Python:抽象类__init__

Question

我试图A用一个默认行为的构造函数声明一个抽象类:所有子类必须初始化一个成员self.n:

from abc import ABCMeta

class A(object):
    __metaclass__ = ABCMeta

    def __init__(self, n):
        self.n = n

但是,我不想让A类被实例化,因为它是一个抽象类.问题是,这实际上是允许的:

a = A(3)

当我期望它应该时,这不会产生任何错误.

那么:如何在为构造函数定义默认行为时定义不可实例化的抽象类？

Answer 1

制作__init__一个抽象的方法:

from abc import ABCMeta, abstractmethod

class A(object):
    __metaclass__ = ABCMeta

    @abstractmethod
    def __init__(self, n):
        self.n = n


if __name__ == '__main__':
    a = A(3)

帮助:

TypeError: Can't instantiate abstract class A with abstract methods __init__

Python 3版本:

from abc import ABCMeta, abstractmethod

class A(object, metaclass=ABCMeta):

    @abstractmethod
    def __init__(self, n):
        self.n = n


if __name__ == '__main__':
    a = A(3)

也适用:

TypeError: Can't instantiate abstract class A with abstract methods __init__

Answer 2

一个不太优雅的解决方案可能是这样的：

class A(object):
  def __init__(self, n):
    if self.__class__ == A:
      raise Exception('I am abstract!')
    self.n = n

用法

class B(A):
  pass
a = A(1)  # Will throw exception
b = B(1)  # Works fine as expected.

Answer 3

您可以重写__new__方法以防止直接实例化。

class A(object):
    __metaclass__ = ABCMeta

    def __new__(cls, *args, **kwargs):
        if cls is A:
            raise TypeError(
                "TypeError: Can't instantiate abstract class {name} directly".format(name=cls.__name__)
            )
        return object.__new__(cls)

输出：

>>> A()
Traceback (most recent call last):
  File "<ipython-input-8-3cd318a12eea>", line 1, in <module>
    A()
  File "/Users/ashwini/py/so.py", line 11, in __new__
    "TypeError: Can't instantiate abstract class {name} directly".format(name=cls.__name__)
TypeError: TypeError: Can't instantiate abstract class A directly