Python使用super()实现方法返回值与跟踪内部状态的问题

问题原因 #

你的代码运行异常和super()的特性无关，属于基础的代码执行顺序错误：
在修改后的Shape.draw()方法中，你将return语句放在了self.has_run = True状态更新代码之前，Python执行到return时会直接终止当前方法的运行，后续的状态更新代码、super().draw()调用永远不会被执行，因此has_run始终保持初始值False，每次调用都会走首次执行的分支。

修正方案 #

调整代码顺序，将状态更新、父类方法调用放在return之前即可：

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
            res = 'a'
        else:
            print('Drawing again.  Setting shape to:', self.shapename)
            res = 'b'
        super().draw()
        return res

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)

运行结果 #

*** 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 again.  Setting shape to: square
['b', 'b', 'b']

内容的提问来源于stack exchange，提问作者bmeyers