Python类__init__覆盖，TypeError：__ init __（）需要5个位置参数，但给出了6个

Question

我正在学习python几天，我正在尝试为现有的类创建一个子类，但我找不到错误。我尝试了一切，但我认为我只是盲目......

输出总是会出现这样的错误

dog = Dog('Max', 120, 50, 'Woff', 'Tom')
TypeError: __init__() takes 5 positional arguments but 6 were given

这是代码，Garfield cat的动物部分工作正常。

class Dog(Animal):
    __owner = ""

    def __init(self, name, height, weight, sound, owner):
        self.__owner = owner
        super(Dog, self).__init__(name, height, weight, sound)

dog = Dog('Max', 120, 50, 'Woff', 'Tom') #<==== HERE ERROR OCCURES

Answer 1

Dog中的方法 init 也应该与__结尾：

class Dog(Animal):
    __owner = ""

# here
def __init__(self, name, height, weight, sound, owner):
    self.__owner = owner
    super(Dog, self).__init__(name, height, weight, sound)

Answer 2

您的dog类的init方法名称错误，因此它不会覆盖基类init，它接受5个参数。

def __init(self, name, height, weight, sound, owner):
    self.__owner = owner
    super(Dog, self).__init__(name, height, weight, sound)

您在功能结束时缺少__。它应该是def __init__(..)