MIPS 汇编、堆栈和基本加法

Question

我目前正在为 class 做作业，它要求我们接受两个变量（a 和 b）的用户输入值并求解方程 [a - ab + 8a - 10b + 19]，同时仅使用堆栈和寄存器 $t0 和 $t1.

我知道我需要做什么，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，我是否应该在弹出和推送时引用特定位置 [例如 4($sp)]？我从谷歌搜索中得到不一致的答案。

很抱歉，您在这里看到的任何奇怪的方法，我已经烦躁了一段时间无济于事。

结果：[a=4，b=3，解=51][a=6，b=4，解=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) 实际上是弹出 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 汇编、堆栈和基本加法

MIPS assembly, stacks and basic addition

assembly

mips