MIPS doubly linked list

# jakotheshadows
#
# CS3421 Lab 3
# Fall 2011
#
# This program contains a function that removes elements from one list that
# occur in another.

    .text
# test program starts here
    la	$a0,list1       # load argument registers
    la	$a1,list1end
    la	$a2,list2
    la	$a3,list2end
    jal	remove          # call the function
    lw  $s0,8($v0)	# get reference to node after header
loop:
    lw  $t0,8($s0)	# see if next pointer is null (trailer)
    li  $t1,-1
    beq $t0,$t1,done    # if so, done
    lw  $a0,0($s0)      # get data from node and print it
    li  $v0,1
    syscall
    la  $a0,space       # print a space
    li  $v0,4
    syscall
    lw  $s0,8($s0)      # get reference to next node
    b   loop            # try again
done:
    li  $v0,10          # exit
    syscall
# the lists
    .data

# the first list
list1:			# header
    .word 0
    .word -1
    .word l1a
l1a:
    .word -1
    .word list1
    .word l1b
l1b:
    .word 4
    .word l1a
    .word l1c
l1c:
    .word 28
    .word l1b
    .word l1d
l1d:
    .word 319
    .word l1c
    .word list1end
list1end:		# trailer
    .word 0
    .word l1d
    .word -1

# the second list
list2:
    .word 0		# header
    .word -1
    .word l2a
l2a:
    .word 0
    .word list2
    .word l2b
l2b:
    .word 4
    .word l2a
    .word l2c
l2c:
    .word 9
    .word l2b
    .word l2d
l2d:
    .word 319
    .word l2c
    .word list2end
list2end:		# trailer
    .word 0
    .word l2d
    .word -1

space:
    .asciiz  " "
    .text

# your remove function starts here:
# $a0 list1, $a2 list2
remove:
	move $s0, $a0 # $s0 = list1.getHead(), $t0 has base address of list1 header
	move $s1, $a2 # $s1 = list2.getHead(), $t1 has base address of list2 header
	lw $t0, 8($s0)# $t0 = $s0.getNext(), $t0 has base address of list1's first element
	lw $t1, 8($s1)# $t1 = $s1.getNext(), $t1 has base address of list2's first element
doWhile:
	li $t2, -1
	lw $t3, 8($t1)
	lw $t3, 8($t3)
	beq $t2, $t3, return
	lw $t2, 0($t0)# $t2 = $t0.getData(), $t2 has data from list1's current node
	lw $t3, 0($t1)# $t3 = $t1.getData(), $t3 has data from list2's current node
	bne $t2, $t3, skip # if($t2==$t3) the node containing identical data is removed from list1
	lw $t2, 4($t0)# $t2 has base of previous
	lw $t3, 8($t0)# $t3 has base of next
	sw $t3, 8($t2)# $t2.setNext($t3)
	sw $t2, 4($t3)# $t3.setPrev($t2), the duplicate has now been removed from list1
	lw $t0, 8($s0)# reset $t0 to first node after the header of list1
	lw $t2, 4($t1)# $t2 has base of previous
	lw $t3, 8($t1)# $t3 has base of next
	sw $t3, 8($t2)# $t2.setNext($t3)
	sw $t2, 4($t3)# $t3.setPrev($t2), the node has been removed from list 2
	lw $t1, 8($s1)# reset $t1 to first node after the header of list2
skip:
	lw $t0, 8($t0)# $t0 = $t0.getNext(), advance through list1
	lw $t2, 8($t0)# $t2 = $t0.getNext(), check if next node is null
	li $t3, -1 
	bne $t3, $t2, doWhile
	lw $t2, 4($t1)# $t2 has base of previous
	lw $t3, 8($t1)# $t3 has base of next
	sw $t3, 8($t2)# $t2.setNext($t3)
	sw $t2, 4($t3)# $t3.setPrev($t2), the node has been removed from list 2
	lw $t1, 8($s1)# reset $t1 to first node after the header of list2
	lw $t0, 8($s1)# reset $t0 to first node after the header of list1
	j doWhile	
return:
	move $v0, $s0
	jr $ra