我的程序不会对大于 130 的数组进行排序
My program won't sort arrays that are larger than 130
所以我的程序应该接受用户输入（10 到 200 之间的整数）并打印出一个 运行dom 数字数组并打印出该数组的排序版本。但是，这仅在我输入 130 或更少时有效。
我不知道我还能做什么。它有效，但只成功了一半。有什么办法可以优化这段代码吗？我已放置线条以帮助显示我在哪个过程中遇到问题。
**** 我 运行 调试器和我在程序抛出异常错误的地方留下了评论。*****
TITLE Program5 (Program5.asm)
INCLUDE Irvine32.inc
; (insert constant definitions here)
MIN_INPUT = 10
MAX_INPUT = 200
LO_RANDOM = 100
HI_RANDOM = 999
.data
; (insert variable definitions here)
intro BYTE "Fun with Arrays! by ", 0
instruction BYTE "This program generates random numbers in the range [100 .. 999], displays the original list, sorts the list, and calculates the median value. Finally, it displays the list sorted in descending order", 0
request DWORD 10
ask_user BYTE "How many numbers should be generated? [10 ... 200]: ", 0
error BYTE "Invalid input", 0
title_1 BYTE "The unsorted random numbers: ", 0
title_2 BYTE "The sorted list: ", 0
space BYTE " ", 0
mult DWORD 0.5
temp DWORD 0
list DWORD MAX_INPUT DUP(?)
.code
main PROC
; (insert executable instructions here)
call randomize
call introduction
push OFFSET request ;passed by reference
call getData
call CrLf
push request ; passed by value
push OFFSET list ; passed by reference
call fillArray
push OFFSET list
push request
push OFFSET title_1
call displaylist
call CrLf
push OFFSET list
push request
call sortList
call CrLf
;
push OFFSET list
push request
push OFFSET title_2
call displaylist
;push OFFSET list
;push request
;call displayMedian
exit ; exit to operating system
main ENDP
; (insert additional procedures here)
introduction PROC
mov edx, OFFSET intro
call WriteString
call CrLf
mov edx, OFFSET instruction
call WriteString
call CrLf
ret
introduction ENDP
getData PROC
;include parameter - request (reference)
push ebp ;Set up stack frame
mov ebp, esp
;get an integer from user
mov ebx, [ebp+8] ;get address of request into ebx
L1:
mov edx, OFFSET ask_user
call WriteString
call ReadDec
cmp eax, MIN_INPUT
jl errorMessage
cmp eax, MAX_INPUT
jg errorMessage
cmp eax, MIN_INPUT
jge endThis
cmp eax, MAX_INPUT
jle endThis
errorMessage:
mov edx, OFFSET error
call WriteString
call CrLf
jmp L1
endThis:
mov [ebx], eax
pop ebp
ret 4 ; remove four more bytes from the stack (after ret @)
getData ENDP
fillArray PROC
;include parameters - request (value), array (reference)
; MAJORITY OF THE FOLLOWING CODE WAS EXTRACTED FROM LECTURE 20 SLIDES
push ebp
mov ebp, esp ;[ebp+4]
mov edi, [ebp+8] ; @list in edi
mov ecx, [ebp+12] ; value of request in ecx
more:
mov eax, HI_RANDOM
sub eax, LO_RANDOM
inc eax
call RandomRange
add eax, LO_RANDOM
mov [edi], eax
add edi, 4
loop more
endmore:
pop ebp
ret 8
fillArray ENDP
;----------------------------------------------------------------------------
;----------------------------------------------------------------------------
;----------------------------------------------------------------------------
sortList PROC
;include parameters - array (reference), request (value)
push ebp
mov ebp, esp ;[ebp+4]
mov edi, [ebp+12] ; @list in edi
mov ecx, [ebp+8] ; value of request in ecx
dec ecx ; request - 1
mov ebx, 0 ; "k"
;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]);
;}
firstLoop:
mov eax, ebx ; "i = k"
mov edx, ebx ; "j = k"
