1 class 继承了 2 个不同的 metaclasses(abcmeta 和用户定义的 meta)

1 class inherits 2 different metaclasses (abcmeta and user defined meta)

我有一个 class1 需要从 2 个不同的 metaclasses 继承,即 Meta1 和 abc.ABCMeta

当前实施:

Meta1 的实现:

class Meta1(type):
    def __new__(cls, classname, parent, attr):
        new_class = type.__new__(cls, classname, parent, attr)
        return super(Meta1, cls).__new__(cls, classname, parent, attr)

实施class1摘要

class class1Abstract(object):
    __metaclass__ = Meta1
    __metaclass__ = abc.ABCMeta

主要实施class

class mainClass(class1Abstract):
    # do abstract method stuff

我知道将 2 个不同的元实现两次是错误的。

我改变了 metclass 的加载方式(尝试了几次),我得到了这个 TypeError: 调用 metaclass bases

时出错
metaclass conflict: the metaclass of a derived class must be a (non-strict) subclass of the metaclasses of all its bases

我运行没主意了...


EDITED 1

我试过这个解决方案,但 mainClass 不是 class1Abstract

的实例
print issubclass(mainClass, class1Abstract) # true
print isinstance(mainClass, class1Abstract) # false

实施class1摘要

class TestMeta(Meta1):
    pass


class AbcMeta(object):
    __metaclass__ = abc.ABCMeta
    pass


class CombineMeta(AbcMeta, TestMeta):
    pass


class class1Abstract(object):
    __metaclass__ = CombineMeta

    @abc.abstractmethod
    def do_shared_stuff(self):
        pass

    @abc.abstractmethod
    def test_method(self):
        ''' test method '''

mainClass 的实现

class mainClass(class1Abstract):
    def do_shared_stuff(self):
        print issubclass(mainClass, class1Abstract) # True
        print isinstance(mainClass, class1Abstract) # False

因为 mainClass 继承自抽象 class python 应该抱怨 test_method 没有在 mainClass 中实现。但它不会抱怨任何事情,因为 print isinstance(mainClass, class1Abstract) # False

dir(mainClass) 没有

['__abstractmethods__', '_abc_cache', '_abc_negative_cache', '_abc_negative_cache_version', '_abc_registry']

求助!


EDITED 2

实施class1摘要

CombineMeta = type("CombineMeta", (abc.ABCMeta, Meta1), {})
class class1Abstract(object):
    __metaclass__ = abc.ABCMeta

    @abc.abstractmethod
    def do_shared_stuff(self):
        pass

    @abc.abstractmethod
    def test_method(self):
        ''' test method '''

mainClass 的实现

class mainClass(class1Abstract):
    __metaclass__ = CombineMeta
    def do_shared_stuff(self):
        print issubclass(mainClass, class1Abstract) # True
        print isinstance(mainClass, class1Abstract) # False

dir(mainClass) 现在有 abstractmethod 的魔术方法

['__abstractmethods__', '__class__', '__delattr__', '__dict__', '__doc__', '__format__', '__getattribute__', '__hash__', '__init__', '__metaclass__', '__module__', '__new__', '__reduce__', '__reduce_ex__', '__repr__', '__setattr__', '__sizeof__', '__str__', '__subclasshook__', '__weakref__', '_abc_cache', '_abc_negative_cache', '_abc_negative_cache_version', '_abc_registry', 'do_shared_stuff', 'test_method']

但是 python 不会警告 test_method 没有被实例化

求助!

无需设置两个元类:Meta1 应继承自 abc.ABCMeta

默认情况下 python 仅在您尝试实例化 class 时抱怨 class 具有抽象方法,而不是在您创建 class 时抱怨。这是因为 class 的 metaclass 仍然是 ABCMeta(或其子类型),所以它允许有抽象方法。

要获得您想要的内容,您需要编写自己的元class,当它发现 __abstractmethods__ 不为空时会引发错误。这样,当不再允许使用 class 抽象方法时,您必须明确说明。

from abc import ABCMeta, abstractmethod

class YourMeta(type):
    def __init__(self, *args, **kwargs):
        super(YourMeta, self).__init__(*args, **kwargs)
        print "YourMeta.__init__"
    def __new__(cls, *args, **kwargs):
        newcls = super(YourMeta, cls).__new__(cls, *args, **kwargs)
        print "YourMeta.__new__"
        return newcls

class ConcreteClassMeta(ABCMeta):
    def __init__(self, *args, **kwargs):
        super(ConcreteClassMeta, self).__init__(*args, **kwargs)
        if self.__abstractmethods__:
            raise TypeError("{} has not implemented abstract methods {}".format(
                self.__name__, ", ".join(self.__abstractmethods__)))

class CombinedMeta(ConcreteClassMeta, YourMeta):
    pass

class AbstractBase(object):
    __metaclass__ = ABCMeta

    @abstractmethod
    def f(self):
        raise NotImplemented

try:
    class ConcreteClass(AbstractBase):
        __metaclass__ = CombinedMeta

except TypeError as e:
    print "Couldn't create class --", e

class ConcreteClass(AbstractBase):
    __metaclass__ = CombinedMeta
    def f(self):
        print "ConcreteClass.f"

assert hasattr(ConcreteClass, "__abstractmethods__")
c = ConcreteClass()
c.f()

