python - 如何在 Python 类中制作抽象的 protected 字段？

Question

我需要编写十几个仅在变量上有所不同的类，这对于类的稳定工作至关重要。没有这些变量，类就无法工作，并且更改变量会破坏它们。我写了一个代码来创建相应的 metaclss:

from abc import ABCMeta, abstractmethod, abstractproperty
import re

def get_field_protector(*names):
    def check_attributes(cls):
        for t in names:
            if not hasattr(cls, t):
                raise Exception('Abstract field "{0}" have to be defined'.format(t))
    def decorate_new_method(inp_method, class_name):
        def inner(cls, *args, **kwargs):
            if cls.__new__.decorated==class_name:
                check_attributes(cls)
            return inp_method(cls, *args, **kwargs)
        return inner
    class result_protector(ABCMeta):
        def __setattr__(self, name, value):
            if name in names:
                raise Exception("Read only class attribute!")
            super(self.__class__, self).__setattr__(name, value)
        def __delattr__(self, name):
            if name in names:
                raise Exception("Read only class attribute!")
            super(self.__class__, self).__delattr__(name)
        def __init__(self, name, bases, attributes):
            super(self.__class__, self).__init__(name, bases, attributes)
            if not hasattr(self.__new__, 'decorated'):
                self.__new__=decorate_new_method(self.__new__, name)
                self.__new__.decorated=name
    return result_protector 

在我开始使用像这样的类之前，它工作得很好:

class test(object, metaclass=get_field_protector("essential_field")):
    essential_field="some_value"
    def __init__(self, val):
        self.val=val

当我尝试创建此类时，我收到:

>>> test(1)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/media/bykov/9016-4EF8/main_architecture.py", line 13, in inner
    return inp_method(cls, *args, **kwargs)
TypeError: object() takes no parameters

Answer 1

布鲁诺的评论是正确的。在 Python 中，您不“保护”父类的属性，您只需用两个前缀下划线命名它。在您的示例中 __essential_field。这意味着将它留给任何后代。