我想重新定义一个 __metaclass__ 但我想回退到如果我没有重新定义就会使用的元类。
class ComponentMetaClass(type):
def __new__(cls, name, bases, dct):
return <insert_prev_here>.__new__(cls, name, bases, dct)
class Component(OtherObjects):
__metaclass__ = ComponentMetaClass
据我了解,默认使用的 __metaclass__ 经历了在类范围内检查定义的过程,然后是基础类,然后是全局类。通常您会在重新定义中使用类型,并且通常是全局类型,但是,我的 OtherObjects 可能已经重新定义了 __metaclass__。所以在使用类型时,我会忽略它们的定义,它们不会运行,对吗?
编辑:请注意,直到运行时我才知道 OtherObjects 是什么
最佳答案
正如@unutbu 所说:“在一个类层次结构中,元类必须是彼此的子类。也就是说,Component 的元类必须是 OtherObjects 的元类的子类。”
这意味着您的问题比您首先要复杂一些 - 不仅您必须从基类调用正确的元类,而且您当前的元类也必须正确地从基类继承。
(破解一些代码,面对奇怪的行为,90 分钟后再回来) 这确实很棘手 - 我必须创建一个类来接收所需的元类作为参数,并且哪个 __call__ 方法动态生成一个新的元类,修改它的基础并向它添加一个 __superclass 属性。
但这应该做你想做的,甚至更多——你只需要从 BaseComponableMeta
继承所有元类,并通过元类“__superclass”属性调用层次结构中的父类(super class):
from itertools import chain
class Meta1(type):
def __new__(metacls, name, bases, dct):
print name
return type.__new__(metacls, name, bases, dct)
class BaseComponableMeta(type):
def __new__(metacls, *args, **kw):
return metacls.__superclass.__new__(metacls, *args, **kw)
class ComponentMeta(object):
def __init__(self, metaclass):
self.metaclass = metaclass
def __call__(self, name, bases,dct):
#retrieves the deepest previous metaclass in the object hierarchy
bases_list = sorted ((cls for cls in chain(*(base.mro() for base in bases)))
, key=lambda s: len(type.mro(s.__class__)))
previous_metaclass = bases_list[-1].__class__
# Adds the "__superclass" attribute to the metaclass, so that it can call
# its bases:
metaclass_dict = dict(self.metaclass.__dict__).copy()
new_metaclass_name = self.metaclass.__name__
metaclass_dict["_%s__superclass" % new_metaclass_name] = previous_metaclass
#dynamicaly generates a new metaclass for this class:
new_metaclass = type(new_metaclass_name, (previous_metaclass, ), metaclass_dict)
return new_metaclass(name, bases, dct)
# From here on, example usage:
class ComponableMeta(BaseComponableMeta):
pass
class NewComponableMeta_1(BaseComponableMeta):
def __new__(metacls, *args):
print "Overriding the previous metaclass part 1"
return metacls.__superclass.__new__(metacls, *args)
class NewComponableMeta_2(BaseComponableMeta):
def __new__(metacls, *args):
print "Overriding the previous metaclass part 2"
return metacls.__superclass.__new__(metacls, *args)
class A(object):
__metaclass__ = Meta1
class B(A):
__metaclass__ = ComponentMeta(ComponableMeta)
# trying multiple inheritance, and subclassing the metaclass once:
class C(B, A):
__metaclass__ = ComponentMeta(NewComponableMeta_1)
# Adding a third metaclass to the chain:
class D(C):
__metaclass__ = ComponentMeta(NewComponableMeta_2)
# class with a "do nothing" metaclass, which calls its bases metaclasses:
class E(D):
__metaclass__ = ComponentMeta(ComponableMeta)
关于python - 查找在 Python 中重新定义之前使用的 __metaclass__,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/8842624/