setattr()不为类成员分配值

问题描述

我正在尝试在Python中建立setattr()的简单测试示例，但是它无法为该成员分配新值.

I'm trying to set up a simple test example of setattr() in Python, but it fails to assign a new value to the member.

class Foo(object):
    __bar = 0
    def modify_bar(self):
        print(self.__bar)
        setattr(self, "__bar", 1)
        print(self.__bar)

在这里，我尝试使用setattr(self, "bar", 1)进行变量赋值，但未成功:

Here I tried variable assignment with setattr(self, "bar", 1), but was unsuccessful:

>>> foo = Foo()
>>> foo.modify_bar()
0
0

有人能解释一下幕后发生的事情吗?我是python的新手，所以请原谅我的基本问题.

Can someone explain what is happening under the hood. I'm new to python, so please forgive my elementary question.

推荐答案

领先的双下划线调用python 名称修改.

A leading double underscore invokes python name-mangling.

所以:

class Foo(object):
    __bar = 0  # actually `_Foo__bar`
    def modify_bar(self):
        print(self.__bar)  # actually self._Foo__bar
        setattr(self, "__bar", 1)
        print(self.__bar)  # actually self._Foo__bar

仅对 进行名称改写仅适用于标识符，而不适用于字符串，这就是为什么setattr函数调用中的__bar不受影响的原因.

Name mangling only applies to identifiers, not strings, which is why the __bar in the setattr function call is unaffected.

class Foo(object):
    _bar = 0
    def modify_bar(self):
        print(self._bar)
        setattr(self, "_bar", 1)
        print(self._bar)

应能按预期工作.

在大多数python代码中，前导双下划线通常不经常使用(因为通常不鼓励使用它们).有一些有效的用例(主要是避免在子类化时发生名称冲突)，但是这种情况很少见，因此通常在野外避免使用名称修饰.

Leading double underscores are generally not used very frequently in most python code (because their use is typically discouraged). There are a few valid use-cases (mainly to avoid name clashes when subclassing), but those are rare enough that name mangling is generally avoided in the wild.

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