[Django]-Wow does six.with_metaclass() work?

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Ok – I think I figured it out. The crux of the matter lies in the 

return meta(name, bases, d)

of the with_metaclass function:

def with_metaclass(meta, *bases):
    """Create a base class with a metaclass."""
    # This requires a bit of explanation: the basic idea is to make a dummy
    # metaclass for one level of class instantiation that replaces itself with
    # the actual metaclass.
    class metaclass(meta):

        def __new__(cls, name, this_bases, d):
            return meta(name, bases, d)
    return type.__new__(metaclass, 'temporary_class', (), {})

Here is how I think it works in sudo code:

(1) with_metaclass takes <<DeclarativeSubBlocksMetaclass>> as meta; and <<BaseStreamBlock>> as bases
(2) class metaclass(meta) --> the class metaclass is then created extending <<DeclarativeSubBlockMetaclass>> as the class type
(3) def __new__(cls, name, this_bases, d): Only rarely will you have to worry about __new__. Usually, you'll just define __init__ and let the default __new__ pass the constructor arguments to it. __new__ takes care of creating the object and assigning memory space to it. This __new__ method is a class method that gets called when you create an instance of the class and it gets called before __init__.  Its main job is to allocate the memory that the object that you are creating uses. It can also be used to set up any aspect of the instance of the class that is immutable Because classes are kind of immutable (they cannot be changed), overloading __new_ is the best place to overload how they are created.
(4) return meta(name, bases, d) --> the class definition ends with returning a <<DeclarativeSubBlockMetaclass>> with the arguments (name, base = BaseStreamBlock, d)

NOTE: We only define the class in 1 - 3; we are not instantiating it this comes below

(5) return type.__new__(metaclass, 'temporary_class', (), {}) --> Here we are using the classic metaclass syntax. This syntax usually looks like this: return type.__new__(cls, name, bases, attrs). We are using this syntax to instantiate the metaclass we defined in (3) and (4). One might think that it is confusing that temporary_class', (), {} are passed on as the 'name', 'bases', and 'attrs' arguments. BUT...
(6) ... when the instantiation arrives at return meta(name,bases,d) we notice that meta doesn't take 'this_bases' as an argument but 'bases'. It derives this value from the arguments which were passed to (1) with_metaclasses. As such bases in this instance == <<BaseStreamBlock>>
(7) Therefore, when we instantiate type.__new__(metaclass, 'temporary_class', (), {}) we essentially execute <<DeclarativeSubBlocksMetaClass>>('temporary_class', <<BaseStreamBlock>>, {})

The step explained in (7) is what the explanation talked about. Essentially what SIX does is to go through the prescribed steps to create a dummy metaclass which it calls temporary_class. Since DeclarativeSubBlocksMetaClass is also a metaclass, it then uses the BaseStreamBlock base to generate a new class.

I hope this makes sense.

Z

👤Mike Nedelko

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