Delphi: string encryption method and base64

Question

Please suggest me a good string encryption method . Not XOR, it isn't strong enough.

Can I use Base64 to represent the encrypted string, but without "=" on the string's end? I can add it manually. Is it normal? That is a user will use Base64 without "=" in a program, and I will add it. I do not want to have a view with '=', it isn't nice :)

Thanks!!!

Answer 1

Here's one encryption library: http://www.cityinthesky.co.uk/opensource/dcpcrypt

Yes, you can show a base64 string without the '=' sign on the end. You just need to make sure that when you pass the value to a method the method is smart enough to add it back on before attempting the decrypt. This is a pretty common scenario.

Answer 2

heres a function (or a couple of functions) to encode and decode strings you can use, you can call it using Base64Encode('string to be encoded') and Base64Decode('string to be decoded') hope this helps.

const
B64: array[0..63] of byte= (65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,
81,82,83,84,85,86,87,88,89,90,97,98,99,100,101,102,103,104,105,106,107,108,
109,110,111,112,113,114,115,116,117,118,119,120,121,122,48,49,50,51,52,53,
54,55,56,57,43,47);

function B64Encode(pInput: pointer; pOutput: pointer; Size: longint): longint;
var
  i, iptr, optr: integer;
  Input, Output: PByteArray;
begin
 Input:= PByteArray(pInput); Output:= PByteArray(pOutput);
 iptr:= 0; optr:= 0;
 for i:= 1 to (Size div 3) do
  begin
    Output^[optr+0]:= B64[Input^[iptr] shr 2];
    Output^[optr+1]:= B64[((Input^[iptr] and 3) shl 4) + (Input^[iptr+1] shr 4)];
    Output^[optr+2]:= B64[((Input^[iptr+1] and 15) shl 2) + (Input^[iptr+2] shr 6)];
    Output^[optr+3]:= B64[Input^[iptr+2] and 63];
    Inc(optr,4); Inc(iptr,3);
  end;
case (Size mod 3) of
 1: begin
    Output^[optr+0]:= B64[Input^[iptr] shr 2];
    Output^[optr+1]:= B64[(Input^[iptr] and 3) shl 4];
    Output^[optr+2]:= byte('=');
    Output^[optr+3]:= byte('=');
end;
 2: begin
    Output^[optr+0]:= B64[Input^[iptr] shr 2];
    Output^[optr+1]:= B64[((Input^[iptr] and 3) shl 4) + (Input^[iptr+1] shr 4)];
    Output^[optr+2]:= B64[(Input^[iptr+1] and 15) shl 2];
    Output^[optr+3]:= byte('=');
 end;
end;
 Result:= ((Size+2) div 3) * 4;
end;


function Base64Encode(const Value: AnsiString): AnsiString;
 begin
  SetLength(Result,((Length(Value)+2) div 3) * 4);
  B64Encode(@Value[1],@Result[1],Length(Value));
end;


function B64Decode(pInput: pointer; pOutput: pointer; Size: longint): longint;
 var
  i, j, iptr, optr: integer;
  Temp: array[0..3] of byte;
  Input, Output: PByteArray;
 begin
  Input:= PByteArray(pInput); Output:= PByteArray(pOutput);
  iptr:= 0; optr:= 0;
  Result:= 0;
  for i:= 1 to (Size div 4) do
  begin
   for j:= 0 to 3 do
    begin
     case Input^[iptr] of
      65..90 : Temp[j]:= Input^[iptr] - Ord('A');
      97..122: Temp[j]:= Input^[iptr] - Ord('a') + 26;
      48..57 : Temp[j]:= Input^[iptr] - Ord('0') + 52;
      43     : Temp[j]:= 62;
      47     : Temp[j]:= 63;
      61     : Temp[j]:= $FF;
    end;
   Inc(iptr);
end;
  Output^[optr]:= (Temp[0] shl 2) or (Temp[1] shr 4);
  Result:= optr+1;
 if (Temp[2]<> $FF) and (Temp[3]= $FF) then
  begin
    Output^[optr+1]:= (Temp[1] shl 4) or (Temp[2] shr 2);
    Result:= optr+2;
    Inc(optr)
 end
  else if (Temp[2]<> $FF) then
   begin
     Output^[optr+1]:= (Temp[1] shl 4) or (Temp[2] shr 2);
     Output^[optr+2]:= (Temp[2] shl 6) or  Temp[3];
     Result:= optr+3;
     Inc(optr,2);
  end;
  Inc(optr);
 end;
end;

function Base64Decode(const Value: AnsiString): AnsiString;
begin
  SetLength(Result,(Length(Value) div 4) * 3);
  SetLength(Result,B64Decode(@Value[1],@Result[1],Length(Value)));
end;