我认为我的交换以及如何访问数组中的元素遇到了问题。现在，所有代码都会运行，但排序后列表不会改变。他是我想要实现的高级类型
for(k=0; k<request-1; k++) {
i = k;
for(j=k+1; j<request; j++) {
if(array[j] > array[i])
i = j;
}
exchange(array[k], array[i]);
}
这是汇编代码。注意:作业是关于将元素压入/弹出堆栈的，因此我无法更改参数。
;this is a library with macros for things like printing numbers and strings
INCLUDE Irvine32.inc
MIN = 10 ;lower range limit
MAX = 200 ;upper range limit
LRANGE = 100
HRANGE = 999
.data
;title, intro, and prompts
title_1 BYTE "PROGRAMMING ASSIGNMENT 5: RANDOM GEN/SORT", 0dh, 0ah, 0
intro_1 BYTE "This program generates random numbers in the range (100 - 999),", 0dh, 0ah
BYTE "displays the original list, sorts the list, and calculates the median value.", 0dh, 0ah
BYTE "Finally, it displays the sorted list in descending order.", 0dh, 0ah, 0
prompt_1 BYTE "How many numbers should be generated? (10 - 200): ", 0
error_1 BYTE "Out of range.", 0dh, 0ah, 0
display_1 BYTE "List of random numbers: ", 0dh, 0ah, 0
display_2 BYTE "The median is: ", 0
display_3 BYTE "The sorted list: ", 0dh, 0ah, 0
;placeholders for user entries and calculated data
randArray DWORD MAX DUP(?)
userNum DWORD ? ;integer to be entered by user
;strings for posting results
goodBye_1 BYTE "Thank you for using the Gen/sort-ulator! Good-bye!", 0
.code
main PROC
call Randomize
;Title Screen
push OFFSET title_1
push OFFSET intro_1
call Intro
;Get and validate user numbers
push OFFSET error_1
push OFFSET prompt_1
push OFFSET userNum
call GetData
;Fill Array with random numbers
push OFFSET randArray
push userNum
call FillArray
;display unsorted results
push OFFSET randArray
push userNum
push OFFSET display_1
call DisplayList
;sort the results
push OFFSET randArray
push userNum
call SortList
;display the median
push OFFSET randArray
push userNum
push OFFSET display_2
call median
;display sorted results
push OFFSET randArray
push userNum
push OFFSET display_3
call DisplayList
;Say "goodbye"
push OFFSET goodBye_1
call Goodbye
exit ; exit to operating system
main ENDP
;-------------------------------------------------------
;Gives an Intro to the program
; Receives parameters on the system stack (in the order pushed):
; Address of the title
; Address of the intro
;post: intro displayed
;registers: none
;-------------------------------------------------------
Intro PROC
pushad
mov ebp, esp
mov edx, [ebp+40]
call writeString
call CrLf
mov edx, [ebp+36]
call writeString
call CrLf
popad
ret 8
Intro ENDP
;-------------------------------------------------------
;Prompts user for an integer, int stores in userNum
; Receives parameters on the system stack (in the order pushed):
; Address of the error message
; Address of the prompt
; Address of return value
;Post: userNum
;registers: none
;-------------------------------------------------------
GetData PROC
pushad
;setup stack and prompt for entry
mov ebp, esp
reenter:
mov edx, [ebp+40]
mov ebx, [ebp+36]
call WriteString
call ReadInt
;validate entry
cmp eax, MIN ;if eax < LOWER
jl badEntry ;jump to summary
cmp eax, MAX ;if eax > UPPER
jg badEntry ;reprompt
jmp goodEntry ;else jump to end, we have a good value
;bad entry reprompt
badEntry:
mov edx, [ebp+44]
call WriteString
jmp reenter
goodEntry:
call CrLf
mov [ebx], eax
popad
ret 12
GetData ENDP
;-------------------------------------------------------
;Fills array with a number of random integers within RANGE
;Recieves parameters on the system stack (in order pushed)
; array
; userNum
;Post: array is filled with userNum number of randoms
;Registers used: none
;-------------------------------------------------------
FillArray PROC
pushad
mov ebp, esp
mov ecx, [ebp+36] ;initialize loop counter with user entry
mov edi, [ebp+40] ;setup array offset
fillLoop:
call nextRand
add edi, 4
loop fillLoop
popad
ret 8
FillArray ENDP
;-------------------------------------------------------
; Procedure nextRand
; adapted from check lecture 20 solutions
; Procedure to get the next random number in the range specified by the user.