inc edx ; "j = k + 1"
push ecx ; pushed the first loop's counter
mov ecx, [ebp+8] ; made the second loop's counter = request
secondLoop:
mov esi, [edi + (edx * 4)] ; array[j] ; EXCEPTION WAS THROWN HERE
cmp esi, [edi + (eax * 4)] ; compare array[j] and array[i]
jg greater
jle lesser
greater:
mov eax, edx
inc edx
loop secondLoop
lesser:
inc edx
loop secondLoop
push edx
push esi
push [edi + (ebx * 4)] ; array[k]
push [edi + (eax * 4)] ; array[i]
call exchangeElements
pop [edi + (eax * 4)]
pop [edi + (ebx * 4)]
pop esi
pop edx
pop ecx ; set the
inc ebx ; increment k in the first loop
loop firstLoop
pop ebp
ret 8
sortList ENDP
exchangeElements PROC
push ebp
mov ebp, esp
mov esi, [ebp+12] ; array[k]
mov edx, [ebp+8] ; array[i]
mov [ebp+8], esi
mov [ebp+12], edx
pop ebp
ret
exchangeElements ENDP
;----------------------------------------------------------------------------
;----------------------------------------------------------------------------
;----------------------------------------------------------------------------
displayMedian PROC
push ebp
mov ebp, esp ;[ebp+4]
mov edi, [ebp+12] ; @list in edi
mov ecx, [ebp+8] ; value of request in ecx
mov eax, ecx
mov ebx, 2
cdq
div ebx
cmp edx, 0
je isEven
cmp edx, 1
je isOdd
;def nlogn_median(l):
;l = sorted(l)
;if len(l) % 2 == 1:
;return l[len(l) / 2]
;else:
;return 0.5 * (l[len(l) / 2 - 1] + l[len(l) / 2])
isEven:
mov esi, [edi + (eax - 1)]
add esi, [edi + (eax)]
mov eax, esi
mov ebx, 2
cdq
div ebx
call WriteDec
isOdd:
mov eax, [edi + (eax*4)]
call WriteDec
pop ebp
ret
displayMedian ENDP
displayList PROC
push ebp
mov ebp, esp ; [ebp+4]
mov ecx, [ebp+12] ; @request
mov edi, [ebp+16] ; @list
mov esi, 10
mov edx, [ebp+8] ; @title
call WriteString
call CrLf
show:
mov eax, [edi]
call WriteDec
mov edx, OFFSET space
call WriteString
add edi, 4
dec esi
cmp esi, 0
je callClear
loopAgain:
loop show
jmp endshow
callClear:
mov esi, 10
call CrLf
jmp loopAgain
endshow:
pop ebp
ret 12
displayList ENDP
END main
更新
如果您查看您的寄存器，您会发现 edx 是 0fA4h，它比崩溃所在行的应有值大。 ecx 是一个负数。这是一个线索，表明你的循环在它应该停止之后正在执行。
问题是 greater 分支会落到 lesser 分支。这将再次递减 ecx，导致它变为负数，并且您的循环将保持 运行，直到您遇到访问冲突。
快速修复是在 greater 标签下的 loop 指令之后放置一个无条件的 jmp。
更好的解决方法是将循环的尾部组合成更简单的条件：
cmp esi, [edi + (eax * 4)] ; compare array[j] and array[i]
jle lesser
mov eax, edx
lesser:
inc edx
loop secondLoop
所以我的程序应该接受用户输入（10 到 200 之间的整数）并打印出一个 运行dom 数字数组并打印出该数组的排序版本。但是，这仅在我输入 130 或更少时有效。
我不知道我还能做什么。它有效，但只成功了一半。有什么办法可以优化这段代码吗？我已放置线条以帮助显示我在哪个过程中遇到问题。
**** 我 运行 调试器和我在程序抛出异常错误的地方留下了评论。*****
TITLE Program5 (Program5.asm)
INCLUDE Irvine32.inc
; (insert constant definitions here)
MIN_INPUT = 10
MAX_INPUT = 200
LO_RANDOM = 100
HI_RANDOM = 999
.data
; (insert variable definitions here)
intro BYTE "Fun with Arrays! by ", 0
instruction BYTE "This program generates random numbers in the range [100 .. 999], displays the original list, sorts the list, and calculates the median value. Finally, it displays the list sorted in descending order", 0
request DWORD 10
ask_user BYTE "How many numbers should be generated? [10 ... 200]: ", 0
error BYTE "Invalid input", 0
title_1 BYTE "The unsorted random numbers: ", 0
title_2 BYTE "The sorted list: ", 0
space BYTE " ", 0
mult DWORD 0.5
temp DWORD 0
list DWORD MAX_INPUT DUP(?)
