Metaclass Mixin或Chaining？

Question

是否可以连接元类？

我有类Model,它使用__metaclass__=ModelBase来处理其命名空间字典.我将继承它并"绑定"另一个元类,因此它不会遮蔽原始元类.

第一种方法是子类class MyModelBase(ModelBase):

MyModel(Model):
    __metaclass__ = MyModelBase # inherits from `ModelBase`

但是,如果没有明确的子类化,是否可以像mixin一样链接它们？就像是

class MyModel(Model):
    __metaclass__ = (MyMixin, super(Model).__metaclass__)

......甚至更好:创建一个将使用__metaclass__它的类的直接父级使用的MixIn :

class MyModel(Model):
    __metaclass__ = MyMetaMixin, # Automagically uses `Model.__metaclass__`

---

原因是:为了更灵活地扩展现有应用程序,我想创建一个全局机制,用于挂钩Django中的,...定义过程Model,Form以便在运行时更改它.

一种常见的机制比使用回调mixin实现多个元类要好得多.

---

在你的帮助下,我终于找到了解决方案:元类MetaProxy.

这个想法是:创建一个元类,调用一个回调来修改正在创建的类的命名空间,然后,在其帮助下__new__,变异为其中一个父类的元类.

#!/usr/bin/env python
#-*- coding: utf-8 -*-

# Magical metaclass
class MetaProxy(type):
    """ Decorate the class being created & preserve __metaclass__ of the parent

        It executes two callbacks: before & after creation of a class, 
        that allows you to decorate them.

        Between two callbacks, it tries to locate any `__metaclass__` 
        in the parents (sorted in MRO). 
        If found — with the help of `__new__` method it
        mutates to the found base metaclass. 
        If not found — it just instantiates the given class.
        """

    @classmethod
    def pre_new(cls, name, bases, attrs):
        """ Decorate a class before creation """
        return (name, bases, attrs)

    @classmethod
    def post_new(cls, newclass):
        """ Decorate a class after creation """
        return newclass

    @classmethod
    def _mrobases(cls, bases):
        """ Expand tuple of base-classes ``bases`` in MRO """
        mrobases = []
        for base in bases:
            if base is not None: # We don't like `None` :)
                mrobases.extend(base.mro())
        return mrobases

    @classmethod
    def _find_parent_metaclass(cls, mrobases):
        """ Find any __metaclass__ callable in ``mrobases`` """
        for base in mrobases:
            if hasattr(base, '__metaclass__'):
                metacls = base.__metaclass__
                if metacls and not issubclass(metacls, cls): # don't call self again
                    return metacls#(name, bases, attrs)
        # Not found: use `type`
        return lambda name,bases,attrs: type.__new__(type, name, bases, attrs)

    def __new__(cls, name, bases, attrs):
        mrobases = cls._mrobases(bases)
        name, bases, attrs = cls.pre_new(name, bases, attrs) # Decorate, pre-creation
        newclass = cls._find_parent_metaclass(mrobases)(name, bases, attrs)
        return cls.post_new(newclass) # Decorate, post-creation



# Testing
if __name__ == '__main__':
    # Original classes. We won't touch them
    class ModelMeta(type):
        def __new__(cls, name, bases, attrs):
            attrs['parentmeta'] = name
            return super(ModelMeta, cls).__new__(cls, name, bases, attrs)

    class Model(object):
        __metaclass__ = ModelMeta
        # Try to subclass me but don't forget about `ModelMeta`

    # Decorator metaclass
    class MyMeta(MetaProxy):
        """ Decorate a class

            Being a subclass of `MetaProxyDecorator`,
                it will call base metaclasses after decorating
            """
        @classmethod
        def pre_new(cls, name, bases, attrs):
            """ Set `washere` to classname """
            attrs['washere'] = name
            return super(MyMeta, cls).pre_new(name, bases, attrs)

        @classmethod
        def post_new(cls, newclass):
            """ Append '!' to `.washere` """
            newclass.washere += '!'
            return super(MyMeta, cls).post_new(newclass)

