ASM Language: Factorial Algorithm Error?

Question

I can't seem to find where I am wrong in my algorithm, my specific problem is that i seem to print the same value for all integer inputted in the console: here is my code

main:

li $v0, 5
syscall
jal factorial
li $v0, 10
syscall

factorial:
    ###preamble###
    subu $sp, $sp, 32
    sw $ra, 28($sp)
    sw $fp, 24($sp)
    addu $fp, $sp, 32
    sw $a0, 20($fp)
    ###preamble###

    lw $v0, 20($fp)
    bgtz $v0, multiply
    li $v0, 1
    j end

multiply:
    lw $v1, 20($fp)
    subu $v0, $v1, 1
    move $a0, $v0
    jal factorial

    lw $v1, 20($fp)
    li $t3, 0
    li $t2, 1
    b multi

    multi:
        beq $t2, $v0, endLOOP
        add $t3, $t3, $v1
        add $t2, $t2, 1
        b multi

    endLOOP:
        move $v0, $t3

end:
    lw $ra, 28($sp)
    lw $fp, 24($sp)
    addu $sp, $sp, 32
    move $a0, $v0
    li $v0, 1
    syscall
    jr $ra

in this code i always seem to print a value of 10 while here in my second code, i always get an 11

main:

li $v0, 5
syscall 
jal factorial
li $v0, 10
syscall

factorial:
    ###preamble###
    subu $sp, $sp, 32
    sw $ra, 28($sp)
    sw $a0, 24($sp)
    li $v0, 1
    ble $a0, $zero, end
    b multiply
    ###preamble###

multiply:
    addi $a0, $a0, -1
    jal factorial
    lw $a0, 24($sp)
    b multi
    multi:
        beq $t2, $v0, endLOOP
        add $t3, $t3, $a0
        add $t2, $t2, 1
        b multi

    endLOOP:
        move $v0, $t3

end:
    lw $ra, 28($sp)
    addu $sp, $sp, 32
    move $a0, $v0
    li $v0, 1
    syscall
    jr $ra

also, please note that we are not allowed to use the mult function so please dont ask why i am not using it. Thank you for your help :D

score 1 · Accepted Answer · answered Jul 15 '13 at 13:54

1

I see a couple of problems with your code:

li $v0, 5
syscall
jal factorial

syscall 5 (read_int) will return the value in $v0, so when you enter factorial for the first time you'll have the argument in $v0. You could fix that by moving the value to $a0 before the jal.

li $v0, 1
syscall
jr $ra

Wouldn't this print make more sense outside of the factorial function? Since $v0 is supposed to hold the function's return value, consider what will happen when you return from factorial(2) to factorial(3) and you've overwritten $v0 with the value 1.

answered Jul 15 '13 at 13:54

Michael

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i got the first part of your answer for my question, but on the second one, i don't quite get it. But to make things much easier for me and for you as well, i'll just ask on how can i print the final value of the factorial in the console? Im not so sure on how to do it – Patrick Leiniel Domingo Jul 15 '13 at 14:11
_"the second one, i don't quite get it"_: I'm saying that you're overwriting the return value with the value 1, so that as you work your way back "out" through the recursion chain, `factorial(n)` will always get the value 1 from `factorial(n-1)` no matter what `n` is. I suggest that you move the print to after the _first_ call to `factorial` (i.e. right before the `exit` `syscall`). (I'm talking about your first implementation now, I didn't look at the second version) – Michael Jul 15 '13 at 14:17
Oh I see. The program is working fine now. Thank you for your help! :D – Patrick Leiniel Domingo Jul 15 '13 at 14:32

ASM Language: Factorial Algorithm Error?

1 Answers1