python - 如何使类装饰器不破坏 isinstance 函数？

Question

我正在创建一个用于测试的单例装饰器，但是当我询问一个对象是否是原始类的实例时，它返回 false。

在示例中，我正在装饰一个计数器类来创建一个单例，因此每次获得值时，无论对象的哪个实例调用它，它都会返回下一个数字。代码几乎可以工作，但函数 isinstance 似乎中断了，我尝试使用 functools.update_wrapper 但我不知道是否可以让 isinstance 函数将 Singleton 识别为 Counter(在下面的代码中)，只要当我要求 Counter 时代码实际上返回 Singleton。

装饰器

def singleton(Class):

    class Singleton:
        __instance = None

        def __new__(cls):
            if not Singleton.__instance:
                Singleton.__instance = Class()

            return Singleton.__instance

    #update_wrapper(Singleton, Class, 
    #    assigned=('__module__', '__name__', '__qualname__', '__doc__', '__annotation__'), 
    #    updated=()) #doesn't seems to work
    return Singleton

装饰类

@singleton
class Counter:
    def __init__(self):
        self.__value = -1
        self.__limit = 6

    @property
    def value(self):
        self.__value = (self.__value + 1) % self.limit
        return self.__value

    @property
    def limit(self):
        return self.__limit

    @limit.setter
    def limit(self, value):
        if not isinstance(value, int):
            raise ValueError('value must be an int.')

        self.__limit = value

    def reset(self):
        self.__value = -1

    def __iter__(self):
        for _ in range(self.limit):
            yield self.value

    def __enter__(self):
        return self

    def __exit__(self,a,b,c):
        pass

测试

counter = Counter()
counter.limit = 7
counter.reset()
[counter.value for _ in range(2)]

with Counter() as cnt:
    print([cnt.value for _ in range(10)]) #1

print([counter.value for _ in range(5)]) #2
print([val for val in Counter()]) #3

print(Counter) #4
print(type(counter)) #5 
print(isinstance(counter, Counter)) #6

输出:

#1 - [2, 3, 4, 5, 6, 0, 1, 2, 3, 4]
#2 - [5, 6, 0, 1, 2]
#3 - [3, 4, 5, 6, 0, 1, 2]
#4 - <class '__main__.singleton.<locals>.Singleton'>
#5 - <class '__main__.Counter'>
#6 - False

(更新包装器未注释)

#1 - [2, 3, 4, 5, 6, 0, 1, 2, 3, 4]
#2 - [5, 6, 0, 1, 2]
#3 - [3, 4, 5, 6, 0, 1, 2]
#4 - <class '__main__.Counter'>
#5 - <class '__main__.Counter'>
#6 - False

Answer 1

您可以使用 singleton Python Decorator Library 中的类装饰器.

它之所以有效，是因为它修改了现有的类(替换 __new__() 方法)，而不是像您问题中的代码中那样用完全独立的类替换它。

import functools

# from https://wiki.python.org/moin/PythonDecoratorLibrary#Singleton
def singleton(cls):
    ''' Use class as singleton. '''

    cls.__new_original__ = cls.__new__

    @functools.wraps(cls.__new__)
    def singleton_new(cls, *args, **kw):
        it =  cls.__dict__.get('__it__')
        if it is not None:
            return it

        cls.__it__ = it = cls.__new_original__(cls, *args, **kw)
        it.__init_original__(*args, **kw)
        return it

    cls.__new__ = singleton_new
    cls.__init_original__ = cls.__init__
    cls.__init__ = object.__init__

    return cls

有了它，我得到以下输出(注意最后一行):

[2, 3, 4, 5, 6, 0, 1, 2, 3, 4]
[5, 6, 0, 1, 2]
[3, 4, 5, 6, 0, 1, 2]
<class '__main__.Counter'>
<class '__main__.Counter'>
True