Bug Squashing: A singly linked list python implementation | Part 3

In the past two episodes in this series, we defined a Node class, complete with __init__ and __str__ methods. However, we came across a bug, and now it's time to squash it.

To remind you of the problem, here's the current code for printing a Node.

# indented within Node() class.
    def __str__(self):
        s = "Node with value {} and next node: {}.".format(self.value, self.next_node)
        return s

Do you see the problem? (I didn't.)
To give you an idea, here's a the result of printing the first node in a list of five nodes.

Node with value 14 and next node: Node with value 27 and next node: Node with value 33 and next node: Node with value 72 and next node: Node with value 38 and next node: None.....

That's a mess. Now do you see what happened? It's the bug at work. It recursively printed all of the nodes in the list. This probably isn't the way we want to print a single node!
At any rate, I wanted to print only the Node's value and the value of the next Node. How can we do that? Here is my solution.

# indented within Node() class.
   def __str__(self):
        s = "Node with value {} and ".format(self.value)
        if not self.next_node:
            s += "no next node"
        else:
            s += "next node has value {}.".format(self.next_node.value)
        return s

Note that I needed the if/else logic because other otherwise, cases where text node's next value is None, self.next_value.value would effectively be calling None.value, which triggers the following type of error.

AttributeError: 'NoneType' object has no attribute 'value'

So that's the first bug of the series fixed.

Next it's on to defining a SinglyLinkedList, but save the reason why for the next post.

The code for this project is available here on GitHub. I'll be updating it gradually as I write this series. I'll be sure to commit at least as often as I post, so you can back up and see that code as it was when I published a given post.