输出:

YourMeta.__new__
YourMeta.__init__
Couldn't create class -- ConcreteClass has not implemented abstract methods f
YourMeta.__new__
YourMeta.__init__
ConcreteClass.f

在Python中,每个class只能有一个metaclass,不多。但是,可以通过混合这些元classes do 来实现类似的行为(比如它有多个元classes)。

让我们从简单的开始。我们自己的 metaclass,只是将新属性添加到 class:

class SampleMetaClass(type):
  """Sample metaclass: adds `sample` attribute to the class"""
  def __new__(cls, clsname, bases, dct):
    dct['sample'] = 'this a sample class attribute'
    return super(SampleMetaClass, cls).__new__(cls, clsname, bases, dct)

class MyClass(object):
  __metaclass__ = SampleMetaClass

print("SampleMetaClass was mixed in!" if 'sample' in MyClass.__dict__ else "We've had a problem here")

这会打印 "SampleMetaClass was mixed in!",所以我们知道我们的基本元 class 工作正常。

现在,另一方面,我们想要一个抽象 class,最简单的是:

from abc import ABCMeta, abstractmethod

class AbstractClass(object):
  __metaclass__ = ABCMeta
  @abstractmethod
  def implement_me(self):
    pass

class IncompleteImplementor(AbstractClass):
  pass

class MainClass(AbstractClass):
  def implement_me(self):
    return "correct implementation in `MainClass`"

try:
  IncompleteImplementor()
except TypeError as terr:
  print("missing implementation in `IncompleteImplementor`")

MainClass().implement_me()

这将打印 "missing implementation in IncompleteImplementor",然后是 "correct implementation in MainClass"。因此,摘要 class 也可以正常工作。

现在,我们有 2 个简单的实现,我们需要将两个元classes 的行为混合在一起。这里有多个选项。

选项 1 - subclassing

可以将 SampleMetaClass 实现为 ABCMeta 的子 class - metaclasses 也是 classes 并且可以继承它们!

class SampleMetaABC(ABCMeta):
  """Same as SampleMetaClass, but also inherits ABCMeta behaviour"""
  def __new__(cls, clsname, bases, dct):
    dct['sample'] = 'this a sample class attribute'
    return super(SampleMetaABC, cls).__new__(cls, clsname, bases, dct)

现在,我们更改 AbstractClass 定义中的元class:

class AbstractClass(object):
  __metaclass__ = SampleMetaABC
  @abstractmethod
  def implement_me(self):
    pass

# IncompleteImplementor and MainClass implementation is the same, but make sure to redeclare them if you use same interpreter from the previous test

和运行我们再次测试:

try:
  IncompleteImplementor()
except TypeError as terr:
  print("missing implementation in `IncompleteImplementor`")

MainClass().implement_me()

print("sample was added!" if 'sample' in IncompleteImplementor.__dict__ else "We've had a problem here")
print("sample was added!" if 'sample' in MainClass.__dict__ else "We've had a problem here")

这仍然会打印出 IncompleteImplementor 未正确实现,MainClass 是,并且两者现在都添加了 sample class 级属性。这里要注意的是,元 class 的 Sample 部分也成功应用于 IncompleteImplementor(好吧,没有理由不这样做)。

正如预期的那样,isinstanceissubclass 仍然按预期工作:

print(issubclass(MainClass, AbstractClass)) # True, inheriting from AbtractClass
print(isinstance(MainClass, AbstractClass)) # False as expected - AbstractClass is a base class, not a metaclass
print(isinstance(MainClass(), AbstractClass)) # True, now created an instance here

选项 2 - 编写元classes

其实问题本身就有这个选项,只需要稍微修改一下就可以了。将新元class 声明为几个更简单的元class 的组合以混合它们的行为:

SampleMetaWithAbcMixin = type('SampleMetaWithAbcMixin', (ABCMeta, SampleMetaClass), {})

如前所述,为 AbstractClass 更改 metaclass(同样,IncompleteImplementorMainClass 不会更改,但如果在同一个解释器中则重新声明它们):

class AbstractClass(object):
  __metaclass__ = SampleMetaWithAbcMixin
  @abstractmethod
  def implement_me(self):
    pass

从这里开始,运行相同的测试应该会产生相同的结果:ABCMeta 仍然有效并确保实现 @abstractmethod-s,SampleMetaClass 添加一个 sample 属性。

我个人更喜欢后一种选择,原因与我通常更喜欢组合而不是继承的原因相同:多个(元)classes 之间最终需要的组合越多 - 组合越简单.

有关元的更多信息classes

最后,我读过的对 metaclasses 最好的解释是这个 SO 答案: What is a metaclass in Python?

在您编辑的代码(1 和 2)中,您几乎完成了。唯一错误的是你如何使用 isinstance。您想要检查 class 实例(在本例中为 self)是否是给定 class(class1Abstract)的实例。例如:

class mainClass(class1Abstract):
def do_shared_stuff(self):
    print issubclass(mainClass, class1Abstract) # True
    print isinstance(self, class1Abstract) # True