如何将头部指针传递给 main.cpp 中的函数

Question

When I try to call the DisplayLinkedList function in main, it prints out nothing. How do i correctly pass the head into the DisplayLinkedList function so that it properly prints out the entire linked list?

void LinkedList::AppendInTheEnd(int numberToAdd)
{
     NodePtr newNode = new Node;
     newNode->nextNode = NULL;
     newNode->data = numberToAdd;

     if(headNode != NULL)
     {
        currentNode = headNode;
        while(currentNode->nextNode != NULL)
        {
            currentNode = currentNode->nextNode;
        }
        currentNode->nextNode = newNode;
     }
     else
     {
         headNode = newNode;
     }
}
void LinkedList::DisplayLinkedList(Node* head)
{

         Node* p;
         p = head;
         cout << "Displaying the list" << p << endl;
         while(p != NULL)
         {
              cout << "Node at " << p << endl;
              cout << " value " << p->data << endl;
              cout << " next: " << p->data << endl;
           }
            p = p->nextNode;
}
int main()
{
   LinkedList::Node* head = NULL;


  LinkedList list;

  list.AppendInTheEnd(9);
  list.AppendInTheEnd(10);

  list.DisplayLinkedList(head);
}

Answer 1

I think you have a design issue here (in addition to what Jabberwocky pointed out). LinkedList::DisplayLinkedList is a member function and the name implies, that it prints the contents of the list. Eg

list.DisplayLinkedList();

would imply, that it prints the content of list .

Then why does it take any parameter? It's a member function, which means it's called on an instance of LinkedList and you can access the members of this instance using this-> . Passing some parameter to print any other list than the current instance just makes not much sense. I suggest the following approach:

void LinkedList::DisplayLinkedList()
{
     Node *p = this->headNode;
     cout << "Displaying the list" << p << endl;
     while(p != NULL)
     {
         cout << "Node at " << p << "\n"; // prints the address
         cout << " value " << p->data << "\n";
         cout << " next: " << p->data << "\n"; // p->nextNode maybe?

         p = p->nextNode;
     }
}

---

For good practice don't spam endl ;)

Answer 2

There is at least a problem here:

     while(p != NULL)
     {
          cout << "Node at " << p << endl;
          cout << " value " << p->data << endl;
          cout << " next: " << p->data << endl;
          p = p->nextNode;   // <------------------+
     }                                             |
                                                   |
     // p = p->nextNode;   // this should be here -|

The p = p->nextNode must be inside the loop.

The other problem I can see is that head is actually never updated. It is initialized once to NULL . So when you call DisplayLinkedList(head) , head is NULL and obviously nothing will be printed.