3.  Review building a complete linked list  List traversal in main ( )  List traversal using a function

4. struct entry { int value; entry* next; }; int main() { entry n1, n2, n3; entry* head = &n1; n1.next = &n2; n2.next = &n3; n3.next = 0; n1.value = 100; n2.value = 200; n3.value = 300; entry * temp; temp= head; // traverse the list while (temp != NULL) { cout value next << endl; temp = temp->next; } return 0; }

5. entry n1, n2, n3; entry* head = &n1; n1.next = &n2; n2.next = &n3; n3.next = 0; n1.value = 100; n2.value = 200; n3.value = 300; n1 n1.valuen1.next n2 n2.valuen2.next n3 n3.valuen3.next 100 200300 (56c) (7fa)(88b) 7fa 88b 0 56c head

6. n1 n1.valuen1.next n2 n2.valuen2.next n3 n3.valuen3.next 100 200300 (56c) (7fa)(88b) 7fa 88b 0 56c head 56c temp entry * temp; temp= head;

7. n1 n1.valuen1.next n2 n2.valuen2.next n3 n3.valuen3.next 100 200300 (56c) (7fa)(88b) 7fa 88b 0 56c head 56c temp while (temp != NULL) { cout value next << endl; temp = temp->next; }

8. n1 n1.valuen1.next n2 n2.valuen2.next n3 n3.valuen3.next 100 200300 (56c) (7fa)(88b) 7fa 88b 0 56c head 56c temp 56c 100 7fa temp temp->value temp->next

9. temp = temp->next; n1 n1.valuen1.next n2 n2.valuen2.next n3 n3.valuen3.next 100 200300 (56c) (7fa)(88b) 7fa 88b 0 56c head 7fa temp

10. while (temp != NULL) { cout value next << endl; temp = temp->next; } n1 n1.valuen1.next n2 n2.valuen2.next n3 n3.valuen3.next 100 200300 (56c) (7fa)(88b) 7fa 88b 0 56c head 7fa temp

11. 56c 100 7fa 7fa 200 88b 88b 300 0 n1 n1.valuen1.next n2 n2.valuen2.next n3 n3.valuen3.next 100 200300 (56c) (7fa)(88b) 7fa 88b 0 56c head 0 temp

12. struct entry { int value; entry* next; }; void printList(const entry*); // prototype int main() { entry n1, n2, n3; n1.next = &n2; n2.next = &n3; n3.next = 0; entry* head = &n1; printList(head); return 0; } void printList(const entry* pointer) { while (pointer != 0 ) { cout value next << endl; pointer = pointer->next; } }

13. struct entry { int value; entry* next; }; void printList(const entry*); // prototype int main() { entry n1, n2, n3; n1.next = &n2; n2.next = &n3; n3.next = 0; entry* head = &n1; printList(head); return 0; } void printList(const entry* pointer) { while (pointer != 0 ) { cout value next << endl; pointer = pointer->next; } } 1 2 3

14. n1 n1.valuen1.next n2 n2.valuen2.next n3 n3.valuen3.next 100 200300 (56c) (7fa)(88b) 7fa 88b 0 56c head 56c pointer void printList(const entry* pointer) { while (pointer != 0 ) { cout value next << endl; pointer = pointer->next; } }

15. Building a complete linked list List traversal in main ( ) List traversal using a function