---

Traversing linked lists

---

• Recall: linked lists
  • Recall that a linked list is chanied togeher using references

  • Example: a list with 3 nodes

    Memory: +---------------+ head: | 8000 |-------------------------------+ +---------------+ | | | | ....... | ....... | +---------------+ | 8000 | data1 | Linked list element 1 <-----+ +---------------+ | next=10000 |-------------------------------------+ +---------------+ | | | | +---------------+ | ....... | ....... | +---------------+ | 9000 | data3 | Linked list element 3 <----+ | +---------------+ | | | next=0 (null) | | | +---------------+ | | ....... | | ....... | | +---------------+ | | 10000 | data2 | Linked list element 2 | <-----+ +---------------+ | | next=9000 |-----------------------------+ +---------------+

  ---

  ---

  ---

  ---

• Structure of our example list
  • We use the following Node class definition in the examples:

    public class Node // Linked list element { int value; // 4 bytes Node next; // 4 bytes }

    You will need to adjust the offsets for different node structures

    The underlying technique on how to traverse the list will remain unchange

    ---

  • Suppose a linked list has been constructed; we just need to traverse the linked list.

    The list list start at the location given by this variable:

    Node head;

  ---

  ---

  ---

  ---

• List traversal example 1: sum the values stored in a list
  • The Java code to find the sum stored in all elements of the list:

    int sum; Node ptr; sum = 0; ptr = head; while ( ptr != null ) { sum = sum + ptr.value; ptr = ptr.next; }

    The flow chart of the above program is:

    In M68000 assembler code:

    move.l #0, sum move.l head, ptr WhileStart: move.l ptr, d0 cmp.l #0, d0 beq WhileEnd movea.l ptr, a0 move.l 0(a0),d0 * d0 = ptr.value add.l d0,sum move.l 4(a0),ptr * ptr = ptr.next bra WhileStart WhileEnd:

  • Program: click here
    Egtapi debug file: click here

  ---

  ---

  ---

  ---

• List traversal example 2: find max of the values stored in a list
  • The Java code to find the maximum of the values stored in a list:

    max = head.value; ptr = head.next; while ( ptr != null ) { if ( ptr.value > max ) max = ptr.value; ptr = ptr.next; }

    The flow chart of the above program is:

    In M68000 assembler code:

    movea.l head, a0 move.l 0(a0), max * max = head.value move.l 4(a0), ptr * ptr = head.next WhileStart: move.l ptr, d0 cmp.l #0, d0 beq WhileEnd movea.l ptr, a0 move.l 0(a0),d0 * d0 = ptr.value cmp.l max, d0 * ptr.value - max ble IfEnd * If ( ptr.value - max <= 0 ) then branch move.l d0, max * max = ptr.value IfEnd: move.l 4(a0),ptr * ptr = ptr.next bra WhileStart WhileEnd:

  • Program: click here
    Egtapi debug file: click here