python - 使用双下划线时什么时候会发生名称重整？

Question

class Test():

    def __init__(self):
        self.__test = "cats"
        print(self.__test)

    def __example(self):
        print("Hello World")

x = Test()
print(x.__dict__)

使用我上面写的代码，打印语句将显示访问变量test，我需要写_Test__test，但是如图所示我可以打印变量如果我之后直接在 __init__ 方法中调用它。所以我的问题是，如果我可以在用它的名称声明后直接访问它，即 self.__test，那么它会在什么时候变得困惑？

Answer 1

通过 .__ 访问的属性在类体中任何地方都被破坏了(但是内部类声明会首先到达它。)。

将其视为语法糖。

在 Test 的上下文中类(class)机构，self.__test是被破坏的名称 self._Test__test 的别名 ;在上下文中，它们的意思完全相同。

演示会使这一点更清楚。首先，一些辅助类。

class PrintAttrAccess:
    def __getattr__(self, name):
        print(name)

class Empty: pass

现在进行演示:

class Test:
    print('IN TEST BODY')
    (lambda: PrintAttrAccess().__in_lambda)()  # Anywhere in the class body works.
    classfoo = Empty()
    classfoo.__foo = 'foo'
    print("Same thing?", classfoo.__foo is classfoo._Test__foo)
    print("vars() of classfoo:", vars(classfoo))

    class Inner:
        print('IN INNER')
        PrintAccess().__inner

    def __init__(self):
        print('IN INIT')
        print("Who am I?", self)
        self.__test = "cats"
        print(self._Test__test)  # It's ALREADY MANGLED!
        # This line means exactly the same as the one above.
        print(self.__test)
        localfoo = Empty()
        localfoo.__spam = 'spam' # "self" isn't special.
        print("vars() of localfoo:", vars(localfoo))


def outside_method(self):
    print('OUTSIDE BODY')
    print("Who am I?", self)
    self.__test = "dogs"
    print(self._Test__test)
    print(self.__test)  # Sugar doesn't apply outside of class body.

Test.outside_method = outside_method  # Add a new method to Test class.

Test().outside_method()  # init and call as method.

输出是:

IN TEST BODY
_Test__in_lambda
Same thing? True
vars() of classfoo: {'_Test__foo': 'foo'}
IN INNER
_Inner__inner
IN INIT
Who am I? <__main__.Test object at 0x000001CCF3048978>
cats
cats
vars() of localfoo: {'_Test__spam': 'spam'}
OUTSIDE BODY
Who am I? <__main__.Test object at 0x000001CCF3048978>
cats
dogs