[英]Determine if a class in python is a metaclass
Is there a way in python to determine if a class object is a metaclass? python中是否有一种方法可以确定类对象是否为元类?
I know that you could check using the equality operator. 我知道您可以使用相等运算符进行检查。 metaclass == type
元类==类型
But that wouldn't cover for user defined metaclasses. 但这不能涵盖用户定义的元类。
Test if the object is a subclass of type
: 测试对象是否是
type
的子type
:
issubclass(MetaClass, type)
This returns True
for all metaclasses, including type
itself. 对于所有元类,包括
type
本身,这将返回True
。
Demo: 演示:
>>> class Meta(type): pass
...
>>> class Foo(object): pass
...
>>> issubclass(Meta, type)
True
>>> issubclass(Foo, type)
False
Most metaclasses are type
subclasses ( issubclass(metaklass, type)
) but not all : 大多数元类都是
type
子类( issubclass(metaklass, type)
), 但不是全部 :
>>> def logging_meta(name, bases, namespace, **kwd):
... print(name, bases, namespace, kwds)
... return type(name, bases, namespace, **kwds)
...
>>> class C(metaclass=logging_meta):
... a = 1
...
C () {'__module__': '__main__', '__qualname__': 'C', 'a': 1} {}
>>> issubclass(logging_meta, type)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: issubclass() arg 1 must be a class
ie, the answer is "any callable eg, a class with a __call__()
method that accepts the same arguments as type()
and returns a class object will do": 即,答案是“任何可调用的对象,例如,具有
__call__()
方法的类,该类接受与type()
相同的参数并返回类对象将可以执行”:
>>> class C(metaclass=lambda *a: lambda *a: None): pass
...
>>> C()
>>> type(C)
<class 'function'>
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