Try Except not catching IndexError

Question

I'm trying to create a linked list for practice.

Here is my code so far:

class Node:
    def __init__(self, data, _next):
        self.data = data
        self._next = _next

test_list = [1,4,9]


def Linked(nodes):
    try:
        linked = []
        for i in range(len(nodes)-1):
            j = Node(nodes[i], nodes[i+1])
            linked.append(j)
        return linked
    except IndexError:
        print('Tail Found')

test = Linked(test_list)


test[2].data

I'm trying to catch the Index Error with the try/except but I'm still getting the following error:

IndexError                                Traceback (most recent call last)
<ipython-input-193-5fa4e0d033a1> in <module>
     20 
     21 
---> 22 test[2].data

IndexError: list index out of range

Why isn't it printing 'Tail Found'?

EDIT:

I changed the code to this:

class Node:
    def __init__(self, data, _next):
        self.data = data
        self._next = _next

test_list = [1,4,9]


def Linked(nodes):
        linked = []
        for i in range(len(nodes)-1):
            j = Node(nodes[i], nodes[i+1])
            linked.append(j)
        return linked


test = Linked(test_list)

try:
    print(test[2].data)
except IndexError:
    print('Tail Found: '+test[1]._next)

And it works now.

If test[2] doesn't exist then I know that test_list[2] is the last one in the list (ie the tail).

Answer 1

Your error are in line 22. There is no indexerror occur in your try catch block

Answer 2

Why "Tail Found"? If you pass a list with length greater than 2 to function Linked, there won't be any exception, so no "Tail Found" !

You gave a list with length 3, then you got a test with length 3-1. Index text allowed only for 0 or 1, test[2] certainly get a result

IndexError: list index out of range

Answer 3

If you print(Linked(test_list)) , you will get:

[<__main__.Node object at 0x02A62220>, <__main__.Node object at 0x02AC1220>]

See? Only 2 elements in the list, so you can't call index 2 (third element) .