Python 3个继承等效语句

Question

从此处的文档中，它声称 super().method(arg) 与以下内容执行相同的操作：super(C, self).method(arg).

https://docs.python.org/3/library/functions.html#super

class shape(object):
    def __init__(self, type):
        self.shapeType = type

class coloredShape1(shape):
    def __init__(self, type, color):
        super().__init__(type)
        self.shapeColor = color

class coloredShape2(shape):
    def __init__(self, type, color):
        super(shape, self).__init__(type)
        self.shapeColor = color

circle = shape("circle")
blueRectangle = coloredShape1("rectangle", "blue")
redSquare = coloredShape2("square", "blue")

创建 blueRectangle 没有问题，但是，redSquare 行抛出以下预期：

Traceback (most recent call last):
  File "test.py", line 17, in <module>
    redSquare = coloredShape2("square", "blue")
File "test.py", line 12, in __init__
  super(shape, self).__init__(type)
TypeError: object.__init__() takes no parameters

我不明白两者之间的区别。有人可以解释我做错了什么吗？

Answer 1

您将错误的 class 传递给了 super()，因此您调用了 object.__init__，而不是 shape.__init__。

class coloredShape2(shape):
    def __init__(self, type, color):
        super(coloredShape2, self).__init__(type)
        self.shapeColor = color

Python 3个继承等效语句

Python 3 inheritance equivalent statements

inheritance

super

python-3.x