Question

我目前正在为类分配，要求我们接受两个变量（a和b）的用户输入值并解决方程式[a - ab + 8a - 10b + 19]，同时仅使用堆栈和寄存器$ t0和$ t1。

我理解我需要做什么，但我无法弄清楚为什么我的程序在编译时显示不正确的解决方案。对于堆栈，我应该在我弹出和推送时引用特定位置[例如4（$ sp）]？我从谷歌搜索得到了不一致的答案。

我为你在这里看到的任何奇怪的方法道歉，我一直无法忍受它一段时间无济于事。

结果：[a = 4，b = 3，溶液= 51] [a = 6，b = 4，溶液= 67]

    .data

aprompt:    .asciiz "Input a value for a: "
bprompt:    .asciiz "Input a value for b: "
eq:         .asciiz "a - ab + 8a - 10b + 19 = "

    .text

main:

    la $a0, aprompt         # print the prompt for a
    li $v0, 4
    syscall

    li $v0, 5               # read input as integer
    syscall
    move $t0, $v0           # store variable a in $t0

    la $a0, bprompt         # print the prompt for b
    li $v0, 4
    syscall

    li $v0, 5               # read input as integer
    syscall
    move $t1, $v0           # store variable b in $t1

    addi $sp,$sp,-4         # decrement stack pointer by 4
    sw $t0,($sp)            # push a on the stack at 0

    mult $t0,$t1            # a*b
    mflo $t0                # store result in $t0
    addi $sp,$sp,-4         # decrement stack pointer by 4
    sw $t0,($sp)            # push a*b on the stack at 4

    li $t0,10               # load 10 into $t0
    mult $t0,$t1            # 10*b
    mflo $t0                # store result in $t0

    lw $t1,($sp)            # pop the stack and store in $t1
    addi $sp,$sp,4          # increment stack pointer by 4

    add $t0,$t0,$t1         # add a*b and 10*b
    lw $t1,($sp)            # pop the stack and store in $t1
    addi $sp,$sp,-4         # decrement stack pointer
    sw $t0,($sp)            # push a*b+10*b on the stack at 0
    addi $sp,$sp,4          # increment stack pointer

    li $t0,8                # store 8 in $t0
    mult $t0,$t1            # 8*a
    mflo $t0                # store in $t0

    addi $t1,$t1,19         # a + 19

    add $t0,$t0,$t1         # store (a + 19) + 8*a in $t0

    lw $t1,($sp)            # pop the stack and store in $t1
    sub $t0,$t0,$t1         # (a + 19 + 8a) - (ab + 10b)

    move $a0,$t0            # print solution as integer
    li $v0,1
    syscall

    li $v0, 10              # exit
    syscall

Answer 1

lw $t1,($sp)            # pop the stack and store in $t1
sub $t0,$t0,$t1         # (a + 19 + 8a) - (ab + 10b)

lw $t1,($sp)实际上是a，而不是ab + 10b。

您可以通过弹出ab + 10b而不是$ sp - 4来修复它。

add $t0,$t0,$t1         # store (a + 19) + 8*a in $t0
addi $sp,$sp,-4         # decrement stack pointer
lw $t1,($sp)            # pop the stack and store in $t1

sub $t0,$t0,$t1         # (a + 19 + 8a) - (ab + 10b)

MIPS组装，堆栈和基本添加

1 个答案: