python - 将一个类标记为抽象类而不定义任何抽象方法

Question

我正在寻找这样的东西:

from abc import ABC

@abstractclass
class A(ABC):
    pass

A()  # FAILS

class B(A):
    pass

B()  # SUCCEEDS

我知道在 A 中用 abstractmethod 装饰 c'tor 或一些虚拟方法的选项，但这需要我在 中覆盖它B。这正是我要避免的。

Answer 1

您可以自定义__new__:

from abc import ABC, abstractmethod

class A(ABC):
    def __new__(cls):
        if cls is A:
            raise TypeError("Can't instantiate abstract class {}".format(A.__name__))
        return super(A, cls).__new__(cls)

class B(A):
    pass

A()  # TypeError
B()  # Ok

你也可以让它成为装饰器风格，例如像这样(只要装饰类没有自定义 __new__ 就可以工作):

def abstractclass(decorated_cls):
    def clsnew(cls):
        if cls is decorated_cls:
            raise TypeError("Can't instantiate abstract class {}".format(decorated_cls.__name__))
        return super(decorated_cls, cls).__new__(cls)
    decorated_cls.__new__ = clsnew
    return decorated_cls

@abstractclass
class A(ABC):
    pass

class B(A):
    pass

A()  # TypeError
B()  # Ok