再次
我正在编写一个MIPS程序，它读取5个整数并将它们存储在数组中。然后它创建一个新数组，其值是初始数组的值乘以其索引。之后我需要找到第二个数组的最大值和最小值。
我是MIPS的新手，这是我写的代码：
.data
Prompt: .asciiz "\n Enter 5 Integers :" #gets number of integers
op: .asciiz "\n Enter Option: \n 1-Find the mult \n 2- Find the max \n 3-Find the min \n 4-Exit"
invalidm: .asciiz "\n Bad Input:"
vec: .space 20
vec2: .space 20
.text
.globl __start
__start:
la $a0,Prompt
li $v0,4
syscall
#Reading integers and store theme in the array
options:
la $a0,op
li $v0,4
syscall
li $v0,5
syscall
blt $v0,1,invalid#
bgt $v0,4,invalid#
beq $v0, 1, multp#
beq $v0, 2, max#
beq $v0, 3, min#
beq $v0, 4, exitpro#
j options
multp:
#multiply every element in the array with its index and store them in the new array vec2
j options
max:
#find the max of the array vec
j options
min:
#find the min of the array vec
j options
invalid:
la $a0,invalidm
li $v0,4
syscall
j options
exitpro:
li $v0,10
syscall
答案 0 :(得分：2)
我想快速记下编码风格：汇编有点恶魔。很有诱惑力地说“哦，我经常做这5行，我应该jal在这里并重复使用它”。这首先是有道理的，但往往导致令人困惑的意大利面条代码，不能跳过3行而不会跳到某处。因此，我的代码有一些代码重复，但没什么不好。
我稍微改变了你的初始条件，数组的长度是硬编码的，但它使用的是堆栈分配的数组，而不是.data标头初始化时静态分配的数组。而是在data部分中分配数组的长度。这只是个人偏好。它的好处是代码应该适用于任何长度＆gt; 0
.data
PromptHead: .asciiz "\n Enter "
PromptTail: .asciiz " integers\n"
Minstr: .asciiz "Min: "
Maxstr: .asciiz "Max: "
newline: .asciiz "\n"
inputs: .word 5
.text
main:
# Output prompt for input
## "\n Enter "
la $a0, PromptHead
li $v0, 4
syscall
## "5"
lw $a0, inputs
li $v0, 1
syscall
## " integers\n"
la $a0, PromptTail
li $v0, 4
syscall
# Backup sp value before allocating array
move $fp, $sp
# calculate size of vector (numinputs * sizeof(int) = numinputs * 4 = numinputs << 2), store in $s1
lw $s0, inputs
sll $s1, $s0, 2
# dynamically grow stack to include array of ints
add $sp, $sp, $s1
# Loop initialization
# i = 0
move $s1, $zero
# while i < numInputs
ReadInput:
slt $s2, $s1, $s0
beq $s2, $zero, ReadInputDone
# Read integer
li $v0, 5
syscall
# Calculate array offset for this loop
sll $s2, $s1, 2
add $s2, $fp, $s2
# Store array value at calculated address
sw $v0, 0($s2)
# i++
addi $s1, $s1, 1
j ReadInput
ReadInputDone:
# Scale(array, length)
move $a0, $fp
move $a1, $s1
jal Scale
# max = FindMax(array, length)
jal FindMax
# Back up return val, print boilerplate
move $t0, $v0
li $v0, 4
la $a0, Maxstr
syscall
move $a0, $t0
li $v0, 1
syscall
la $a0, newline
li $v0, 4
syscall
# min = FindMin(array,length)
move $a0, $fp
jal FindMin
# Back up return val, print boilerplate
move $t0, $v0
li $v0, 4
la $a0, Minstr
syscall
move $a0, $t0
li $v0, 1
syscall
la $a0, newline
li $v0, 4
syscall
# Exit program
li $v0, 10
syscall
# Scale(array, length)
# Scales each array element by index+1. This does not alter any s or a registers,
# Alters the array in place.
Scale:
