Linked list insertion at end

#1

in function insertAtTail why it is written (tail->next!=NULL) why we can’t simply write tail!=NULL as before already we pointed node *tail=head

#2

hello @shubhangis248

to insert at end we need to reach to the end node of current existing list right?

and how we can identify end node ?
value of next of end node will be NULL right?
i.e end_node->next will be NULL.
thats is the reason why we put tail->next!=NULL condition in the while condition,
the while loop will break when we will have tail->next==NULL
which means tail is the end node of current list

#3

if we already declared tail as of type node it will have two types int data and next node as per class so why to declare tail->next particularly why not just tail!=NULL

#4

if we will use this condition then our loop will stop when tail will become NULL right?

but what we want ? we want last node of the list.
what unique property last node has?
there is no node next to last node right?
that means tail->next is NULL that is the reason we r using tail->next!= NULL in place of tail!=NULL

#5

If that’s the case,then while printing linked list why we declare head!=NULL not head->next!=NULL in that case also,we are going till end of linked list

#6

1->2->3->NULL

just do a dry run u will get difference

#7

understood what u r saying