x86程序集-需要对两个给定数组进行冒泡排序的程序的帮助
x86程序集-需要对两个给定数组进行冒泡排序的程序的帮助
提问于 2022-10-19 21:22:22
我正在制作一个x86汇编程序，它利用气泡排序算法详细在欧文汇编教科书(为x86系统)按升序对两个给定的数组进行排序。我的第二个过程搜索每个数组并输出每个数组中最大的值。我知道我忘记了一个跳转命令(sortAscending过程)，虽然我不知道该把它放在哪里来完成我需要做的事情。当我用VS2022构建我的ASM文件时，我会收到以下错误：
(x86)\MSBuild\Microsoft.Cpp\v4.0\V140\BuildCustomizations\masm.targets(50,5)：1>C:\Program MSB3721:命令"ml.exe /c /nologo /Sg /Zi /Fo"Debug\Template.obj“/Fl”Project.lst /I "C:\irvine“/W3 /errorReport:prompt /TaTemplate.asm”与代码1一起退出。
我不确定是否有其他错误，但如果有人注意到了，请指出。我的代码如下所示。
TITLE Sort Arrays, Version 1 (SortArrays.asm)
; This program sifts through and
; sorts arrays (in place) in ascending
; order and outputs the greatest value
; in each array.
; Name: Kasizah
; Date: 10-19-2022
INCLUDE Irvine32.inc
.data
Array1 DWORD 0C0D12AFh, 00030256h, 0FFAABBCCh, 0F700F70h, 00000000h, 0E222111Fh, 0ABCDEF01h, 01234567h
Array2 DWORD 61A80000h, 024F4A37h, 0EC010203h, 0FAEEDDCCh, 2C030175h, 84728371h, 63AA5678h, 0CD454443h, 22222222h, 61B1C2D3h, 7A4E96C2h, 81002346h, 0FDB2726Eh, 65432100h, 0FFFFFFFFh
; message strings
largestUnsignedS BYTE "The largest unsigned value in the array is: ",0
largestUnsignedF BYTE ".",0
.code
main PROC
mov esi,OFFSET Array1 ; point to start of Array1
mov ecx,LENGTHOF Array1 ; get length of Array1
call sortAscending ; sort Array1
; display sorted array using DumpMem
mov esi,OFFSET Array1 ; starting OFFSET
mov ecx,LENGTHOF Array1 ; number of units
mov ebx,TYPE Array1 ; doubleword format
call DumpMem
; get greatest value in Array1
mov esi,OFFSET Array1 ; point to start of Array1
mov ecx,LENGTHOF Array1 ; get length of Array1
call Crlf ; skip line
call getLargest ; display largest value in Array1
call Crlf ; skip line
mov esi,OFFSET Array2 ; point to start of Array2
mov ecx,LENGTHOF Array2 ; get length of Array2
call sortAscending ; sort Array2
; display sorted array using DumpMem
mov esi,OFFSET Array2 ; starting OFFSET
mov ecx,LENGTHOF Array2 ; number of units
mov ebx,TYPE Array2 ; doubleword format
call DumpMem
; get greatest value in Array2
mov esi,OFFSET Array2 ; point to start of Array2
mov ecx,LENGTHOF Array2 ; get length of Array2
call Crlf ; skip line
call getLargest ; display largest value in Array2
call Crlf ; skip line
exit ; exit the program
main ENDP
;-------------------------------------------------------
sortAscending PROC
;
; Sorts an array of 32-bit signed integers in ascending
; order using the bubble sort algorithm.
; Receives: pointer to array, array size
; Returns: nothing
;-------------------------------------------------------
mov edi,esi ; duplicate "point to first value"
; (used to reset ESI)
dec ecx ; decrement ECX (count) by 1
outer_loop:
push ecx ; save outer loop count
mov esi,edi ; point to first value
sort:
mov eax,esi ; get current array value
cmp [esi+4],eax ; compare [ESI] and [ESI+4]
jg next ; if [ESI] <= [ESI+4], no swap
xchg eax,[esi+4] ; else swap pair
mov [esi],eax
next:
add esi,4 ; move both pointers forward
loop sort ; inner loop
pop ecx ; retrieve outer loop count
loop outer_loop ; else repeat outer loop
quit:
ret
sortAscending ENDP
;-------------------------------------------------------
getLargest PROC
;
; Searches an array for its largest
; value.
; Receives: ESI - pointer
; Returns: statement of largest value in array
;-------------------------------------------------------
sift:
mov eax,[esi] ; move [ESI] into EAX
cmp [esi+4],eax ; compare EAX with [ESI+4]
jg next ; if EAX >= [ESI+4] don't replace
mov eax,[esi+4] ; mov [ESI+4] into EAX
next:
add esi,4 ; move pointer forward
cmp ecx,esi ; make sure that esi isn't at the end of array
je quit ; jump to quit if at the end of array
loop sift ; else return to sift loop
quit:
mov ebx,eax ; move EAX into EBX temporarily
mov eax,largestUnsignedF ; move largestUnsignedF string into EAX
call WriteString ; display largestUnsignedF string
mov eax,ebx ; move EBX into EAX
call WriteInt ; display EAX
mov eax,largestUnsignedS ; move largestUnsignedS string into EAX
call WriteString ; display largestUnsignedS string
ret
getLargest ENDP
END main回答 1
Stack Overflow用户
回答已采纳
发布于 2022-10-19 21:56:33
mov eax，esi；获取当前数组值
您忘记了方括号来加载存储在address ESI中的值：
mov eax, [esi]getLargest代码可以简单地获取最后一个数组元素，因为数组得到了优先排序。
mov eax, [esi + ecx * 4 - 4]您的消息还谈到了无符号结果，但是排序算法将这些元素视为签名的dword！对于未签名的，请使用ja (JumpIfAbove)。
您的sift代码有错误：
cmp ecx,esi ; make sure that esi isn't at the end of array
je quit ; jump to quit if at the end of array
loop sift ; else return to sift loop
quit:cmp ecx, esi将计数与地址进行比较！不能工作！把这两行都放下。
如果您在启动循环之前先对ECX进行减量，并通过loop获取结果，那么loop指令就足够了。
找到最大元素的代码如下所示：
mov eax, 80000000h ; Smallest signed dword
More:
cmp [esi], eax
jng Skip
mov eax, [esi]
Skip:
add esi, 4
dec ecx
jnz More
mov ebx, eax ; The signed maximum页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/74132050
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