Delphi:将字符串放入编辑框时是否存在访问冲突？

Question

好吧,我正在学习Delphi中的一些内联汇编,并且汇编加密例程一切都很顺利,直到我尝试将ShortString解析为Textbox.

我得到的违规行为如下: 错误http://img828.imageshack.us/img828/2410/43832949.png

完整代码在这里:

procedure TForm2.Button1Click(Sender: TObject);

var
len,keylen:integer;
name, key:ShortString;

begin

name :=  ShortString(Edit1.Text);
key := '_r <()<1-Z2[l5,^';
len := Length(name);
keylen := Length(key);

nameLen := len;
serialLen := keyLen;

asm

  XOR EAX,EAX
  XOR ESI,ESI
 XOR EDX,EDX
 XOR ECX,ECX


  @loopBegin:

        MOV EAX,ESI
        PUSH $019
        CDQ
        IDIV DWORD PTR DS:[serialLen]
        MOV EAX,ESI
        POP EBX
        LEA ECX,DWORD PTR DS:[key+EDX]
        CDQ
        IDIV DWORD PTR DS:[nameLen]
        LEA EAX,DWORD PTR DS:[name]
        MOVZX EAX,BYTE PTR DS:[name+EDX]
        MOVZX EDX,BYTE PTR DS:[ECX]
        XOR EAX,EDX
        CDQ
        IDIV EBX
        ADD DL,$041
        INC ESI
        CMP ESI,DWORD PTR DS:[serialLen]
        MOV BYTE PTR DS:[ECX],DL

        JL @loopBegin


end;

edit2.Text:= TCaption(key);


end;

如果我在"edit2.Text:= TCaption(key);"行上放置一个断点 我可以看到ShortString"key"确实已经被正确加密了,但它背后也有许多奇怪的字符.

前16个字符是真正的加密.

加密http://img831.imageshack.us/img831/365/29944312.png

bigger version: http://img831.imageshack.us/img831/365/29944312.png

谢谢!

Answer 1

代码的作用

对于那些不会说汇编程序的人来说,这就是代码可能应该在Pascal中执行的操作."可能"因为原文包含一些错误:

procedure TForm14.Button1Click(Sender: TObject);
var KeyLen:Integer;
    Name, Key:ShortString;
    i:Integer;
    CurrentKeyByte:Byte;
    CurrentNameByte:Byte;
begin
  Name := ShortString(Edit1.Text);
  Key := '_r <()<1-Z2[l5,^';
  keyLen := Length(key);

  asm int 3 end; // This is here so I can inspect the assembler output in the IDE
                 // for the "Optimised" version of the code

  for i:=1 to Length(Name) do
  begin
    CurrentKeyByte := Byte(Key[i mod KeyLen]);
    CurrentNameByte := Byte(Name[i]);
    CurrentNameByte := ((CurrentKeyByte xor CurrentNameByte) mod $019) + $041;
    Name[i] := AnsiChar(CurrentNameByte);
  end;

  Caption := Name;

end;

启用优化后,由此生成的汇编代码实际上比建议的代码更短,不包含冗余代码,我愿意打赌更快.以下是我在Delphi生成的代码中注意到的一些优化(与OP提出的汇编代码相比):

• 德尔福扭转了循环(下降0).这节省了一条"CMP"指令,因为编译器可以简单地"DEC ESI"并循环零标志.
• 使用"XOR EDX"和"DIV EBX"进行第二次分割,节省了一些微小的周期.

为什么提供的汇编程序代码失败？

这是原始汇编代码,带有注释.例程结束时的错误,在"CMP"指令处 - 它将ESI与KEY的长度进行比较,而不是NAME的长度.如果KEY比NAME长,则"加密"继续在NAME的顶部,覆盖内容(被覆盖的内容之一是字符串的NULL终止符,导致调试器在正确的字符后显示有趣的字符).

