Coursal
Coursal

Reputation: 1387

Assembly MIPS: Initializing and suming up an array

Let's say we have an array of 10 numbers and we want to sum those numbers to a variable like this:

int arr[10]= {1, 15, 0, -3, 99, 48, -17, -9, 20, 15};

sum = 0;

for(i=0; i<10; i++)

   sum = sum + arr[i];

When I try to "assembly mips" this, a particular line goes off:

.data
arr: .word 1, 12, 0, -3, 99, 48, -17, -9, 20, 15

.text
.globl main
main:

add $t0, $zero, $zero   #counter i
add $t1, $zero, $zero   #sum

la  $t2, arr      #loading the address of the array to a register

loop:
slti    $t3, $t0, 10        #i<10
beq $t3, $zero, EXIT

add $t1, $t1, $t2($t0)     #wrong wrong very wrong

addi    $t0, $t0, 1        #i++
j loop
EXIT:

li  $v0, 10
syscall

I have a problem understanding how to express the "arr[i]" point to Assembly.

What is the right expression for it?

Do I have to take another registry for each number of the array?

Upvotes: 3

Views: 17037

Answers (1)

Coursal
Coursal

Reputation: 1387

Thanks to Jester from above, the correct code for this is the following:

.data
arr: .word 1, 12, 0, -3, 99, 48, -17, -9, 20, 15

.text
.globl main
main:

add $t0, $zero, $zero   #counter i
add $t1, $zero, $zero   #sum

la  $t2, arr

loop:
slti    $t3, $t0, 10        #i<10
beq $t3, $zero, EXIT

lw  $t4, ($t2)  #$t4 = arr[i]
addi    $t2, $t2, 4

add $t1, $t1, $t4   #sum = sum + arr[i]

addi    $t0, $t0, 1 #i++
j loop
EXIT:

add $a0, $zero, $t1     #moving the sum to $a0 register for printing

li  $v0, 1
syscall

li  $v0, 10
syscall

Upvotes: 5

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