汇编 MIPS : Initializing and suming up an array

Question

假设我们有一个包含 10 个数字的数组，我们想将这些数字相加到一个变量中，如下所示:

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];

当我尝试“组装 mips”时，特定的一行出现了:

.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

我无法理解如何将“arr[i]”指向程序集。

正确的表达方式是什么？

我是否必须为阵列的每个数字进行另一个注册表？

Answer 1

感谢上面的 Jester，正确的代码如下:

.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