box

Answer 1

__class__ is a data descriptor object. Many attributes on Python core objects are implemented as descriptors. You should see that as an implementation detail, nothing more.

__class__ is a descriptor because Python needs to be able to validate new values you assign to it; there are certain limitations to assigning to __class__ that need to be honoured, and making __class__ a descriptor is the most efficient method of doing so.

Descriptor objects are automatically found and invoked on the type when you try to access the attribute on an object. instance.__class__ will find and execute the __class__ descriptor on the class (by searching through all classes in the inheritance graph), typically ending up with object.__dict__['__class__'].__get__(instance, type(instance)) (where object is usually the first class on which the __class__ attribute is found in the type(instance).__mro__ sequence); This happens because Python will always use the type's __getattribute__ method to find attributes, and that method knows how to handle descriptors found on the class and bases, as well as look at the object.__dict__ attributes. So they don't live on the object __dict__ itself, they live on the object type, by design.

Now, class objects are also callable objects. That's how you create an instance; for a given class Foo, you create an instance by calling it, so Foo(). instance.__class__ is just a reference to the class object, just like class_obj = Foo would create a reference to a class. Calling the class object produces a new instance, whatever reference you used to get to it.

Finally, type.__class__ is just a reference to type() itself:

>>> type.__class__ is type
True
>>> type.__class__
<class 'type'>
>>> type(type)
<class 'type'>

That's because type is its own type. The parent-child relationships of the Python type system must stop somewhere, and type is that point.

In your example, a.__class__ is a reference to the Foo class. And type.__class__ is the same object as type, so you essentially did this:

Foo == type(a)

which is indeed true, the type of a is Foo.

Perloff

Thank you for your answer. Now, you said: instance.__class__ will find and execute type(instance).__dict__['__class__'].__get__(instance, type(instance)) but if do: type(instance).__dict__ there is not a __ class__ descriptor, which gets me to my first question, and assume that the descriptor __ class__ exist only in the object type and is passed to all their instance objects (all classes). ? – Box

@maxfraguas: sorry, I made things a little more confusing than I meant to. It's object.__dict__['__class__'] that is the descriptor object; all classes inherit from that. The whole class hierarchy is searched for such attributes (so next(t.__dict__['__class__'] for t in type(instance).__mro__ if '__class__' in t.__dict__).__get__(instance, type(instance)). – Perloff

Thanks @Matijn Pieters. After a little of playing around, I found __class__ inside the object's __dict__. – Box

Can this method be overriden? – Compony

@LexPodgorny: It's not a method, it's a data descriptor. Yes, you can create your own __class__ data descriptor, e.g. by using a property with a setter. – Perloff