Just for the understanding purpose i have posted two functions & code is in C++. Trying to insert each node as head and link that node to the previous one but while loop goes infinite while printing.

Is there something wrong with print function or insert function?
My print function is working fine when i'm dealing with nodes inserted at the end i can provide that code too if anyone needed. THANKS

int insert_first(int n)
    if(start == NULL)
        curr  =  new node;
        curr  -> data = n;
        start =  curr;
        temp  =  curr;
        cout  << "Head added: " << curr->data <<endl;
        curr =  new node;
        curr -> data = n;
        curr-> link = start;
        start = curr;

        cout<<"First node Added: "<<start->data<<endl;

void print()

    node* print;
    print = start;

        print = print->link ;

1 Answer 1


I think you have a basic misconception. Here is what your linked list should look like:

[head] -> [value1, link1] -> [value2, link2] -> ...

The head should only be a link (i.e. pointer) to the first node. You correctly coded, that if (head == NULL) create a new node and make start (i.e. head) point to that node. So you would have:

[head] -> [value1, link1]

Now on your else block, which runs when there is already a head, you create a new node, you give it a value, and you make its link point to head. And here are your mistakes.

Firstly, no node will point to this node, so you effectively haven't added it to the list. Moreover, even if you wanted it to be the head, by the time you make this node the head, you lose the other head and effectively all the other nodes behind it. And you also make this node, point to itself. See here:

curr-> link = start;
start = curr;

Here is how the linked list should work. Say we already have the following list:

[head] -> [value1, link1] -> [value2, link2]
  1. To append at the end of the list, you must create a new node (let's call it node3).
  2. Give a value to this node's value variable.
  3. Then make link3 point to NULL, to be sure it's initialized.
  4. Then make link2 point to node3.

Then you would have the following:

[head] -> [value1, link1] -> [value2, link2] -> [value3, link3]

Now say you want to add a node at the beginning of the list. Here's what you should do:

  1. Create a new node (let's call it node4).
  2. Give a value to this node's value variable.
  3. Make link4 point to node1, so that you don't lose all the members of the list.
  4. Make head point to node4

Then you would have the following:

[head] -> [value4, link4] -> [value1, link1] -> [value2, link2] -> [value3, link3]

Tip #1

The word print is in some languages a reserved word or a function, so you should better rename the variable in your print() function. You should also not use the same name for both a function and a variable. It makes your code unreadable. Rename the function to printList() and the variable to nodeToPrint or currentNode.

If this answers your question please accept it by clicking the gray check-mark to the left, so that it becomes green. You can also vote it up by pressing the up arrow above the check-mark. And don't forget to keep coding!

  • in adding at front how we would get know that this is node1 and we need to link our node4 to that? Commented Sep 8, 2015 at 4:45
  • another thing after i resolve this issue & understand it is this enough to go for binary trees? Commented Sep 8, 2015 at 4:53
  • If you want to add to the beginning, since head now points to node1, make link4 point to where head points and then make the head point to node4. If it seems too complicated better re-watch the lectures. A binary tree would still use a value but probably two links (pointers) One for left and one for right. It's got some other differences but it shouldn't be much harder.
    – ChrisG
    Commented Sep 8, 2015 at 6:52

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