返回值和跟踪内部状态时如何使用 Python super()？

Question

我正在尝试在 Python (3.7.6) 中构造一个调用 super() 的类继承的玩具示例。我想跟踪链中中间类的状态，并且我想让委托的类方法返回一个值。我认为这应该很简单，但我得到了一个奇怪的结果。

（我的最小示例基于this tutorial。）

我可以设置一个中产阶级跟踪状态的例子，没问题：

In [1]: class Root:
   ...:     def draw(self):
   ...:         # the delegation chain stops here
   ...:         assert not hasattr(super(), 'draw')
   ...: 
   ...: class Shape(Root):
   ...:     def __init__(self, shapename, **kwds):
   ...:         self.shapename = shapename
   ...:         self.has_run = False
   ...:         super().__init__(**kwds)
   ...:     def draw(self):
   ...:         if not self.has_run:
   ...:             print('Drawing for the first time.  Setting shape to:', self.shapename)
   ...:             self.has_run = True
   ...:         else:
   ...:             print('Drawing again.  Setting shape to:', self.shapename)
   ...: 
   ...:         super().draw()
   ...: 
   ...: class ColoredShape(Shape):
   ...:     def __init__(self, color, **kwds):
   ...:         self.color = color
   ...:         super().__init__(**kwds)
   ...:     def draw(self):
   ...:         print('Drawing.  Setting color to:', self.color)
   ...:         super().draw()
   ...: 
   ...: cs = ColoredShape(color='blue', shapename='square')
   ...: print('*** first pass')
   ...: cs.draw()
   ...: print('*** second pass')
   ...: cs.draw()
*** first pass
Drawing.  Setting color to: blue
Drawing for the first time.  Setting shape to: square
*** second pass
Drawing.  Setting color to: blue
Drawing again.  Setting shape to: square

但是，一旦我修改了返回值的方法，我就失去了跟踪状态的能力：

In [2]: class Root:
   ...:     def draw(self):
   ...:         # the delegation chain stops here
   ...:         assert not hasattr(super(), 'draw')
   ...: 
   ...: class Shape(Root):
   ...:     def __init__(self, shapename, **kwds):
   ...:         self.shapename = shapename
   ...:         self.has_run = False
   ...:         super().__init__(**kwds)
   ...:     def draw(self):
   ...:         if not self.has_run:
   ...:             print('Drawing for the first time.  Setting shape to:', self.shapename)
   ...:             return 'a'
   ...:             self.has_run = True
   ...:         else:
   ...:             print('Drawing again.  Setting shape to:', self.shapename)
   ...:             return 'b'
   ...:         super().draw()
   ...: 
   ...: class ColoredShape(Shape):
   ...:     def __init__(self, color, **kwds):
   ...:         self.color = color
   ...:         super().__init__(**kwds)
   ...:     def draw(self):
   ...:         print('Drawing.  Setting color to:', self.color)
   ...:         foo = super().draw()
   ...:         return [foo] * 3
   ...: 
   ...: cs = ColoredShape(color='blue', shapename='square')
   ...: print('*** first pass')
   ...: out = cs.draw()
   ...: print(out)
   ...: print('*** second pass')
   ...: out = cs.draw()
   ...: print(out)
*** first pass
Drawing.  Setting color to: blue
Drawing for the first time.  Setting shape to: square
['a', 'a', 'a']
*** second pass
Drawing.  Setting color to: blue
Drawing for the first time.  Setting shape to: square
['a', 'a', 'a']

我觉得我缺少有关 super() 功能的一些基本知识，如果有人可以提供任何帮助，我将不胜感激！

Answer 1

这是一个非常愚蠢的错误。在链中的中产阶级，我是在设置内部状态变量之前返回的。这可以通过在任何返回语句之前更新状态变量来纠正。

以下更正代码：

class Root:
    def draw(self):
        # the delegation chain stops here
        assert not hasattr(super(), 'draw')

class Shape(Root):
    def __init__(self, shapename, **kwds):
        self.shapename = shapename
        self.has_run = False
        super().__init__(**kwds)
    def draw(self):
        if not self.has_run:
            print('Drawing for the first time.  Setting shape to:', self.shapename)
            out = 'a'
            self.has_run = True
        else:
            print('Drawing again.  Setting shape to:', self.shapename)
            out = 'b'
        super().draw()
        return out

class ColoredShape(Shape):
    def __init__(self, color, **kwds):
        self.color = color
        super().__init__(**kwds)
    def draw(self):
        print('Drawing.  Setting color to:', self.color)
        foo = super().draw()
        return [foo] * 3

cs = ColoredShape(color='blue', shapename='square')
print('*** first pass')
out = cs.draw()
print(out)
print('*** second pass')
out = cs.draw()
print(out)