Question

我正在尝试访问一个类的实例（称为“ a”），该实例传递给第二个类（称为“ b”）。我的意图是通过装饰实例b的类来访问实例“ a”，以便我可以在“ a”上设置一些后台线程任务，同时仍将“ b”用于其他更重要的任务。这可能吗？

import threading    
import inspect

Class doStuff():
    def __init__(self, somePropertyFromAnotherClass):
        self.lock = threading.Lock()
        self.prop = somePropertyFromAnotherClass

    def doCoolThreadingStuff():
        print("do threading stuff with {}".format(self.prop))


def someDecorator(cls):
    def wrapper(cls):
        print(inspect.getrgspec(cls.__init__))
        #ds = doStuff()  ## this is the bit that i can't figure out!
    wrapper(cls)
    return cls


Class A():
    def __init__(self):
        self.obj = "i'm an object" 


@someDecorator
Class B():
    def __init__(self, obj):
        self.obj = obj

    def doSomethingWithObj():
        print('doing something with obj')


if __name__ == "__main__":
    a = A()
    b = B(a)

Answer 1

装饰器的格式不正确

def someDecorator(cls_obj):
    def wrapper(*args):
        ds = doStuff(args[0]) # positional : / 
        ds.doCoolThreadingStuff() 
        return cls_obj(*args)
    return wrapper

我能够使用上面的装饰器定义来访问A类（传递给B类）的实例

python（2.7）：使用类装饰器访问类属性

1 个答案: