CODEWITHSUNDEEP
c++classlinked-list
Euxitheos
Euxitheos

Reputation: 325

Add node to Front of Linked List then count number of nodes in Linked List

I would like to add a node to the front of the list by calling an object and print the length of the list. So far I keep getting a segmentation fault. I want something basic enough that I can build ideas from:

#include <iostream>

using namespace std;

class Node {
  public:
    int data;
    Node *next; //creating a pointer to store the address of the next node or null (at the end of a linked list)
   int key;

    Node(int d){
        data = d;
        next = NULL;
        key = -1;
    }
};

class LinkedList {
public:
    Node *head;
    LinkedList (){
        head = NULL;

    }

    void addNodetoFront(int data){
        Node *n = new Node(data);//allocate memory on the heap with address that is returned to n (pointer)


        n->next=head;
        head=n;


        if(n != NULL){
            n->key = n->next->key+1;
        }
        else{
            n->key = 1;
        }
    }

};

int main(){
    LinkedList linkedlist;

    linkedlist.addNodetoFront(2);
    linkedlist.addNodetoFront(3);
    linkedlist.addNodetoFront(4);
    linkedlist.addNodetoFront(5);
    linkedlist.addNodetoFront(26);
    linkedlist.addNodetoFront(27);
    linkedlist.addNodetoFront(9);
    linkedlist.addNodetoFront(45);
    linkedlist.addNodetoFront(87);

    return 0;
}

Expected is linked list of 9 nodes, and the program prints 9 elements.

Upvotes: 2

Views: 183

Answers (1)

paddy
paddy

Reputation: 63481

Your head starts off as NULL. When you add the first node, then that node becomes the head, and its next pointer becomes NULL. Now, look at this:

if(n != NULL){
    n->key = n->next->key+1;  // But n->next is NULL :(
}

Above, n->next->key is undefined behavior, as you are dereferencing a NULL pointer. Furthermore, while it's actually impossible for n to be NULL, the following logic for if n hypothetically could be NULL is crazy:

else{
    n->key = 1;  // But you said n was NULL :(
}

The only way any of this makes sense is if you actually had a typo in your test and you meant:

if(n->next != NULL)

Putting this all together, makes your function look like this:

void addNodetoFront(int data)
{
    Node *n = new Node(data);

    n->next = head;
    head = n;

    if (n->next != NULL) {
        n->key = n->next->key + 1;
    } else {
        n->key = 1;
    }
}

Alternatively, the following more compact style achieves exactly the same:

void addNodetoFront(int data)
{
    Node *n = new Node(data);
    n->next = head;
    n->key = (head ? head->key + 1 : 1);
    head = n;
}

Upvotes: 2

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