; Preconditions: LRANGE < HRANGE
; Registers used: eax, edi
;-------------------------------------------------------
nextRand PROC
mov eax, HRANGE
sub eax, LRANGE
inc eax ;add 1 to get the number of integers in range
call RandomRange
add eax, LRANGE ;eax has value in [LOW - HIGH]
mov [edi],eax
ret
nextRand ENDP
;-------------------------------------------------------
;Sorts the contents of an integer array
; Receives parameters on the system stack (in order pushed)
; Array
; Array Size
;registers: none
;-------------------------------------------------------
sortList PROC
pushad
mov ebp, esp
mov ecx, [ebp+36]
mov edi, [ebp+40]
dec ecx ;ecx < request-1
mov ebx, 0 ;ebx=k
;for(k=0; k<request-1; k++)
outerLoop:
mov eax, ebx ;eax=i=k
mov edx, eax
inc edx ;edx=j=k+1
push ecx
mov ecx, [ebp+36] ;ecx < request
;for(j=k+1; j<request; j++)
innerLoop:
mov esi, [edi+edx*4]
cmp esi, [edi+eax*4]
jle skip
mov eax, edx
skip:
inc edx
loop innerLoop
;swap elements
lea esi, [edi+ebx*4]
push esi
lea esi, [edi+eax*4]
push esi
call exchange
pop ecx
inc ebx
loop outerLoop
popad
ret 8
sortList ENDP
;-------------------------------------------------------
; Exchange k and i
; Receives parameters on the system stack (in order pushed)
; array[k]
; array[i]
;registers: none
;-------------------------------------------------------
Exchange PROC
pushad
mov ebp,esp
mov eax, [ebp+40] ;array[k] low number
mov ecx, [eax]
mov ebx, [ebp+36] ;array[i] high number
mov edx, [ebx]
mov [eax], edx
mov [ebx], ecx
popad
ret 8
Exchange ENDP
;-------------------------------------------------------
;Displays the median of an integer array
; Receives parameters on the system stack (in order pushed)
; Array
; Array Size
; display string
;registers: none
;-------------------------------------------------------
Median PROC
pushad
mov ebp, esp
mov edi, [ebp+44]
;display string
mov edx, [ebp+36]
call writeString
;calculate median element
mov eax, [ebp+40]
cdq
mov ebx, 2
div ebx
shl eax, 2
add edi, eax
cmp edx, 0
je isEven
;Array size is odd, so display the middle value
mov eax, [edi]
call writeDec
call CrLf
call CrLf
jmp endMedian
isEven:
;Array size is even so average the two middle values
mov eax, [edi]
add eax, [edi-4]
cdq
mov ebx, 2
div ebx
call WriteDec
call CrLf
call CrLf
endMedian:
popad
ret 12
Median ENDP
;-------------------------------------------------------
;Displays the contents of an integer array, 10 per row
; Receives parameters on the system stack (in order pushed)
; Array
; Array Size
; display string
;registers: none
;-------------------------------------------------------
DisplayList PROC
pushad
mov ebp, esp
;display string
mov edx, [ebp+36]
call writeString
call CrLf
mov ecx, [ebp+40]
mov edi, [ebp+44]
mov ebx, 0
;display array contents
listloop:
inc ebx ;counter for 10 items per row
mov eax, [edi]
call writeDec
add edi, 4
cmp ebx, 10
jne noReturn ;jump if 10 items are not yet printed
call CrLf
mov ebx, 0
jmp noTab ;this skips adding a tab on a new row
noReturn:
mov al, TAB
call writeChar
noTab:
loop listloop
call CrLf
popad
ret 12
DisplayList ENDP
;-------------------------------------------------------
;Says good-bye to the user
; Receives parameters on the system stack:
; Address of string
;registers: none
;-------------------------------------------------------
Goodbye PROC
pushad
mov ebp, esp
mov edx, [ebp+36]
call writeString
call CrLf
popad
ret 4
Goodbye ENDP
END main
最佳答案
为了交换元素，您应该传递它们的地址(换句话说，指向每个元素的指针)。您所做的只是交换作为参数传递的值，这些值也立即被释放。您的代码相当于此 C 函数:
void Exchange(int x, int y)
{
int eax = x;
int ebx = y;
x = ebx;
y = eax;
}
你需要这样的东西:
void Exchange(int* x, int* y)
{
int eax = *x;
int ebx = *y;
*x = ebx;
*y = eax;
}
在 asm 中，这可能看起来像:
Exchange PROC
pushad
mov eax, [esp+40]
mov ecx, [eax]
mov ebx, [esp+36]
mov edx, [ebx]
mov [eax], edx
mov [ebx], ecx
popad
ret 8
Exchange ENDP
要调用此函数，您将使用以下命令:
lea esi, [edi+ebx*4]
push esi
lea esi, [edi+eax*4]
push esi
call Exchange
请注意，您的 Exchange 函数有一条指令 mov eax, [edi]，我无法理解它。
更新:是的，您可以通过手动计算来模拟LEA的功能。例如，lea esi, [edi+ebx*4] 变为:
mov esi, ebx
shl esi, 2 ; esi=ebx*4
add esi, edi ; esi=edi+ebx*4
关于sorting - assembly 中的选择排序过程，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/13407151/