Question

我需要编写许多类，它们之间的区别仅在于变量，这对于稳定类的工作至关重要。没有这样的变量，类就无法工作，改变变量会破坏它们。我写了一个用于创建相应metaclss的代码：

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

在我开始使用类似这样的类之前，一切都很好：

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，您无需“保护”父类的属性，只需用两个带前缀的下划线命名即可。在您的示例__essential_field中。这意味着将它留给任何后裔。

如何在Python类中创建抽象的受保护字段？

1 个答案: