反转字符串并用汇编语言打印

Question

我在这里试图颠倒这个字符串，但我真的卡住了。

我从变量letter和length中取出字符串，抛出代码，将它们发送到reversing_loop，并像代码中那样做，如果计数器达到与长度相同的值，则退出并打印结果。我甚至不确定这两行

sw $a2, result
la $a0 , $a2

代码如下：

.data

letters: .asciiz "abcd"
result: .asciiz ""

.text

.globl main
main:
la $a0, letters    # reading chars letters
li $a1, 4     # store length
li $a3, 0          # counter
add $a0, $a0, $a1  # address of last char

reversing_loop:
beq $a3, $a1, Exit
lb $t0,($a0)
sb $t0,($a2)
addi $a0,$a0, -1   # i -=1
addi $a2,$a2, 1  # move address
addi $a3, $a3, 1 # increment counter
j reversing_loop

Exit:
sw $a2, result
la $a0 , $a2
li $v0 , 4
syscall
li $v0 , 10
syscall

.end main

Answer 1

我最终完成了代码，它可以正常运行。

.data

letters: .byte 'a','b','c','d','e'
result: .asciiz ""

.text

.globl main
main:
la $a0, letters    # reading chars letters
li $a1, 5          # store length
la $a2 , result        # intialize a2
li $a3, 0          # counter
addi $a0, $a0, 4  # address of last char

reversing_loop:
beq $a3, $a1, Exit
lb $t0, 0($a0)
sb $t0, 0($a2)
addi $a0,$a0, -1   # i -=1
addi $a2,$a2, 1  # move address
addi $a3, $a3, 1 # increment counter
j reversing_loop

Exit:
addi $a2,$a2,-5
add $a0,$0, $a2

li $v0 , 4
syscall
li $v0 , 10
syscall

.end main