ast wrote:
> 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
>>> class T(type):
... def __new__(*args, **kw): return type.__new__(*args)
... def __prepare__(*args, **kw):
... print(kw)
... return {}
... def __init__(*args, **kw):
... pass
...
>>> class A(metaclass=T, answer=42):
... pass
...
{'answer': 42}
Adapted from
<http://martyalchin.com/2011/jan/20/class-level-keyword-arguments/>.
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