对于C中具有阵列的循环到具有偏移的MIPS

Question

So I have an assignment problem which asks for a for loop in C to be converted into MIPS. My professor moves extremely fast so I couldn't catch half of what he said. Here is the code:

for (i=0; i<10; i++){
    a[i] = b[i] + c[i];
}

This fragment is stored in memory starting from location 00000100 Hex. Convert this code to MIPS and provide numeric offsets for each branch or jump instruction used.

I don't quite understand the use of offsets. From the lecture slides given to us, it seems load word and store word commands are used for the offsets, which are also used for the arrays but I'm not sure how to go about it. Below is something I put together based on other solutions I saw, but am of course open to changes. I'm hoping it's at least going in the right direction. Any help would be appreciated.

#t0 = i
#s0 = a
#s1 = b
#s2 = c
#t3, t4, t5, t6, t7 = free

loop:
   bgt $t0,9,exit    #exit before i reaches 10
   addi $t3,$s1,$t0  #temp reg $t3 = address of b[i]
   addi $t4,$s2,$t0  #temp reg $t4 = address of c[i]
   lw $t5,0($t3)     #temp reg $t5 = c[i]
   lw $t6,0($t4)     #temp reg $t6 = a[i]
   add $t3,$t5,$t6   #temp reg $t10 = b[i] + c[i]
   addi $t7,$s0,$t0  #temp reg $t7 = address of a[i]
   sw $t3,0($t7)   #store word a[i] = b[i] + c[i]
   addi $t0,$t0,1    #increment i by 1
   j loop            #jump to start of loop
exit:

Answer 1

From what I can see, you are storing it in the same index each time. In MIPS, every 4 bytes is a word, so you must store it in 0, 4, 8, etc. Also, you need to allocate memory for your array before you start using one. Here's an example.

li $v0, 9       # create an array, start address in $v0
li $a0, 80      # allocate 80 bytes, or 20 words
syscall
move $t0, $v0   # move from $v0 (temp) to $t0

Check out this tutorial and see if it helps at all.