CODEWITHSUNDEEP
pythonpython-3.xlinked-listnonetype
Ajinkya Gawali
Ajinkya Gawali

Reputation: 107

Updating 'None' value does not reflect in the object

I have linked list with the following structure.

class Node:
  def __init__(self,val):
    self.val = val
    self.next = None

n1 = Node(1)
n2 = Node(2)
n3 = Node(3)
n4 = Node(4)

n1.next = n2
n2.next = n3
n3.next = n4

The last node's next is None , but when update it , it does reflect in the linked list.

a = n4.next 
a = Node(5)
print(n4.next)
>> None

But when I directly update (without temp variable), it works

n4.next = Node(5)
print(n4.next)
>> <<Node Object>>

Why does this happen? I thought None is also an object in python and object assignments just copy references.

Upvotes: 2

Views: 689

Answers (1)

Mitch
Mitch

Reputation: 3418

I think you're confusing the special Python None singleton object with a normal Python class instance.

And you are confusing the classic programming language "variable" term (a container for value(s)) with references (variables that refer, or point to, containers of values)

In Python all variable names are references to objects.

So say you do something like 

a = n1
a.val = 10

This would be reflected in n1 object (a Node class instance) because ais a reference to n1.

Trying to do something like 

n4.next = None 
a = n4.next
a = Node(3)

will make a refer to None first, and then changes a to refer to a different object, a new Node instance.

If a refers to a Node instance, you can change the instance via a

n4.next = Node(2) 
a = n4.next
a.val = 5 #this would alter n4.next

But if a refers to the special object None, None will not act like a Node instance and hence cannot be changed

Upvotes: 3

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