我使用异或算法将２个字符串加密，形成一个新的字符串，出现不可显示的字符怎么处理？(100分)

我发现生成的新字符串中,偶而会形成不可显示的字符.这时我就无法将该字符串拷贝到别的
编辑窗口中.
有什么办法处理这种问题?

base64再编码。 ：）

有算法没有?
谢谢指教

只让密码字节的 $9F(1001 1111) 这6位参与加密。

這情況一般出現在網頁的傳輸上...不知你是發生在什麼情況?

小雨哥已经说了好法子了。

其实你可以用MD5加密，或不要直接传送进字符串，传送到内存流。[]

procedure TForm1.Button1Click(Sender: TObject);
const Str='Button1Click';
var S:String;
I:Integer;
begin
S:='Private declarations';
for I:=1 to Length(S) do
S:=Chr(Ord(S) xor Ord(Str[((I-1) mod Length(Str))+1]));
S:=TIdEncoderMIME.EncodeString(S);
Caption:=S;
end;

procedure TForm1.Button2Click(Sender: TObject);
const Str='Button1Click';
var S:String;
I:Integer;
begin
S:=Caption;
s:=TIdDecoderMIME.DecodeString(S);
for I:=1 to Length(S) do
S:=Chr(Ord(S) xor Ord(Str[((I-1) mod Length(Str))+1]));
Caption:=S;
end;

TIDDecoderMIME.DecodeString 是什么？

Base64编码函数：
function Base64_Encode(var str:string):boolean;
const
EncodeBase64:string='ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/';

var
Byte1,Byte2,Byte3:char;
LenMod:integer;
LenDiv:integer;
i:integer;
EncodeStr:string;
begin
EncodeStr:='';
LenDiv:=length(str) div 3;
LenMod:=length(str) mod 3;
for i:=1 to LenDiv do
begin
Byte1:=str[i*3-2];
Byte2:=str[i*3-1];
Byte3:=str[i*3];
EncodeStr:=EncodeStr+EncodeBase64[(Byte(Byte1) shr 2)+1];
EncodeStr:=EncodeStr+EncodeBase64[(((Byte(Byte1) shl 4) or (Byte(Byte2) shr 4)) and $3f)+1];
EncodeStr:=EncodeStr+EncodeBase64[(((Byte(Byte2) shl 2) or (Byte(Byte3) shr 6)) and $3f)+1];
EncodeStr:=EncodeStr+EncodeBase64[((Byte(Byte3) and $3f))+1];
end;{end of for}

if LenMod=1 then
begin
Byte1:=str[length(str)];
EncodeStr:=EncodeStr+EncodeBase64[(Byte(Byte1) shr 2)+1];
EncodeStr:=EncodeStr+EncodeBase64[((Byte(Byte1) shl 4) and $3f)+1];
EncodeStr:=EncodeStr+'=';
EncodeStr:=EncodeStr+'=';
end;

if LenMod=2 then
begin
Byte1:=str[length(str)-1];
Byte2:=str[length(str)];
EncodeStr:=EncodeStr+EncodeBase64[(Byte(Byte1) shr 2)+1];
EncodeStr:=EncodeStr+EncodeBase64[(((Byte(Byte1) shl 4) or (Byte(Byte2) shr 4)) and $3f)+1];
EncodeStr:=EncodeStr+EncodeBase64[((Byte(Byte2) shl 2) and $3f)+1];
EncodeStr:=EncodeStr+'=';
end;

Str:=EncodeStr;
end;

解码函数：
function Base64_Decode(var str:string):boolean;
const
DecodeBase64:array[0..127] of integer
=(-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,
-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,
-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,62,-1,-1,63,
52,53,54,55,56,57,58,59,60,61,-1,-1,-1,-1,-1,-1,
-1,0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,
15,16,17,18,19,20,21,22,23,24,25,-1,-1,-1,-1,-1,
-1,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,
41,42,43,44,45,46,47,48,49,50,51,-1,-1,-1,-1,-1);

var
i,LenDiv:integer;
DecodeStr:string;
B1,B2,B3,B4:char;
begin
LenDiv:=length(str) div 4;
DecodeStr:='';
for i:=1 to LenDiv do
begin
B1:=str[i*4-3];
B2:=str[i*4-2];
B3:=str[i*4-1];
B4:=str[i*4];
DecodeStr:=DecodeStr+char((DecodeBase64[ord(B1)] shl 2)
or (DecodeBase64[ord(B2)] shr 4));
if B3<>'=' then
DecodeStr:=DecodeStr+char((DecodeBase64[ord(B2)] shl 4)
or (DecodeBase64[ord(B3)] shr 2));
if B4<>'=' then
DecodeStr:=DecodeStr+char((DecodeBase64[ord(B3)] shl 6)
or DecodeBase64[ord(B4)]);
end;{end of for}

Str:=DecodeStr;
end;

thanx

谁知道 Base24 算法? 也就是微软序列号的转换算法,另开帖送分

除此之外，用byte数组也可以啊