Why is MIPS printing memory locations on the second time around - mips

Question

[Solved]
We were tasked with writing a bubble sort in MIPS and printing both the unsorted and sorted arrays
I've written the following code to accomplish this.
But when I go to run it, the original array prints as I would expect
Original Array
[34, 23, -1, -2, 0]
But then it prints this for the sorted array.
Sorted Array
[173496791, 1818324585, 1920090400, 686433, 1919898378]
(the Unicode squares do print)
array: .word 34, 23, -1, -2, 0, 89
length: .word 6
original: .asciiz "Original Array\n"
sorted: .asciiz "\nSorted Array\n"
.text
# main()
main:
la $a2, length
lw $a2, 0($a2) # a2 <= n_element
add $t9, $0, $0 # t9 : counter
la $a1, array # a1 = &array[0]
addi $t0, $0, 0 # for t0=0 to
addi $a2, $a2, -1 # because we do [i+1]
#says prints original array
la $a0, original
li $v0, 4
syscall
#prints array
jal print
#printing the string saying sorted
la $a0, sorted
li $v0, 4
syscall
#after printing original we reset the values?
la $a2, length
lw $a2, 0($a2) # a2 <= n_element
add $t9, $0, $0 # t9 : counter
add $t8, $0, $0 # t10 : counter 2
la $a1, array # a1 = &array[0]
addi $t0, $0, 0 # for t0=0 to
addi $a2, $a2, -1 # because we do [i+1]
#see if that worked it did
li $s0, 1 #boolean swapped = false 0=false 1=true
outerLoop:
beqz $s0, exit #if flag is false exit
li $s0, 0
addi $a1, $a1, 4 # next array offset
addi $t9, $t9, 1 # counter++
move $t8 $0
sub $s2, $a2, $t9
innerLoop:
bge $t8, $s2, exitInner
lw $t7, 0($a1) # t7 = array[i]
lw $t8, 4($a1) # t8 = array[i+1]
ble $t7, $t8, skip #if array[i]<= arr[i+1] skip the swapping
sw $t8, 0($a1) #a[i+1] = a[1]
sw $t7, 4($a1) #a[i] = a[i+1]
li $s0, 1#flag update
skip:
addi $t8, $t8, 1 #count
addi $a1, $a1, 4
j innerLoop
exitInner:
j outerLoop
exit:
la $a2, length
lw $a2, 0($a2) # a2 <= n_element
add $t9, $0, $0 # t9 : counter
addi $a2, $a2, -1 # because we do [i+1]
jal print
j quit#end Program
#print array
print:
lw $t7,0($a1) # t7 = array[i]
lw $t8,4($a1) # t8 = array[i+1]
add $a0, $t7, 0
addi $v0, $0, 1 #print int using system call the value stored in a0
syscall
addi $a0, $0, 0x20 # ascii 0x20 = ' '
addi $v0, $0, 11
syscall
addi $a1, $a1, 4 # next array offset
addi $t9, $t9, 1 # counter++
beq $t9, $a2, return #return to the where we were we were at beforehand
j print
return:
jr $ra
quit:
addi $v0, $0, 10
syscall
I think these are memory locations? I'm not sure why it's printing that as it is because I reset my counter values before trying to print. So the question is. Why am I printing memory locations? if that is what is happening
I was sorting in place so I had to reset the array before the print function.