我试图找出子串和第一次出现的索引。但有些不对劲。我比较模式数组的每个元素和字符串数组的每个元素，直到指针到达'\ 0'。有什么问题。算法完全错了？
#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
答案 0 :(得分：0)
我将你的代码用于类似项目，在inner_loop中你没有合适的bne
我刚刚放了一个bne，现在只打印子串的字符串..
答案 1 :(得分：0)
.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"