.code
main PROC
; (insert executable instructions here)
call randomize
call introduction
push OFFSET request ;passed by reference
call getData
call CrLf
push request ; passed by value
push OFFSET list ; passed by reference
call fillArray
push OFFSET list
push request
push OFFSET title_1
call displaylist
call CrLf
push OFFSET list
push request
call sortList
call CrLf
;
push OFFSET list
push request
push OFFSET title_2
call displaylist
;push OFFSET list
;push request
;call displayMedian
exit ; exit to operating system
main ENDP
; (insert additional procedures here)
introduction PROC
mov edx, OFFSET intro
call WriteString
call CrLf
mov edx, OFFSET instruction
call WriteString
call CrLf
ret
introduction ENDP
getData PROC
;include parameter - request (reference)
push ebp ;Set up stack frame
mov ebp, esp
;get an integer from user
mov ebx, [ebp+8] ;get address of request into ebx
L1:
mov edx, OFFSET ask_user
call WriteString
call ReadDec
cmp eax, MIN_INPUT
jl errorMessage
cmp eax, MAX_INPUT
jg errorMessage
cmp eax, MIN_INPUT
jge endThis
cmp eax, MAX_INPUT
jle endThis
errorMessage:
mov edx, OFFSET error
call WriteString
call CrLf
jmp L1
endThis:
mov [ebx], eax
pop ebp
ret 4 ; remove four more bytes from the stack (after ret @)
getData ENDP
fillArray PROC
;include parameters - request (value), array (reference)
; MAJORITY OF THE FOLLOWING CODE WAS EXTRACTED FROM LECTURE 20 SLIDES
push ebp
mov ebp, esp ;[ebp+4]
mov edi, [ebp+8] ; @list in edi
mov ecx, [ebp+12] ; value of request in ecx
more:
mov eax, HI_RANDOM
sub eax, LO_RANDOM
inc eax
call RandomRange
add eax, LO_RANDOM
mov [edi], eax
add edi, 4
loop more
endmore:
pop ebp
ret 8
fillArray ENDP
;----------------------------------------------------------------------------
;----------------------------------------------------------------------------
;----------------------------------------------------------------------------
sortList PROC
;include parameters - array (reference), request (value)
push ebp
mov ebp, esp ;[ebp+4]
mov edi, [ebp+12] ; @list in edi
mov ecx, [ebp+8] ; value of request in ecx
dec ecx ; request - 1
mov ebx, 0 ; "k"
;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]);
;}
firstLoop:
mov eax, ebx ; "i = k"
mov edx, ebx ; "j = k"
inc edx ; "j = k + 1"
push ecx ; pushed the first loop's counter
mov ecx, [ebp+8] ; made the second loop's counter = request
secondLoop:
mov esi, [edi + (edx * 4)] ; array[j] ; EXCEPTION WAS THROWN HERE
cmp esi, [edi + (eax * 4)] ; compare array[j] and array[i]
jg greater
jle lesser
greater:
mov eax, edx
inc edx
loop secondLoop
lesser:
inc edx
loop secondLoop
push edx
push esi
push [edi + (ebx * 4)] ; array[k]
push [edi + (eax * 4)] ; array[i]
call exchangeElements
pop [edi + (eax * 4)]
pop [edi + (ebx * 4)]
pop esi
pop edx
pop ecx ; set the
inc ebx ; increment k in the first loop
loop firstLoop
pop ebp
ret 8
sortList ENDP
exchangeElements PROC
push ebp
mov ebp, esp
mov esi, [ebp+12] ; array[k]
mov edx, [ebp+8] ; array[i]
mov [ebp+8], esi
mov [ebp+12], edx
pop ebp
ret
exchangeElements ENDP
;----------------------------------------------------------------------------
;----------------------------------------------------------------------------
;----------------------------------------------------------------------------
displayMedian PROC
push ebp
mov ebp, esp ;[ebp+4]
mov edi, [ebp+12] ; @list in edi
mov ecx, [ebp+8] ; value of request in ecx
mov eax, ecx
mov ebx, 2
cdq
div ebx
cmp edx, 0
je isEven
cmp edx, 1
je isOdd
;def nlogn_median(l):
;l = sorted(l)
;if len(l) % 2 == 1:
;return l[len(l) / 2]
;else:
;return 0.5 * (l[len(l) / 2 - 1] + l[len(l) / 2])
isEven:
mov esi, [edi + (eax - 1)]
add esi, [edi + (eax)]
mov eax, esi
mov ebx, 2
cdq
div ebx
call WriteDec
isOdd:
mov eax, [edi + (eax*4)]
call WriteDec
pop ebp
ret
displayMedian ENDP
displayList PROC
push ebp
mov ebp, esp ; [ebp+4]
mov ecx, [ebp+12] ; @request
mov edi, [ebp+16] ; @list
mov esi, 10
mov edx, [ebp+8] ; @title
call WriteString
call CrLf
show:
mov eax, [edi]
call WriteDec
mov edx, OFFSET space
call WriteString
add edi, 4
dec esi
cmp esi, 0
je callClear
loopAgain:
loop show
jmp endshow
callClear:
mov esi, 10
call CrLf
jmp loopAgain
endshow:
pop ebp
ret 12
displayList ENDP
END main
更新
如果您查看您的寄存器，您会发现 edx 是 0fA4h，它比崩溃所在行的应有值大。 ecx 是一个负数。这是一个线索，表明你的循环在它应该停止之后正在执行。
问题是 greater 分支会落到 lesser 分支。这将再次递减 ecx，导致它变为负数，并且您的循环将保持 运行，直到您遇到访问冲突。
快速修复是在 greater 标签下的 loop 指令之后放置一个无条件的 jmp。
更好的解决方法是将循环的尾部组合成更简单的条件：
cmp esi, [edi + (eax * 4)] ; compare array[j] and array[i]
jle lesser
mov eax, edx
lesser:
inc edx
loop secondLoop