[Resolved]adding even or odd numbers in array, MIPS32

[voidflux]

Hello everyone. I go this program to add up all the elements in the array and dispaly its value, but now i'm trying to add up only even numbers if the user enters 0. If they don't enter 0, then it will add up all the odd and dispaly the sum. Store result in register $t1.

Here is the code i have to add up all the elements in the array:

.data
.align 2
item: .word 1
.word 2
.word 3
.word 4
.word 5
.word 6
.word 7
.word 8
.word 9
.word 10
.align 0
str: .asciiz "The answer is "
cr: .asciiz "\n"
.text
.align 2
.globl main
main: ori $2, $0, 5
syscall
move $t0, $2


li $t2, 0            #i=0

loopsum:
    bge $t2, 10, done    #while (i < 10)
    mul $t3, $t2, 4        #4*i
    la $t4, item        #calculate &item[i] address
    add $t3, $t3, $t4
    lw $t5, 0($t3)        #load item[i] into $t5
    

    add $t1,$t1,$t5        #$t1 = $t1 + item[i]
    add $t2, $t2, 1        #i++
b    loopsum




done: ori $2, $0, 4
la $4, str
syscall
ori $2, $0, 1
add $4, $t1, $0
syscall
ori $2, $0, 4
la $4, cr
syscall 

This will print The answer is: 55.

Now i'm having alot of trouble on how i can implement this in asm. What i want is:

if(userinput == 0)
loop through array if(item[i] % 2 == 0) { $t1 = $t1 + item[i] };
else
loop through array if(item[i] % 2 != 0) { $t1 = $t1 + item[i] };

So if the user inputs 0, that means sum up all the even numbers in the array
else
sum up all the odd numbers in the array.

Will this work to check to see if the remainder is 0? (if it is 0, then u konw its an even number.)

lw $t5, 0($t3)        #load item[i] into $t5
    
    ori $t6, $0, $t5    #t6 = $t5
    div $t6,2        #$HI = $s % $t    
beq $HI,0,loopeven    # if ($t6 % 2) == 0, even number 

THanks!

[CornedBee]

Can't help you with your exact problem, but generally I would do it like this: (pseudo-assembly)

set $r1 if $userinput eq 0 (is there a seq instruction?)
loop:
  load next num into $r2
  div $r2 by 2, remainder into $r3
  jump to loop if $r3 eq $r1 (in other words, if the remainder is 1 and the user entered
                            0, skip the counting step. Same if rem is 0 and ui is != 0)
  add $r2 to $total
  jump to loop unconditionally

You still need loop termination in this, of course.

[voidflux]

Thanks for the reply, this is how I ended up doing it:

.data
	.align 2
item: 	.word 1
	.word 2
	.word 3
	.word 4
	.word 5
	.word 6
	.word 7
	.word 8
	.word 9
	.word 10
	.align 0
str: 	.asciiz "The answer is "	
cr:	.asciiz "\n"
	.text
	.align 2
	.globl main
main:	ori $2, $0, 5
	syscall
	move $t0, $2

	la $t2, item			#load base address of item array into $t2
	beq $0, $t0, even		#if user inputed 0, go to even loop
oloop:	lw $t3, ($t2)		#load item[i] into $t3
	add $t1, $t1, $t3		#add t1 = t1+t3
	beq $t3, 9, done		#if t3 == 9, jump to done
	addi $t2, $t2, 8		#t2 = t2 + 8, jump to next odd word, which is 8 positons away 
	j oloop
even:	addi $t2, $t2, 4
eloop:	lw $t3, ($t2)
	add $t1, $t1, $t3
	beq $t3, 10, done
	addi $t2, $t2, 8
	j eloop

done: 	ori $2, $0, 4
	la $4, str
	syscall
	ori $2, $0, 1
	add $4, $t1, $0
	syscall
	ori $2, $0, 4
	la $4, cr
	syscall

[CornedBee]

Ah, but of course that works only for this particular sequence. What about a random sequence of numbers?

[voidflux]

hah i know i took the easy way out, if i kept working at it, i think i would have gotten it. But i asked the professor and she said don't mess
div or $HI register yet, it was an exercise in understanding that words are 4 bits long, not 1.

So if you want to go to the next memory location, you must jump +4 and not +1 or else you'll end up just moving 1 bit over. If i had more time I would work on it.

[CornedBee]

bytes, not bits

[voidflux]

oops, that would be bad if i wrote that on the exam, thanks!