python - 普通类中的抽象方法

Question

我正在阅读官方 python documentation .

在提到的链接中，第二行指出:

Using this decorator requires that the class’s metaclass is ABCMeta or is derived from it.

但是，我成功地定义了下面给定的类。

from abc import abstractmethod

class A(object):
    def __init__(self):
        self.a = 5
    @abstractmethod
    def f(self):
        return self.a

a = A()
a.f()

所以，上面的代码运行良好。而且，我能够创建一个子类

class B(A):
    def __init__(self):
        super(B, self).__init__() 

b = B()
b.f()

不覆盖上面定义的抽象方法。

所以，这基本上是否意味着如果我的基类的 metaclass 不是 ABCMeta(或派生自它)，该类的行为就不像抽象类，即使我有一个抽象方法吗？

这意味着文档需要更加清晰？

或者，这种行为是否有用，但我没有捕获要点。

Answer 1

So, basically does this mean that if my base class's metaclass is not ABCMeta(or derived from it), the class does not behave like an abstract class even though I have an abstract method in it?

正确。

所有 abstractmethod 所做的就是用 __isabstractmethod__ = True 标记方法。 ABCMeta 完成所有实际工作。 Here是abstractmethod的代码:

def abstractmethod(funcobj):
    """A decorator indicating abstract methods.
    Requires that the metaclass is ABCMeta or derived from it.  A
    class that has a metaclass derived from ABCMeta cannot be
    instantiated unless all of its abstract methods are overridden.
    The abstract methods can be called using any of the normal
    'super' call mechanisms.
    Usage:
        class C(metaclass=ABCMeta):
            @abstractmethod
            def my_abstract_method(self, ...):
                ...
    """
        funcobj.__isabstractmethod__ = True
        return funcobj