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POComplications in MIPS code of Recursive Reversal of a string
 

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  1. POComplications in MIPS code of Recursive Reversal of a string
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    <p>I have written this MIPS code of recursive reversal of a string. However, the output is coming out to be the same that has been input by the user. Can someone please help me out and indicate where am I going wrong? Please reply as soon as possible.</p> <pre><code># Program to reverse a string input by the user .data .align 2 array: .space 50 input: .asciiz "Enter a string: " output: .asciiz "\nThe reversed string is: " .text .globl main main: addi $s0, $zero, 50 addi $t0, $zero, 0 la $a0, input li $v0, 4 syscall la $a0, array li $v0, 8 syscall initiate: add $t0, $a0, $zero # initial address add $t1, $zero, $zero # count=0 add $t2, $zero, $zero # i=0 la $t0, array # base address of the array add $t3, $t0, $t2 # &amp; array[i] loop: lb $t3, 0($t3) # fetch array[i] beqz $t3, EndOfString # loop exits if it is a null character;array[i] !='\0' bne $t3, $0, continue # otherwise loop continues add $t1, $t1, 1 # count++ continue: add $t2, $t2, 1 # i++ j loop addi $a1, $zero, 50 jal StringReversal EndOfString: la $a0, output li $v0, 4 syscall la $a0, array li $v0, 4 syscall li $v0, 10 syscall StringReversal: add $t0, $a0, $zero # initial address add $t4, $zero, $zero # j = start = 0 addi $t5, $a1, -1 # k = end-1 SwapLoop: add $t6, $t0, $t4 lb $t7, 0($t6) # load byte array[start] add $t8, $t0, $t5 lb $t9, 0($t8) # load byte array[end-1] sb $t7, 0($t8) # array[end-1] = array[start] sb $t9, 0($t6) # array[start] = array[end-1] addi $t4, $t4, 1 # j++ addi $t5, $t5, -1 # k-- slt $t9, $t5, $t4 beqz $t9, SwapLoop jr $ra </code></pre>
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    1. COPlease indent all your code one more time so your entire code is inside a single code block.
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    2. COI have done the necessary indentation
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