使用此程序一段时间后,它会在创建输出文件后继续停止。我使用的是Visual Basic 2010,我仍然是初学者。作业问题是:
描述(对称加密):
编码
解码
我可以弄清楚如何加密文本,但是在使用位于此教科书地址的库创建输出文件时遇到问题: http://www.kipirvine.com/asm/examples/index.htm
我将在下面包含我的代码,并通过评论材料显示我在此处尝试了多少次。这本书不能很好地解释我,所以如果我能看到它试图说它将是非常有帮助的!
提前致谢!
INCLUDE Irvine32.inc
BUFMAX = 128 ; maximum buffer size
KEYMAX = 128 ; maximum buffer size
BUFFER_SIZE = 5000
.data
sPrompt BYTE "Enter some text message: ", 0
keyPrompt BYTE "Enter a private key [1-255]: ", 0
cFile BYTE "Enter a filename for cypher text: ", 0
sEncrypt BYTE "Cypher text ", 0
sDecrypt BYTE "Decrypted: ", 0
error BYTE "The key must be within 1 - 255! ", 0
buffer BYTE BUFMAX + 1 DUP(0)
bufSize DWORD ?
keyStr BYTE KEYMAX + 1 DUP(0)
keySize DWORD ?
key DWORD ?
filename BYTE "newfile.txt ", 0
fileHdl DWORD ?
bufFile BYTE BUFFER_SIZE DUP (?)
.code
main PROC
call InputTheString ; input the plain text
call InputTheKey ; input the security key
call CypherFile ; input a cypher filename
;call TranslateBuffer ; encrypt the buffer
;mov edx, OFFSET sEncrypt ; display encrypted message
;call DisplayMessage
;call TranslateBuffer ; decrypt the buffer
;mov edx, OFFSET sDecrypt ; display decrypted message
;call DisplayMessage
exit
main ENDP
InputTheKey PROC
pushad ; save 32-bit registers
LK: mov edx, OFFSET keyPrompt ; display a prompt
call WriteString ; Enter a private key [1-255]
call Crlf ; start a new line
call ReadInt ; read int into system
mov key, eax ; store int into keyStr
cmp eax, 255 ; compare newly read int
ja LC ; jump if above 255 to LC
cmp eax, 1 ; compare newly read int
jb LC ; jump if below 1 to LC
jmp LR ; if between range jump to LR
LC: mov edx, OFFSET error ; The key must be within 1 - 255!
call WriteString ; Display the error
call Crlf ; start a new line
loop LK ; loop back to enter the security key
LR: popad ; restore the registers
ret
InputTheKey ENDP
CypherFile PROC
pushad
mov edx, OFFSET cFile ; "Enter a filename for cypher text
call WriteString ; Enter a name for encrypted file
call Crlf
call ReadString ; Store the filename in eax
;mov filename, eax
mov edx, OFFSET filename
;push eax
;mov eax, fileHdl
;mov edx, OFFSET bufFile
;mov ecx, BUFFER_SIZE
;mov edx, "C:\outputtext.txt"
call CreateOutputFile
;mov edx, OFFSET filename
;mov ecx, SIZEOF filename
;push eax
;mov eax, bufSize
call WriteToFile
pop eax
;call CloseFile
ret
CypherFile ENDP
InputTheString PROC
pushad ; save 32-bit registers
mov edx, OFFSET sPrompt ; display a prompt
call WriteString ; "Enter some text message"
call Crlf ; start a new line
mov ecx, BUFMAX ; maximum character count
mov edx, OFFSET buffer ; point to the buffer
call ReadString ; input the string
mov bufSize, eax ; save the length
popad
ret
InputTheString ENDP
COMMENT !
DisplayMessage PROC
pushad
call WriteString
mov edx, OFFSET buffer ; display the buffer
call WriteString
call Crlf
call Crlf
popad
ret
DisplayMessage ENDP
TranslateBuffer PROC
pushad
mov ecx, bufSize ; loop counter
mov esi, 0 ; index 0 in buffer
mov edi, 0 ; index 0 in the key
L1:
mov al, keyStr[edi] ; get a character from encryption key
xor buffer[esi], al ; translate a byte
inc esi ; point to next byte
inc edi ; go to next position in key
cmp edi, keySize ; compare if equal to size of the key
jb L2
mov edi, 0 ; reset to beginning of the key
L2: loop L1
popad
ret
TranslateBuffer ENDP
!
END main
答案 0 :(得分:2)
从哪里开始?
您是否错过了ReadString的一些参数?也许是指向存储输入文件名的位置的指针?要接收文件名的缓冲区的大小?
CypherFile
您使用pushad
将所有寄存器压入堆栈,但最后只有pop eax
。那里的大问题应该是popad
就目前而言,它没有向输出文件写入任何内容,因为WriteToFile的参数被注释掉了。
编辑 - 反对我的“只是给代码” 您需要告诉ReadString保存输入的文件名和缓冲区大小的位置。 然后将其传递给CreateOutputFile - 在CyperFile proc -
mov edx, offset buffer ; you are missing a buffer for filename
mov ecx, BUFMAX ; buffer size
call ReadString
mov edx, offset buffer ; Pass this to CreateOutputFile
call CreateOutputFile
现在正在阅读CreateOutputFile
的来源,它说它在成功时返回一个文件句柄,将其保存在某处。完成写入文件后,将其与CloseFile
一起使用。如果未成功创建文件,则会返回INVALID_HANDLE_VALUE