CODEWITHSUNDEEP
cpointersnodessingly-linked-list
Riddhiman Saikia
Riddhiman Saikia

Reputation: 31

Linked list delete node. Free(pointer) prints 0 in next node

Below is a delete node in Linklist code that takes head pointer and position to delete as arguments (position index start at ZERO in Linklist). And after deleting, returns a pointer to head.

Node* delete(Node* head, int position) 
{
    Node *p = head;
    if(!position)
    {
        p = p->next;
    }
    else
    {
        while(position--)
        {
            if(!position) head->next = head->next->next; 
            head = head->next;
        }
    }
    free(head);
    return p; 
}

Suppose list: 20-2-19-7-3-6. And position to delete is 2 (node 19 to delete, since index start from zero).

After deleting, and printing, it says: 20-2-0-3-6.(i.e. the node directly next to the one deleted prints 0)

But if I remove the line "free(head)", then it prints: 20-2-7-3-6 (correctly).

Please help and explain why.

PS: There is no issue when deleting head node or tail node. But any other node in between shows 0 in next node.

Upvotes: 1

Views: 618

Answers (1)

dodobhoot
dodobhoot

Reputation: 503

This is the dry run of the code:

20 --> 2 --> 19 --> 7 --> 3 --> 6
^
head

while(position--) // position == 2
{
    if(!position) // position == 1, condition is false
        head->next = head->next->next; 
    head = head->next;
}

20 --> 2 --> 19 --> 7 --> 3 --> 6
       ^
       head

while(position--) // position == 1
{
    if(!position) // position == 0, condition is true
        head->next = head->next->next;
    head = head->next;
}

            /-----\
20 --> 2 --/ 19 --> 7 --> 3 --> 6    // 2's next is pointing to 7 now
                    ^
                    head

Now free(head) will be executed which will delete the node containing number 7. Now, when you print, you'll probably get:

20 -> 2 -> (reference to deleted node) -> 3 -> 6

I think this is undefined behavior that you are referencing the deleted node and it's printing 0.

Upvotes: 2

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