C - can't prepend a linked list

Question

I'm following the tutorial over here to build a linked list. I can't get the "Adding an item to the beginning of the list (pushing to the list)" part to work.
My code:
node_t* prepend(node_t **head, int val) {
    //create new node pointer
    node_t *new_node = (node_t*) malloc(sizeof(node_t));
    new_node->val = val;
    new_node->next = *head; //set its next to existing head (pointer of pointer)
    //update existing head to point to new node
    *head = new_node;
    return *head;
}

int my_first_ll() {

//define a local variable called head that will point to the first node
    node_t *head = NULL;
    head = (node_t*) malloc(sizeof(node_t));

//check for null pointer
    if (head == NULL) {
        return 1;
    }

//note how because head is a pointer we're using -> rather than dot notation to access attributes
    head->val = 1;
    head->next = (node_t*) malloc(sizeof(node_t));
    head->next->val = 2;
    head->next->next = NULL; //last item should point to a NULL

head = prepend(head, 0);
    print_list(head);
}

It prints:
Currently at node 0

Instead of
Currently at node 0
Currently at node 1
Currently at node 2

So it seems when I insert the new head I fail to link to the previous one - but I just can't figure out how.

megaftw · Answer

You can pass **head to the function, so you basically pass a pointer to a pointer:
void prepend(node_t **head, int data){
    node_t *new = malloc(sizeof(node_t));
    //set data
    new->data = data;  
    //set the next pointer of new to current head                
    new->next = *head; 
    //now set the newly created node to be the new head                     
    *head = new;                            
}

But the you'd have to call the function like this:
int main()
{
    node_t *head = NULL;
    prepend(&head, 3);
    prepend(&head, 6);
    printlist(head);
    deltelist(&head);
    return 0;
}

And theres no reason to return the *head in your function because you basically change the head node you created in main.
So if I now print my list, it prints
6
3

And never forget to delete the list after you used it.
Hope this helped you :)

ilmoi · Answer

I found a solution: instead of passing **head to prepend like they do in the tutorial, passing *head solves it. Thus the final code:
node_t* prepend(node_t *head, int val) {
    //create new node pointer
    node_t *new_node = (node_t*) malloc(sizeof(node_t));
    new_node->val = val;
    new_node->next = head; //set its next to existing head (pointer of pointer)
    //update existing head to point to new node
    head = new_node;
    return head;
}

Could someone explain why one works and the other doesn't? And why did they use two stars in the tutorial? And if two stars is actually correct, then why is it failing for me?

C - can't prepend a linked list

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