Question

***CODE MUST BE IN C++***

Using the linked list in "basic linked list" which has a STRUCTURE for each node, write FUNCTION which starts at the head and outputs the value for each node until the last node is reached. Note: your function should work with the structure that is in this program. Please do not use the example which is for a class, but this example canbe helkpful.  Also note that the function must work no matter how many nodes are in the list.

// ILLUSTRATES LINKED LIST
// NOTE USING POINTER TO ACCESS 
// without naming variable

#include<iostream>
using namespace std;

struct ListNode
  {
       double value;  // data member
       ListNode *next; // pointer to next node 
                       // NOTE this is pointer to same struct type
       ListNode *prev;
  };
  
  int main()
  {
      ListNode *head;  // define pointer to be used as list head
      
       // create first node with data value 12.5
       head = new ListNode;  // Allocate new node 
       head->value = 12.5; // store the value
       head->next = NULL; // signify end of list
       
       // create second node with data value 13.5
       ListNode *secondPtr = new ListNode;
       secondPtr->value = 13.5;
       secondPtr->next = NULL; // second node is end of list
       head->next = secondPtr; // first node points to second node
       
       //  print the list
       cout<< "First item is "<< head->value<<endl;
       cout<< "Second item is "<<head->next->value<<endl;
       
       system("pause");
       return }

Answer 1

// ILLUSTRATES LINKED LIST
// NOTE USING POINTER TO ACCESS
// without naming variable

#include<iostream>

using namespace std;

struct ListNode {
    double value;  // data member
    ListNode *next; // pointer to next node
    // NOTE this is pointer to same struct type
    ListNode *prev;
};

void printList(ListNode * head) {
    cout << "List is: ";
    while (head != NULL) {
        cout << head->value << " ";
        head = head->next;
    }
    cout << endl;
}

int main() {
    ListNode *head;  // define pointer to be used as list head

    // create first node with data value 12.5
    head = new ListNode;  // Allocate new node
    head->value = 12.5; // store the value
    head->next = NULL; // signify end of list

    // create second node with data value 13.5
    ListNode *secondPtr = new ListNode;
    secondPtr->value = 13.5;
    secondPtr->next = NULL; // second node is end of list
    head->next = secondPtr; // first node points to second node

    //  print the list
    cout << "First item is " << head->value << endl;
    cout << "Second item is " << head->next->value << endl;

    printList(head);

    system("pause");
    return 0;
}