Dynamic Linked List Lesson xx

1. Dynamic Linked List Lesson xx

2. Objectives • Memory from the heap • Dynamic memory allocation using the new operator • Build a dynamic linked list • Another way of traversing a linked list • Deallocating memory using delete operator • Dynamic linked list variation

3. Memory Setup Operating System Program Stack Heap

4. Operator new Operating System int * ptr; ptr = new int; Program Stack ptr 9ab Heap 9ab int

5. Program to Create a Dynamic Linked List (Part 1) struct entry { int value;   entry* next; }; int main() {   entry* temp, * head = 0; // initially empty list /*allocate the first node*/ head= new entry; temp = head; // create 4 more nodes   for (inti = 0; i < 4; i++)   {cin >>temp‑>value; /*put address of next cell in .next member */ temp‑>next = new entry; temp = temp‑>next; //temp now points to last node   } temp‑>value = 0; /*last node points to nothing*/ temp‑>next = 0; /*it's a dummy node*/

6. (Part 2) Program to Create a Dynamic Linked List   // traverse the list for (entry* p = head; p; p = p->next)   { cout << "node address: " << p       << " value " << p->value       << " next " << p->next << endl;   }   // deallocate memory   while (head)   {     entry* n = head->next;     delete head;     head = n;   }   return 0; }

7. Starting the Linked List struct entry { int value;   entry* next; }; int main() {   entry* temp, * head = 0; // initially empty list temp head 0

8. Create 1st Node /*allocate the first node*/ head= new entry; temp = head; (76c) temp 76c head 76c

9. Loop for Input and Creating Additional Nodes   for (inti = 0; i < 4; i++)   {cin >>temp‑>value; /*put address of next cell in .next member */ temp‑>next = new entry; temp = temp‑>next; //temp now points to last node   } (76c) 76c 42 temp head 76c

10. Adding Additional Nodes   for (inti = 0; i < 4; i++)   {cin >>temp‑>value; /*put address of next cell in .next member */ temp‑>next = new entry; temp = temp‑>next; //temp now points to last node   } 76c temp (76c) (88b) 42 88b head 76c

11. Advancing temp to Next Node   for (inti = 0; i < 4; i++)   {cin >>temp‑>value; /*put address of next cell in .next member */ temp‑>next = new entry; temp = temp‑>next; //temp now points to last node   } 88b temp (76c) (88b) 42 88b head 76c

12. Repeating the Loop   for (inti = 0; i < 4; i++)   {cin >>temp‑>value; /*put address of next cell in .next member */ temp‑>next = new entry; temp = temp‑>next; //temp now points to last node   } (76c) (88b) (4af) 97 4af 42 88b head 76c 4af temp

13. End of the Loop head 42 97 35 17 temp

14. Terminating the Linked List  temp‑>value = 0; /*last node points to nothing*/ temp‑>next = 0; /*it's a dummy node*/ head 17 0 0 42 97 35 temp

15. Printing the Linked List   // traverse the list for (entry* p = head; p; p = p->next)   { cout << "node address: " << p       << " value " << p->value       << " next " << p->next << endl;   } head 17 0 0 42 97 35 p temp

16. Output node address: 76c value 42 next 88b node address: 88b value 97 next 4af node address: 4af value 35 next 57d node address: 57d value 17 next a6e node address: a6e value 0 next 0 (57d) (a6e) (76c) (88b) (4af) 17 a6e 0 0 97 4af 42 88b 35 57d head 76c a6e 0 temp p

17. Deleting the Linked List // deallocate memory   while (head)   {     entry* n = head->next;     delete head;     head = n;   } head 17 0 0 42 97 35 n

18. After Deleteing 1st Node // deallocate memory   while (head)   {     entry* n = head->next;     delete head;     head = n;   } head 17 0 0 97 35 n

19. Variations of a Linked List // create 4 more nodes   for (inti = 0; i < 4; i++)   {cin >>temp‑>value; temp‑>next = new entry; temp = temp‑>next;   } int n; cout << “How many nodes ? “; cin >> n; // create n-1 more nodes   for (inti = 0; i < n-1; i++)   {cin >>temp‑>value; temp‑>next = new entry; temp = temp‑>next;   }

20. Summary Memory from the heap Dynamic memory allocation using the new operator Build a dynamic linked list Another way of traversing a linked list Deallocating memory using delete operator Dynamic linked list variation