我一直得到 0 作为我的 mips 代码的答案，我应该在其中找到两个数字的 gcd

问题描述

.data
    n1:.asciiz"Enter the first number:"
    n2: .asciiz"Enter the second number:"

.text
.globl main
main:
    li $v0,4
    la $a0,n1
    syscall
    
    li  $v0,5      # get input from user
    syscall
    move $a0,$s0
      
    li $v0,n2
    syscall
    
    li  $v0,5      # get second input from user
    syscall
    move $a0,$s1
   
    jal calcGCD # call function calcGCD

    add $a0,$v0,$zero 
    li $v0,1
    syscall # print result
    li $v0,10 # exit program 
    syscall

calcGCD:
    #GCD(n1,n2)
    # n1 = $a0
    # n2 = $a1

    addi $sp,$sp,-12
    sw $ra,0($sp) # save function into stack
    sw $s0,4($sp) # save value $s0 into stack 
    sw $s1,8($sp) # save value $s1 into stack 

    add $s0,$a0,$zero # s0 = a0 ( value n1 ) 
    add $s1,$a1,$zero # s1 = a1 ( value n2 ) 

    addi $t1,$zero,0 # $t1 = 0
    beq $s1,$t1,return # if s1 == 0 return

    add $a0,$s1 # make a0 = $s1
    div $s0,$s1 # n1/n2
    mfhi $a1 # reminder of n1/n2 which is equal to n1%n2

    jal calcGCD

exitGCD:
    lw $ra,0 ($sp)  # read registers from stack
    lw $s0,4 ($sp)
    lw $s1,8 ($sp)
    addi $sp,12 # bring back stack pointer
    jr $ra
return:
    add $v0,$s0 # return $v0 = $s0 ( n1)
    j exitGCD

解决方法

您的代码的主要问题在于您将用户输入存储在寄存器中的部分。在移动 $a0,$s0 中，当用户输入存储在 $v0 中时，您将 $s0 处的值移动到 $a0 中，因此，它应该是移动 $a0,$v0，并且从您的函数中，您似乎有将第二个输入存储在 $a1 中，但在您的代码中，两个输入都存储在同一个寄存器中，因此下一个命令应该是 move $a1,$v0。这是您的代码的一个版本。

.data
n1:.asciiz"Enter the first number:"
n2: .asciiz"Enter the second number:"

.text
.globl main
main:
    li $v0,4
    la $a0,n1
    syscall
    
    li  $v0,5      # get input from user
    syscall
    move $t0,$v0 #temporarily store the user input in another register because $a0 is having another value stored in it in the next command
      
    li $v0,n2
    syscall
    
    li  $v0,5      # get second input from user
    syscall
    move $a1,$v0
    move $a0,$t0    #transfer the first user input into $a0 to be used in the function
   
    jal calcGCD # call function calcGCD

    add $a0,$v0,$zero 
    li $v0,1
    syscall # print result
    li $v0,10 # exit program 
    syscall

calcGCD:
    #GCD(n1,n2)
    # n1 = $a0
    # n2 = $a1

    addi $sp,$sp,-12
    sw $ra,0($sp) # save function into stack
    sw $s0,4($sp) # save value $s0 into stack 
    sw $s1,8($sp) # save value $s1 into stack 

    add $s0,$a0,$zero # s0 = a0 ( value n1 ) 
    add $s1,$a1,$zero # s1 = a1 ( value n2 ) 

    addi $t1,$zero,0 # $t1 = 0
    beq $s1,$t1,return # if s1 == 0 return

    add $a0,$s1 # make a0 = $s1
    div $s0,$s1 # n1/n2
    mfhi $a1 # reminder of n1/n2 which is equal to n1%n2

    jal calcGCD

exitGCD:
    lw $ra,0 ($sp)  # read registers from stack
    lw $s0,4 ($sp)
    lw $s1,8 ($sp)
    addi $sp,12 # bring back stack pointer
    jr $ra
return:
    add $v0,$s0 # return $v0 = $s0 ( n1)
    j exitGCD