Question

我有一个“抽象”类，例如：

class A:
    def do_some_cool_stuff():
        ''' To override '''
        pass

    def do_some_boring_stuff():
        return 2 + 2

B类，继承A：

class B(A):
    def do_stuff()
        return 4

有没有办法声明方法A.do_some_cool_stuff 必须被覆盖，并且可能在尝试创建B类对象时应该引发一些警告，当B有未实施A.do_some_cool_stuff？

Answer 1

是的，将A定义为ABC (Abstract Base Class)：

from abc import ABCMeta, abstractmethod

class A(object):
    __metaclass__ = ABCMeta

    @abstractmethod
    def do_some_cool_stuff():
        ''' To override '''
        pass

    def do_some_boring_stuff():
        return 2 + 2

您可以继承A，但如果do_some_cool_stuff()方法具有具体实现，则只能创建此类子类的实例：

>>> from abc import ABCMeta, abstractmethod
>>> class A(object):
...     __metaclass__ = ABCMeta
...     @abstractmethod
...     def do_some_cool_stuff():
...         ''' To override '''
...         pass
...     def do_some_boring_stuff():
...         return 2 + 2
... 
>>> class B(A):
...     def do_stuff():
...         return 4
... 
>>> B()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: Can't instantiate abstract class B with abstract methods do_some_cool_stuff

是否有可能在Python中声明必须覆盖该方法？

1 个答案: