MIPS assembly - removing vowels from an input string - string

Question

The following is my code for a MIPS assembly program that is intended to remove vowels from an input string and then print the new string. As it stands, the program simply does not remove the vowels and just reprints the same string that was input.
.text
.globl main
main:
# display prompt
li $v0, 4
la $a0, prompt
syscall
# accept input string
li $v0, 8
la $a0, str
li $a1, 82
syscall
li $t0, 0 # add a null to the bottom of the stack
subu $sp, $sp, 4
sw $t0, ($sp)
li $t1, 0 # initiate index
pushloop:
lbu $t0, str($t1) # (I think the program is placing a zero in $t0 here, thus causing it to branch immediately to poploop, which then branches to the end since only the null has been pushed onto the stack)
beqz $t0, poploop # once we reach null char, finish
subu $sp, $sp, 4
sw $t0, ($sp)
addiu $t1, $t1, 1
j pushloop
nop
# $t1 is not reset
poploop:
lw $t0, ($sp)
addu $sp, $sp, 4
beqz $t0, done # once we reach null char, finish
nop
# check if vowel
li $t2, 0x61 # a
beq $t0, $t2, vowel
nop
li $t2, 0x65 # e
beq $t0, $t2, vowel
nop
li $t2, 0x69 # i
beq $t0, $t2, vowel
nop
li $t2, 0x6F # o
beq $t0, $t2, vowel
nop
li $t2, 0x75 # u
beq $t0, $t2, vowel
nop
# if not a vowel, store it at current index in string
sb $t0, str($t1)
j decrement
nop
vowel: # if vowel, remove character
li $t0, 0
sb $t0, str($t1)
decrement:
addiu $t1, $t1, -1 # decrement index
j poploop
nop
done:
li $v0, 4
la $a0, str
syscall
li $v0, 10 # exit program
syscall
nop
.data
str: .space 82
prompt: .asciiz "Input a string:\n"

Answer 1

So. I took a look at what you've writen and I've fixed it.
My first thought is I don't know what you are doing with the stack and stack pointer ($sp). It didn't seem necessary so I took it out.
Next is that the approach is wrong. Your approach is to search the string and replace every lower-case vowel (or at least 'a', 'e', 'i', 'o' and 'u') with a 0. This will not work.
If you think about a typical C string, they are delimeted by a 0, so if you take the string My name is Jeoff and apply your algorithm you will get My n\0m\0 \0s J\0\0ff which of course will print as My n.
So instead, I chose a separate algthm that chooses to not store anything in the case of a vowel and instead shift all following characters over by 1. In so doing we can easily remove all vowels from a string without requiring a secondary buffer.
Take a look below:
.text
.globl main
main:
# display prompt
li $v0, 4
la $a0, prompt
syscall
# accept input string
li $v0, 8
la $a0, str
li $a1, 82
syscall
li $t1, 0 # initiate index
li $t3, 0 # vowel count
poploop:
lb $t0 str($t1)
# check if vowel
li $t2, 'a' # a
beq $t0, $t2, vowel
nop
li $t2, 'e' # e
beq $t0, $t2, vowel
nop
li $t2, 'i' # i
beq $t0, $t2, vowel
nop
li $t2, 'o' # o
beq $t0, $t2, vowel
nop
li $t2, 'u' # u
beq $t0, $t2, vowel
nop
# if not a vowel, store it at current index in string less vowel count
sub $t2, $t1, $t3
sb $t0, str($t2)
j next
nop
vowel: # if vowel, inc count
addi $t3, $t3, 1
next:
addi $t1, $t1, 1
beqz $t0, done # once we reach null char, finish
nop
j poploop
nop
done:
li $v0, 4
la $a0, str
syscall
li $v0, 10 # exit program
syscall
nop
.data
str: .space 82
prompt: .asciiz "Input a string:\n"