python

Question

我正在看官方pythondocumentation.

在提到的link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 class。

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 subclass

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

b = B()
b.f()

没有覆盖上面定义的抽象方法。

所以，基本上这是否意味着如果我的基础 class 的 metaclass 不是 ABCMeta（或派生自它），则 class 不会表现像一个抽象 class 即使我有一个抽象方法在里面？

也就是说，文档需要更加清晰？

或者，这种行为是否有用，我没抓住重点。

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。

abstractmethod 所做的只是用 __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

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