10-01-2009, 07:54 AM
OK, so you want to change the copyright info in 2.0???
Don't get how the 0123 encoding/decoding works???
I made a program to help you.
>>Download here<<
NOT LIKELY TO BE UPDATED!!!
Credits to Silva for the table of characters!!!
Don't get how the 0123 encoding/decoding works???
I made a program to help you.
>>Download here<<
Screenshot (Click to View)
NOT LIKELY TO BE UPDATED!!!
Credits to Silva for the table of characters!!!
Source Code (Click to View)
Declarations:
Encoding:
Decoding:
VBNET-Code:
Shared Table() As Char = New Char() {" "c, "!"c, """"c, "#"c, "$"c, "%"c, "&"c, "'"c, "("c, ")"c, "*"c, "+"c, ","c, "-"c, "."c, "/"c, "0"c, "1"c, "2"c, "3"c, "4"c, "5"c, "6"c, "7"c, "8"c, "9"c, ":"c, ";"c, "<"c, "="c, ">"c, "?"c, "@"c, "A"c, "B"c, "C"c, "D"c, "E"c, "F"c, "G"c, "H"c, "I"c, "J"c, "K"c, "L"c, "M"c, "N"c, "O"c, "P"c, "Q"c, "R"c, "S"c, "T"c, "U"c, "V"c, "W"c, "X"c, "Y"c, "Z"c, "["c, "\"c, "]"c, "^"c, "_"c, "`"c, "a"c, "b"c, "c"c, "d"c, "e"c, "f"c, "g"c, "h"c, "i"c, "j"c, "k"c, "l"c, "m"c, "n"c, "o"c, "p"c, "q"c, "r"c, "s"c, "t"c, "u"c, "v"c, "w"c, "x"c, "y"c, "z"c, "{"c, "|"c, "}"c, "~"} Shared Trim() As String = New String() {Environment.NewLine, vbCrLf, vbCr, vbLf} |
Encoding:
VBNET-Code:
Shared Function Encode0123(ByVal Contents As String) As String Dim Result As String = "" Dim NewLineCount As Integer = -1 Dim Lines() As String = Contents.Split(Environment.NewLine) Try For i2 As Integer = 0 To Lines.Length - 1 NewLineCount += 1 If NewLineCount > 0 Then Result &= Environment.NewLine End If Dim LineBuffer As String = Lines(i2) For Each s As String In Trim LineBuffer = LineBuffer.Trim(s) Next For i3 As Integer = 0 To LineBuffer.Length - 1 Dim c As Char = LineBuffer(i3) If (Not c = Environment.NewLine) AndAlso (Array.IndexOf(Table, c) >= 0) Then Dim i As Integer = Array.IndexOf(Table, c) i = (i + (i3 Mod 4)) Mod (Table.Length) Result &= Table(i) End If Next Next Catch ex As Exception MessageBox.Show("Invalid decrypted string!!!", "Invalid String", MessageBoxButtons.OK) End Try Return Result End Function |
Decoding:
VBNET-Code:
Shared Function Decode0123(ByVal Contents As String) As String Dim Result As String = "" Dim NewLineCount As Integer = -1 Dim Lines() As String = Contents.Split(Environment.NewLine) Try For i2 As Integer = 0 To Lines.Length - 1 NewLineCount += 1 If NewLineCount > 0 Then Result &= Environment.NewLine End If Dim LineBuffer As String = Lines(i2) For Each s As String In Trim LineBuffer = LineBuffer.Trim(s) Next For i3 As Integer = 0 To LineBuffer.Length - 1 Dim c As Char = LineBuffer(i3) If (Not c = Environment.NewLine) AndAlso (Array.IndexOf(Table, c) >= 0) Then Dim i As Integer = Array.IndexOf(Table, c) Dim k As Integer = i3 Mod 4 i -= k i = i Mod (Table.Length) Do While i < 0 i += (Table.Length) Loop Result &= Table(i) End If Next Next Catch ex As Exception MessageBox.Show("Invalid encrypted string!!!", "Invalid String", MessageBoxButtons.OK) End Try Return Result End Function |