python - 在 python 3 中使用链表进行三个(或更多)不同变量赋值的操作顺序

Question

我正在编写简单的代码来反转链表，并意识到赋值可以在一行上完成，我觉得这很酷:

    def Reverse(head):
      prev_node = None
      curr_node = head

      while curr_node:
        prev_node, curr_node.next, curr_node = curr_node, prev_node, curr_node.next

      return prev_node

    def Reverse(head):
      prev_node = None
      curr_node = head
      print(curr_node.data)
      print(curr_node.next.data)
      print(prev_node)

      while curr_node:
        prev_node, curr_node, curr_node.next = curr_node, curr_node.next, prev_node

      return prev_node

输入是

1 2 3 4

输出是

1
2
None

但以下错误是从 while 循环中产生的(仅在第二个代码块上；第一个运行正常)

      prev_node, curr_node, curr_node.next = curr_node, curr_node.next, prev_node
    AttributeError: 'NoneType' object has no attribute 'next'

我能找到的关于这个主题的最接近的讨论是 here .这表示首先评估 RHS，从左到右。我认为这意味着存储 curr_node，然后是 prev_node，然后是 curr_node.next。然后分别赋给prev_node、curr_node.next和curr_node。我看不出第一个和第二个例子之间的区别。我错过了一些简单的东西吗？

有谁知道为什么第一个示例运行而第二个示例产生错误？

Answer 1

是的，元组赋值首先计算右侧(从左到右)，然后执行赋值，也是从左到右。

来自Assignment statements documentation :

An assignment statement evaluates the expression list (remember that this can be a single expression or a comma-separated list, the latter yielding a tuple) and assigns the single resulting object to each of the target lists, from left to right.

在第二种情况下，您将 None 分配给 curr_node，因此下一次对 curr_node.next 的分配失败。

换句话说，这是可行的:

prev_node, curr_node.next, curr_node = curr_node, None, None

(假设在执行 curr_node.next = None 时 curr_node 仍然是一个带有 next 属性的 Node 实例)，但是如果你交换参数:

prev_node, curr_node, curr_node.next = curr_node, None, None

现在 curr_node = None 在 curr_node.next = prev_node 之前执行。

这是一个简化的演示，向您展示分配事项的顺序:

>>> class Node:
...     next = None
...
>>> curr_node = Node()
>>> curr_node.next, curr_node = None, None  # first attribute, then name
>>> curr_node = Node()
>>> curr_node, curr_node.next = None, None  # first the name, no attribute anymore
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
AttributeError: 'NoneType' object has no attribute 'next'