虽然不允许覆盖EBX和ESI,但这并不是导致AV代码的原因,可能是因为周围的Delphi代码没有使用EBX或ESI(只是试过这个).

asm

 XOR EAX,EAX ; Wasteful, the first instruction in Loop overwrites EAX
 XOR ESI,ESI
 XOR EDX,EDX ; Wasteful, the first CDQ instruction in Loop overwrites EDX
 XOR ECX,ECX ; Wasteful, the first LEA instruction overwrites ECX


 @loopBegin:
       ; Etering the loop, ESI holds the index for the next char to be
       ; encrypted.

       MOV EAX,ESI ; Load EAX with the index for the next char, because
                   ; we intend to do some divisions (setting up the call to IDIV)
       PUSH $019   ; ? pushing this here, so we can pop it 3 lines later... obfuscation
       CDQ         ; Sign-extend EAX (required for IDIV)
       IDIV DWORD PTR DS:[serialLen] ; Divide EAX by the length of the key.
       MOV EAX,ESI ; Load the index back to EAX, we're planning on an other IDIV. Why???
       POP EBX     ; Remember the PUSH $019?
       LEA ECX,DWORD PTR DS:[key+EDX] ; EDX is the result of "ESI mod serialLen", this
                                      ; loads the address of the current char in the
                                      ; encryption key into ECX. Dividing by serialLen
                                      ; is supposed to make sure we "wrap around" at the
                                      ; end of the key
        CDQ ; Yet some more obfuscation. We're now extending EAX into EDX in preparation for IDIV.
            ; This is obfuscation becasue the "MOV EAX, ESI" instruction could be written right here
            ; before the CDQ.
        IDIV DWORD PTR DS:[nameLen] ; We divide the current index by the length of the text
                                    ; to be encrypted. Once more the code will only use the reminder,
                                    ; but why would one do this? Isn't ESI (the index) always supposed to
                                    ; be LESS THEN nameLen? This is the first sign of trouble.
        LEA EAX,DWORD PTR DS:[name] ; EAX now holds the address of NAME.
        MOVZX EAX,BYTE PTR DS:[name+EDX] ; EAX holds the current character in name
        MOVZX EDX,BYTE PTR DS:[ECX]      ; EDX holds the current character in Key
        XOR EAX,EDX ; Aha!!!! So this is an obfuscated XOR loop! EAX holds the "name[ESI] xor key[ESI]"
        CDQ         ; We're extending EAX (the XOR result) in preparation for a divide
        IDIV EBX    ; Divde by EAX by EBX (EBX = $019). Why????
        ADD DL,$041 ; EDX now holds the remainder of our previous XOR, after the division by $019;
                    ; This is an number from $000 to $018. Adding $041 turns it into an number from
                    ; $041 to $05A (ASCII chars from "A" to "Z"). Now I get it. This is not encryption,
                    ; this is a HASH function! One can't un-encrypt this (information is thrown away at
                    ; the division).
        INC ESI     ; Prep for the next char


        ; !!! BUG !!!
        ;
        ; This is what's causing the algorithm to generate the AV. At this step the code is
        ; comparing ESI (the current char index) to the length of the KEY and loops back if
        ; "ESI < serialLen". If NAME is shorter then KEY, encryption will encrypt stuff beyond
        ; then end of NAME (up to the length of KEY). If NAME is longer then KEY, only Length(Key)
        ; bytes would be encrypted and the rest of "Name" would be ignored.
        ;
        CMP ESI,DWORD PTR DS:[serialLen]


        MOV BYTE PTR DS:[ECX],DL ; Obfuscation again. This is where the mangled char is written
                                 ; back to "Name".

        JL @loopBegin            ; Repeat the loop.

我的2美分的建议

汇编程序应该用于SPEED优化而不是其他任何东西.它看起来好像OP试图使用Assembler来模糊代码所做的事情.没有帮助,我花了几分钟才弄清楚代码究竟在做什么,我不是一个汇编专家.

Answer 2

首先,您需要保留EDI和ESI.只有EAX,EDX和ECX可以在没有保存的情况下使用(除非您加载它并需要保存它).

尝试在您的代码周围添加一些PUSH EDI,PUSH ESI和POP ESI,POP EDI.