# Backup return address and fp on stack.
# Not strictly necessary here, but usually good to do this by habit
# when you're learning
sw $fp, 0($sp)
sw $ra, 4($sp)
addi $fp, $sp, 8
move $sp, $fp
# Load arguments into scratch registers
move $t0, $a0
move $t1, $a1
# Loop initialization as above
# i = 0
move $t2, $zero
# while i < 5
ScaleInput:
slt $t3, $t2, $t1
beq $t3, $zero, ScaleInputDone
# Calculate element offset, store address in $t3
sll $t3, $t2, 2
add $t3, $t0, $t3
# Load array element at $t3, multiply it by current index+1
# Grab it from the multiplication register (assume no overflow)
# then store the result back in the array
lw $t5, 0($t3)
addi $t4, $t2, 1
mult $t5, $t4
mflo $t5
sw $t5, 0($t3)
# i++
addi $t2, $t2, 1
j ScaleInput
ScaleInputDone:
# Unwind stack, restore frame pointer and
# return address. Again, not necessary here, but good
# practice
lw $ra, -4($fp)
move $sp, $fp
lw $fp, -8($fp)
jr $ra
# int Max(array, length); result returned in $v0,
# No s or a registers are altered
FindMax:
# Backup return address and fp on stack.
sw $fp, 0($sp)
sw $ra, 4($sp)
addi $fp, $sp, 8
move $sp, $fp
# Load arguments into scratch registers
move $t0, $a0
move $t1, $a1
# set currMin = array[0]
lw $v0, 0($t0)
# Loop initialization
# i = 1
li $t2, 1
# while i < 5
MaxLoop:
slt $t3, $t2, $t1
beq $t3, $zero, MaxLoopDone
# Calculate element offset, store address in $t3
sll $t3, $t2, 2
add $t3, $t0, $t3
# Load array element at $t3, check if it's the new max
lw $t4, 0($t3)
sgt $t5, $t4, $v0
beq $t5, $zero, notGreater
# If so, set return value to it
move $v0, $t4
notGreater:
# i++
addi $t2, $t2, 1
j MaxLoop
MaxLoopDone:
# Unwind stack
lw $ra, -4($fp)
move $sp, $fp
lw $fp, -8($fp)
jr $ra
# int Min(array, length); result returned in $v0,
# No s or a registers are altered
FindMin:
# Backup return address and fp on stack.
sw $fp, 0($sp)
sw $ra, 4($sp)
addi $fp, $sp, 8
move $sp, $fp
# Load arguments into scratch registers
move $t0, $a0
move $t1, $a1
# set currMin = array[0]
lw $v0, 0($t0)
# Loop initialization
# i = 1
li $t2, 1
# while i < 5
MinLoop:
slt $t3, $t2, $t1
beq $t3, $zero, MinLoopDone
# Calculate element offset, store address in $t3
sll $t3, $t2, 2
add $t3, $t0, $t3
# Load array element at $t3, check if it's the new min
lw $t4, 0($t3)
slt $t5, $t4, $v0
beq $t5, $zero, notLesser
# If so, set return value to it
move $v0, $t4
notLesser:
# i++
addi $t2, $t2, 1
j MinLoop
MinLoopDone:
# Unwind stack
lw $ra, -4($fp)
move $sp, $fp
lw $fp, -8($fp)
jr $ra
正如我在代码中提到的那样，存储$ra和$fp时的痴迷有点偏执。如果我真的是偏执狂，我也会存储所有s个寄存器。但是，这通常是一种很好的做法，当您决定在函数中间添加“函数调用”时，可以省去很多麻烦。
这基本上就是汇编的方法，你在头脑中用C编写程序，然后将其翻译成汇编。因此，为什么我将缩放，查找最大值等视为“函数”。
一些成员注释：我反复使用sll $register, $register, 2而不是乘以4（MIPS32上的字大小）。这是因为执行此操作的指令较少，因为无需调用li后跟mult后跟mflo。你可以这样做，而且我曾经这样做过，但是一旦你习惯了sll和其他一些小小的操作，那么使用sll以及更容易阅读就更简洁了。
我确实使用了一些pseudoinstructions，这不是什么大问题，我知道支持它们的大多数现代MIPS汇编器和模拟器（包括SPIM）。我在分支指令之后与线路危险地跳舞（理论上它总是执行而不管分支结果如何），但在这种情况下它通常并不重要，并且大多数模拟器和汇编器通常会在它无论如何时为你注入无操作。
我知道汇编可能有点难以阅读，所以请随时询问您是否有任何问题。