find substring and indices in mips - mips

Question

im trying find out substring and first occurrence indices. but something wrong. im comparing each element of pattern array and each element of string array until pointer reach to '\0'. whats the problem. algorithm is totaly wrong ?
#Note: $v0 is a symbolic name used by the assember for $2.
# $a0 is a symbolic name used by the assember for $4.
.data
prompt_str: .asciiz "Please type a text string: "
prompt_ptr: .asciiz "Please type a pattern string: "
print_yes: .asciiz "Yes, there is a match."
print_no: .asciiz "No, there is no match."
text_str: .asciiz "Text string : "
pattern_str: .asciiz "Pattern string : "
print_out: .asciiz "Output to be produced :"
print_dash: .asciiz "----------------------"
print_index: .asciiz "Starting index :"
print_msg : .asciiz "Length of longest partial match = "
nl: .asciiz "\n"
print_outer: .asciiz "please enter string"
str : .space 81
ptr : .space 81
tmp : .space 81
.text
main: la $a0, prompt_str
li $v0, 4 #print_string command.
syscall
la $a0,str #read string
li $a1,81
li $v0,8
syscall
la $t0,str #move string to $t0
la $a0,prompt_ptr
li $v0,4 #print pattern command
syscall
la $a0,ptr #read pattern
li $a1,81
li $v0,8
syscall
la $t1,ptr #move pattern to $t1
lb $t2,0($t0) #pointer first element array of string
move $t4,$t2 #address pointer of $t2
lb $t3,0($t1) #pointer first element array of pattern
outer_loop : beq $t2,$0,end_outer_loop
j inner_loop
inner_loop : beq $t2,$0,end_inner_loop
beq $t3,$0,end_inner_loop
beq $t2,$t3,end_inner_loop
addiu $t2,$t2,1
addiu $t3,$t3,1
j inner_loop
end_inner_loop :bne $t3,$0,inc_ptr
j print_match
inc_ptr : add $t2,$t4,1
j outer_loop
end_outer_loop :la $a0,print_outer
li $v0,4
syscall
print_match : la $a0,text_str #print string
li $v0,4
syscall
move $a0,$t0
li $v0,4
syscall
la $a0,nl #print newline character
li $v0,4
syscall
la $a0,pattern_str #print pattern string
li $v0,4
syscall
move $a0,$t1
li $v0,4
syscall
la $a0,nl #print newline character
li $v0,4
syscall
la $a0,print_out #print output line and newline character
li $v0,4
syscall
la $a0,nl
li $v0,4
syscall
la $a0,print_dash
li $v0,4
syscall
la $a0,print_yes
li $v0,4
syscall
la $a0,print_index #print starting index
li $v0,4
syscall
li $v0,10
syscall
end_loop : li $v0,10
syscall

Answer 1

I used your code for a similar project, in inner_loop you don't have a proper bne
I just put one bne and now it print just the string whit the substring..

Answer 2

.text
.globl main
main:
li $v0, 4
la $a0, msg1
syscall
li $v0, 8
la $a0, strMain
li $a1, 99
syscall
li $v0, 4
la $a0, msg2
syscall
li $v0, 8
la $a0, strSub
li $a1, 99
syscall
la $a0,strMain
jal findLengthString
move $a2, $v0
la $a0, strSub
jal findLengthString
move $a3, $v0 # M
sub $a2, $a2, $a3 # N-M
la $a0, strMain
la $a1, strSub
jal subStringMatch
move $t1, $v0
li $v0, 1
move $a0, $t1
syscall
exit:
li $v0, 10
syscall
lb $t9, endline
findLengthString:
li $t0, -1
move $s0, $a0
loop_fls:
lb $t1, 0($s0)
beq $t1, $t9, foundLength
addi $t0, $t0, 1
addi $s0, $s0, 1
j loop_fls
foundLength:
move $v0, $t0
jr $ra
subStringMatch:
li $t0, 0 #i
loop1:
bgt $t0,$a2, loop1done
li $t1, 0 #j
loop2:
bge $t1, $a3, loop2done
add $t3, $t0, $t1
add $t4, $a0, $t3
lb $t3, 0($t4) # main[i+j]
add $t4, $a1, $t1
lb $t4, 0($t4) # sub[j]
# if a0[i + j] != a1[j]
bne $t3, $t4, break1
addi $t1, $t1, 1
j loop2
loop2done:
beq $t1, $a3, yesReturn
j break1
yesReturn:
move $v0, $t0
jr $ra
break1:
addi $t0, $t0, 1
j loop1
loop1done:
li $v0, -1
jr $ra
.data
msg1: .asciiz "Enter Main String: "
msg2: .asciiz "Enter String to Check SubString: "
strMain: .space 100
strSub: .space 100
endline: .asciiz "\n"