Simple Array Manipulation MIPS Assembly - arrays

Question

I'm new to MIPS assembly and I'm trying to learn how to use arrays. I understand that arrays are declared in the .data segment and then you load the address of the array to the register. However, I'm confused on accessing elements of the the array and changing the values. Here is a simple program I wrote in C that I'm trying to convert into assembly. Any explanation/code is greatly appreciated!
void addArray (int arrA[], int arrB[], int i, int j) {
arrB[j] = arrA[i] + arrA[i + 1];
printf("%d", arrB[j]);
}
int main(void) {
int arrA [] = {1,2,3,4,5,6};
int arrB [] = {1,1,1,1,1,1};
int i = 2;
int j = 3;
addArray(arrA, arrB, i, j);
return 0;
}

Answer 1

Try this code
.data
arrA: .word 1,2,3,4,5,6
arrB: .word 1,1,1,1,1,1
.text
.globl main
main:
la $a0,arrA #load address of arrA into $a0 register (first parameter)
la $a1,arrB #load address of arrB into $a1 register (second parameter)
li $a2,4 # load $a2 register with 3 (threeth parameter)
li $a3,3 # load $a3 register with 3 (fourth parameter)
jal addArray # call addArray function
exit: # syscall to exit the programm
li $v0, 10
syscall
addArray:
addi $sp,$sp,-4 #allocate space in stack
sw $ra,0($sp)
move $t0,$a2 #copy a content of $a2 register into $t0
move $t1,$a3 #copy a content of $a3 register into $t1
mulu $t0,$t0,4 # multiplication by 4 with the value of $t0, because each number have 4 bytes as offset
addu $a0,$a0,$t0 #
lw $t3,($a0) # arrA[i]
addi $a0,$a0,4 # i+1
lw $t4,($a0) # arrA[i+1]
add $t5,$t3,$t4 # arrA[i] + arrA[i + 1]
mul $t1,$t1,4 # j
addu $a1,$a1,$t1
sw $t5,($a1) # arrB[j] = arrA[i] + arrA[i + 1];
print_out:
lw $s0,($a1)
li $v0, 1 #print the input number
move $a0, $s0
syscall
lw $ra,0($sp) # free stack and return to main
addi $sp,$sp,4
jr $ra