私有变量在python类中的实际实现

问题描述

我有一个问题，当我阅读python文档中的私有变量。

 <$ c $ 
 c>>>>> class a（）：
 def __init __（self）：
 self.public = 11 
 self._priv = 12 
>>> b = a（）
>>>> print b._priv 
>>>> 12 
  

我想知道是否有一种方法可以使变量

解决方案

• Guido van Rossum认为： em>。


• 扩展版本：我们都是成年人。

您可以隐藏 em>如果你必须，但你真的shoudn't：

  class NonAdult（object）：
 def __init __ self）：
 self .__ private_number ='42'
 def __getattr __（self，name）：
如果name.startswith（'__ private'）：
 raise AttributeError 
 
如果__name__ =='__main__'：
 na = NonAdult（）
 print（na .__ private_number）＃raisees AttributeError 
 print（na .__ dict __ ['_ NonAdult__private_number']）仍然会返回42 
  

I had a question when I was reading the python documentation on private variables link.

So the documentation tells that it is a convention to name private variables with an underscore but python does not make the field private.

>>> class a():
      def __init__(self):
         self.public = 11
         self._priv = 12
>>> b = a()
>>> print b._priv
>>> 12

I wanted to know if there is a way in which I can make a variable "truly" private in python.

解决方案

• Guido van Rossum put it: we are all adults.
• The extended version: We are all adults. Feel free to shoot yourself in the foot if you must.

You can hide it a bit if you must, but you really shoudn't:

class NonAdult(object):
    def __init__(self):
        self.__private_number = '42'
    def __getattr__(self, name):
        if name.startswith('__private'):
            raise AttributeError

if __name__ == '__main__':
    na = NonAdult()
    print(na.__private_number) # raises AttributeError
    print(na.__dict__['_NonAdult__private_number']) # would still return 42

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