What does this MIPS program do? - mips

Question

I know the program loads the address of .word 5 which is the initial value, I'm getting confused as to what the program is actually doing. Is it comparing stacks of the word or adding.
.data
arg: .word 5
.text
.globl main
main:
la $t3, arg
lw $t2, 0($t3)
lw $t3, 0($t3)
addi $t1, $zero, 0
beqz $t2, fin
fori:
add $t1, $t1, $t2
addi $t3, $t3, -1
bnez $t3, fori
fin:
li $v0, 10
syscall

Answer 1

Looks like it does addition based on the first item in the space as it doesn't access the array from inside the loop.
So for instance if you input a character representing the ascii value of 5 (not '5' itself iirc) you might see something like this:
int t2 = 5, t3 = 5, t1 = 0;
do {
t1 += t2;
t3 -= 1;
} while(t3 > 0);
Someone has already answered but I shall provide the translation I did:
.data
arg: .word 5
.text
.globl main
main:
la $t3, arg # load-address of arg into t3
lw $t2, 0($t3) # load-word from the address in t3 + 0 offset, like t3[0]
lw $t3, 0($t3) # same but to t3
addi $t1, $zero, 0 # initialize t1 to 0? weird they don't use li
beqz $t2, fin # if t2 is zero jump to fin:
fori:
add $t1, $t1, $t2 # t1 = t1 + t2
addi $t3, $t3, -1 # t3 -= 1
bnez $t3, fori # if(t3 != 0) goto fori
fin:
li $v0, 10 # load immediate 10 into v0, which is the syscall to terminate the program
syscall

Answer 2

Disclaimer: I don't have any direct MIPS experience, I just have been around a number of assembly languages.
That said, I think what the program does is calculating the square of 'arg' by repeated addition - in this case 5 * 5 = 25. However, the result in $t1 doesn't seem to be stored anywhere, instead the program just exits.