__prepare__ metaclass's method

[ast]

Hi

On python doc here: 

https://docs.python.org/3.4/reference/datamodel.html

it is said about __prepare__ metaclass's method:

If the metaclass has a __prepare__ attribute, it is called as 
namespace = metaclass.__prepare__(name, bases, **kwds) 
where the additional keyword arguments, if any, come from
the class definition.

I don't understand what they call the "class definition".

So I took their example and add a print(kwds)

class OrderedClass(type):

     @classmethod
     def __prepare__(metacls, name, bases, **kwds):
        print(kwds)
        return collections.OrderedDict()

     def __new__(cls, name, bases, namespace, **kwds):
        result = type.__new__(cls, name, bases, dict(namespace))
        result.members = tuple(namespace)
        return result

class A(metaclass=OrderedClass):
    def one(self): pass
    def two(self): pass
    def three(self): pass
    def four(self): pass

but print(kwds) outputs an empty dictionnary {}

So what kwds is supposed to contains ?

Thx