    # Here goes the inheritance...
    class MyModel(Model):
        __metaclass__ = MyMeta
        a=1
    class MyNewModel(MyModel):
        __metaclass__ = MyMeta # Still have to declare it: __metaclass__ do not inherit
        a=2
    class MyNewNewModel(MyNewModel):
        # Will use the original ModelMeta
        a=3

    class A(object):
        __metaclass__ = MyMeta # No __metaclass__ in parents: just instantiate
        a=4
    class B(A): 
        pass # MyMeta is not called until specified explicitly



    # Make sure we did everything right
    assert MyModel.a == 1
    assert MyNewModel.a == 2
    assert MyNewNewModel.a == 3
    assert A.a == 4

    # Make sure callback() worked
    assert hasattr(MyModel, 'washere')
    assert hasattr(MyNewModel, 'washere')
    assert hasattr(MyNewNewModel, 'washere') # inherited
    assert hasattr(A, 'washere')

    assert MyModel.washere == 'MyModel!'
    assert MyNewModel.washere == 'MyNewModel!'
    assert MyNewNewModel.washere == 'MyNewModel!' # inherited, so unchanged
    assert A.washere == 'A!'

Answer 1

一个类型只能有一个元类,因为元类只是简单地说明了类语句的作用 - 拥有多个元类是没有意义的.出于同样的原因,"链接"毫无意义:第一个元类创建了类型,那么第二个应该做什么？

您将必须合并两个元类(就像与任何其他类一样).但这可能很棘手,特别是如果你真的不知道他们做了什么.

class MyModelBase(type):
    def __new__(cls, name, bases, attr):
        attr['MyModelBase'] = 'was here'
        return type.__new__(cls,name, bases, attr)

class MyMixin(type):
    def __new__(cls, name, bases, attr):
        attr['MyMixin'] = 'was here'
        return type.__new__(cls, name, bases, attr)

class ChainedMeta(MyModelBase, MyMixin):
    def __init__(cls, name, bases, attr):
        # call both parents
        MyModelBase.__init__(cls,name, bases, attr)
        MyMixin.__init__(cls,name, bases, attr)

    def __new__(cls, name, bases, attr):
        # so, how is the new type supposed to look?
        # maybe create the first
        t1 = MyModelBase.__new__(cls, name, bases, attr)
        # and pass it's data on to the next?
        name = t1.__name__
        bases = tuple(t1.mro())
        attr = t1.__dict__.copy()
        t2 = MyMixin.__new__(cls, name, bases, attr)
        return t2

class Model(object):
    __metaclass__ = MyModelBase # inherits from `ModelBase`

class MyModel(Model):
    __metaclass__ = ChainedMeta

print MyModel.MyModelBase
print MyModel.MyMixin

正如你所看到的,这已经涉及一些猜测,因为你并不真正知道其他元类的功能.如果两个元类都非常简单,那么这可能会有效,但我不会对这样的解决方案有太多的信心.

为合并多个基础的元类编写元类是留给读者的练习;-P

Answer 2

我不认为你可以像那样链接它们，我也不知道这会如何运作。

但是您可以在运行时创建新的元类并使用它们。但这是一个可怕的黑客。:)

zope.interface 做了类似的事情，它有一个顾问元类，它只会在构建后对类做一些事情。如果已经有一个元类，它会做的一件事是在它完成后将先前的元类设置为元类。

（但是，除非您必须这样做，或者认为这很有趣，否则请避免做这些事情。）

Answer 3

我不知道"混合"元类的任何方法,但你可以像普通类一样继承和覆盖它们.

说我有一个BaseModel:

class BaseModel(object):
    __metaclass__ = Blah

现在你想在一个名为MyModel的新类中继承它,但是你想在元类中插入一些额外的功能,但是原来的功能保持不变.要做到这一点,你会做类似的事情:

class MyModelMetaClass(BaseModel.__metaclass__):
    def __init__(cls, *args, **kwargs):
        do_custom_stuff()
        super(MyModelMetaClass, cls).__init__(*args, **kwargs)
        do_more_custom_stuff()

class MyModel(BaseModel):
    __metaclass__ = MyModelMetaClass