如何获得实例引用的名称？

Question

是否可以在函数中获取正在调用函数的实例的引用名称？在由实例的引用调用的函数中，应识别调用该函数的引用的名称。这可能吗？谢谢。

redCar = Car()
blackCar = Car()

redCar.carControl()
blackCar.carControl()


def carControl():

    if trafficJam == True:
        redCar.changeSpeed(500)
        redCar.warningLights(On)

        #is something like the following possible?
        "GetNameOfReferenceWhichCalledTheFunction".changeSpeed(500)
        "GetNameOfReferenceWhichCalledTheFunction".warningLight(On)

Answer 1

实例不调用函数。您可以调用method of实例，例如， redCar.carControl()。因为carControl是一种方法，所以需要在类内部定义。

但是，在该方法中，您可以访问redCar-因为它作为参数传递，并且您需要一个参数来接收它。按照惯例，我们将名称self用于此参数。

请仔细研究示例：

traffic_jam = True

class Car:
    def control(self):
        if traffic_jam:
            # The display message will show some debug information that
            # identifies the object. It doesn't have the name `red_car`
            # built into it - it cannot, because you can use multiple names
            # for the same thing - but it is identified uniquely.
            print(self, "slowing down and putting on warning lights")

red_car = Car()
# When this call is made, the `red_car` will be passed to the method,
# which knows it as `self`.
red_car.control()

Answer 2

使用类时，可以使用自动填充的第一个参数（通常称为self）访问调用该方法的实例。>

class Car:
    def carControl(self):
        if trafficJam == True:
            self.changeSpeed(500)
            self.warningLights(On)

    def changeSpeed(self, value):
        self.speed = value


redCar = Car()
redCar.carControl()
redCar.changeSpeed(250)

Answer 3

class Car:
    def __init__(self, name):
        self.name = name

    def change_speed(self, speed):
        print(self.name, ": speed =", speed)

    def car_control(self, traffic_jam):
        if traffic_jam:
            self.change_speed(0)
        else:
            self.change_speed(50)


red_car = Car("red")
black_car = Car("black")

red_car.car_control(True)
black_car.car_control(False)