在 MIPS 汇编中反转字符串

Question

我试图提示用户输入字符串的长度，为该字符串分配空间，然后反向打印出来。

对于我的一生，我不明白为什么这不起作用..

Sample Output:
(spim) run    
Please enter an integer:    
7    
string
(spim)

现在“字符串”的长度应该是 6 对吗？ + 应该使它成为 7 的空终止字符。任何人都可以发现我的方法哪里出了问题吗？

.data
    nl: .asciiz "\n"
    inputPrompt: .asciiz "Please enter an integer:\n"

    theString: .space 32
    theInteger: .word 1

.text
main: 
    la $a0, inputPrompt #load address a0 with prompt
    li $v0, 4       #load system call, print string into v0
    syscall 

    li $v0, 5       #load system call, read int into v0
    syscall
    sw $v0, theInteger  #store saved int into $t0

    li $v0, 8           #load system call, read string with mem address
    la $a0, theString   #load address of reserved string space
    lw $a1, theInteger  #load address of saved int length for string    
    syscall

    lw $t0, theInteger
    add $a1,$zero,$t0   #pass lenght of string
    jal stringreverse   #reverse the string

stringreverse:
    add $t0,$a0,$zero   #starting address
    add $t1,$zero,$zero     #i = 0
    addi $t2,$a1,-1     #j = length-1

loop:
    add $t3,$t0,$t1
    lb $t4,0($t3)   #the lb string[i]
    add $t5,$t0,$t2
    lb $t6,0($t5)   #the lb string[j]
    sb $t4,0($t5)   #string[j] = string[i]
    sb $t6,0($t3)   #string[i] = string[j]
    addi $t1,$t1,1  #i++
    addi $t2,$t2,-1     #j--

    slt $t6,$t2,$t1
    beqz $t6,loop

exit:
    li $v1, 4       #system call to print reversed string
    la $a2, 0($a1)
    syscall

    li $v0, 10
    syscall         # Exit program

Answer 1

索引中有一个小错误...并且我没有在旧的之上重新写入，而是使用了名为反向的新内存空间...

stringreverse:
    add $t0,$a0,$zero   #starting address
    add $t1,$zero,$zero     
    add $t3,$zero,$zero     #i = 0
    addi $t2,$a1,-2     #j = length-1

loop:
    add $t5,$t0,$t2
    lb $t6,0($t5)   #the lb string[j]
    sb $t6,reverse($t3)
    addi $t2,$t2,-1     #j--
    addi $t3,$t3,+1     #i++

    slt $t7,$t2,$t1
    beqz $t7,loop