我试图使用一个链接列表程序来显示用户输入的反向字符串.以下是我的程序来反转用户输入的字符串.它还给出了字符串的长度：
TITLE ReadStringProc (ReadStringProc.asm)
include irvine32.inc
ListNode STRUCT
NodeData DWORD ?
NextPtr DWORD ?
ListNode ENDS
NULL = 0
Counter = 0
.data
input byte 100 dup(0)
stringinput byte "Enter any string: ",0
totallength byte "The total length is: ",0
reverse byte "The reverse string is: ",0
LinkedList LABEL DWORD
REPT input
Counter = Counter + 1
ListNode <Counter,($+ Counter * SIZEOF ListNode)>
ENDM
ListNode <0,0> ; tail node
.code
stringLength proc
push ebp
mov ebp,esp
push ebx
push ecx
mov eax,0
mov ebx,[ebp+8]
L1:
mov ecx,[ebx] ;you can use ecx,cx,ch,cl
cmp ecx,0 ;you can use ecx,cl
JE L2
add ebx,1
add eax,1
jmp L1
L2:
pop ecx
pop ebx
mov ebp,esp
pop ebp
ret 4
stringLength endp
swap MACRO first,last
push eax
mov ah,first
mov al,last
xor al,ah ;x
xor ah,al ;y
xor al,ah ;x
mov last,al
mov first,ah
pop eax
endM
stringReverse proc
push ebp
mov ebp,esp
push OFFSET input
call stringLength
mov edx,[ebp+8] ;edx = offset string to reverse
mov esi,offset 0
dec eax
mov ebx,edx ;ebx stores the pointer to the first character
add ebx,eax ;now ebx store the pointer to the last character before the '$'
reverseloop:
push edx
push ebx
swap [edx],[ebx]
inc edx ;increment of the right-most pointer
dec ebx ;decrement of the right-most pointer
cmp edx,ebx ;compares the left-most pointer to the right-most
jb reverseloop
jmp stopEnd ;"ja",there is no need to check a condition twice
stopEnd:
mov esp,ebp
pop ebp
ret 4
stringReverse endp
main proc
call clrscr
mov edx,offset stringinput
call writeString
mov edx,offset input
call writeString
call stringLength
mov edx,offset input
mov ecx,sizeof input
call readstring
call crlf
mov edx,offset totallength
call writestring
call writedec
call crlf
mov edx,offset reverse
call crlf
call writeString
push offset input
call stringReverse
mov edx,offset input
call writeString
call crlf
exit
main endp
end main
我的目标是找到一种方法来使用在Kip Irvine的Assembly x86书中找到的这段代码,并将其与我所使用的代码结合起来,以便我可以使用链接列表来显示反向字符串：
ListNode STRUCT
NodeData DWORD ?
NextPtr DWORD ?
ListNode ENDS
TotalNodeCount = 15
NULL = 0
Counter = 0
.data
LinkedList LABEL DWORD
REPT TotalNodeCount
Counter = Counter + 1
ListNode <Counter,0> ; tail node
.code
main PROC
mov esi,OFFSET LinkedList
; Display the integers in the NodeData members.
NextNode:
; Check for the tail node.
mov eax,(ListNode PTR [esi]).NextPtr
cmp eax,NULL
je quit
; Display the node data.
mov eax,(ListNode PTR [esi]).NodeData
call WriteDec
call Crlf
; Get pointer to next node.
mov esi,(ListNode PTR [esi]).NextPtr
jmp NextNode
quit:
exit
有人可以指导我走正确的道路.谢谢.
解决方法
您不需要链接列表来反转字符串.不,认真
但是,如果你坚持…
stringReverse proc; <-- Warning: EBX and ESI are not preserved!
push ebp;
mov ebp,esp;
mov ebx,[ebp + 8]; <-- EBX = target string pointer
push ebx;
call stringLength; <-- we trust that stringLength() preserves EBX
; mov ebx,[ebp + 8]; <-- uncomment this if it actually does not
mov esi,ebx;
xor edx,edx;
add ebx,eax;
neg eax;
@fill:
mov cl,[ebx + eax];
sub esp,8; <-- allocating a new linked list item on stack
mov [esp],cl; <-- 1st DWORD / 1st BYTE = current character
mov [esp + 4],edx; <-- 2nd DWORD = next item pointer
mov edx,esp; <-- EDX = current item pointer
inc eax;
jne @fill;
@roll:
mov cl,[edx];
mov [esi],cl;
inc esi;
mov edx,[edx + 4]; <-- next item,here we go!
test edx,edx; <-- what if we are done?
jne @roll;
mov esp,ebp; <-- discarding the allocated stack
pop ebp;
ret 4;
这样的东西
它应该工作,但我没有真正测试.
