reverse last 5 nodes of a linked list

list.cxx

// Here's what the linked list looks like. 
// I've assumed int values, so you can see that it works
class Node 
{
public:
    Node() : next(NULL), value(0) {}
    Node(int v) : next(NULL), value(v){}
    Node* next;
    int value;
};

int length(Node* head)
{
    int n = 0;
    for (Node* curr = head; curr != NULL; curr = curr->next)
        n++;
    return n;
}

Node* reverse(Node* ptr)
{
    Node* temp;
    Node* prev = NULL;
    while (ptr != NULL) 
    {
        temp = ptr->next;
        ptr->next = prev;
        prev = ptr;
        ptr = temp;
    }
    return prev;
}

void reverse_last(Node* head, n)
{
    int n = length(head);
    Node* e = head;
    for (int i = 0; i < n-5-1; i++)
        e = e->next;
    e->next = reverse(e->next);
}

int main()
{
    // initialize a list of length n.
    int n = 10;
    Node* head = new Node(0);
    Node* curr = head;
    
    for (int i = 1; i < n; i++)
    {
        Node* t = new Node(i);
        curr->next = t;
        curr = t;
    }
    
    // do the reversal
    reverse_last(head, 5);

    // print what we've got
    for (Node* curr = head; curr != NULL; curr = curr->next)
        std::cout << curr->value << std::endl;

    return 0;
}