Question

I am trying to create my own datatype that is like a vector or an array.

I am having troubles with my print function; When I go to print the list, it only prints the last item in the list.

// LinkedListClass.cpp : Defines the entry point for the console application.

#include "stdafx.h"
#include <iostream>

class Node
{
public:
 int value;
 Node* next;

 Node::Node(int val)
 {
  value = val;
 };
};

class List
{
public:
 Node* firstNode;
 Node* currentNode;
 int size;

 List::List()
 {
  firstNode = NULL;
  currentNode = firstNode;
  size = 0;
 };

 void push(Node* node)
 {
  if(firstNode == NULL)
  {
   firstNode = node;
   firstNode->next = currentNode;
   size++;
  }
  else
  {
   currentNode = node;
   currentNode = currentNode->next;
   size++;
  }
 };

 void print()
 {
  if(firstNode != NULL)
  {
   Node* printNode = firstNode;
   while(printNode->next != NULL)
   {
    std::cout << "List Item " << printNode->value << std::endl;
    printNode = printNode->next;
   }
  }
 };
};

int _tmain(int argc, _TCHAR* argv[])
{
 List ll = List();
 for(int i = 0; i < 10; ++i)
 {
  Node val = Node(i);
  ll.push(&val);
 }
 std::cout << ll.firstNode->value << std::endl;
 ll.print();
 std::cout << "Size " << ll.size << std::endl;
 std::cin.ignore();
 return 0;
}

/* Output

9
Size 10

*/

I know this is nowhere near completed, but if you have any other pointers (lol), please feel free to suggest.

Answer 1

There are three important errors:

push() — fixed

void push(Node* node)
 {
  if(firstNode == NULL)
  {
   firstNode = node;
   currentNode = node;
   // firstNode->next = currentNode; --> this does nothing useful!
   size++;
  }
  else
  {
   currentNode->next = node;
   currentNode = node;
   //currentNode = node;               -|
   //currentNode = currentNode->next;  -|----> why? what? Do explain.
   size++;
  }
 }

I think by assigning firstNode->next = currentNode; you expected the next time currentNode was updated, it would update firstNode->next as well.

It doesn’t work that way.

firstNode->next = currentNode; implies that the address stored in currentNode is now in firstNode->next. So next time you store something in currentNode = node; you’re not storing it in firstNode->next. So you have a broken linked list — which is why your output didn’t go very far.

Also, this is really bad. By setting currentNode=node before setting the current node’s next pointer to node, you’ve broken the list again. You should first point currentNode->next to node and then set the currentNode as node (node being the node which you’re pushing onto your list).

Node val = Node(i);

The scope of val is only within that iteration of your loop. Once you loop around, it’s off the stack and doesn’t exist anymore. But you’ve copied the pointer of val to your list — so now with the right push method, you’re just adding a dangling pointer.

Node *val = new Node(i);
ll.push(val);

You need to put it on the heap so it stays on till you don’t need it anymore.

… which leads us to your destructor!

Since you’ve allocated a node, you’ll need to deallocate it. So do that in your destructor — traverse your list and deallocate all those nodes.