by Jacob Steenhagen (3 Submissions)
Category: Encryption
Compatability: VB 6.0
Difficulty: Unknown Difficulty
Originally Published: Tue 22nd August 2000
Date Added: Mon 8th February 2021
Rating: 
(1 Votes)
Encryption use an EncryptKey
API Declarations
' if that's what you really want).
VBcoders
'
' When I wrote this encryption module I had a couple of goals:
' 1. Must be at least moderately tough to crack
' 2. All characters must be able to be displayed by Windows
' (For sending via e-mail and using copy/paste to decrypt)
'
' Usage:
' Useing these functions is as easy as you'd assume it'd be,
' simply call the function as follows...
' ReturnString = [De|En]cryptString("Text to be de/encrypted", "Crypt Key")
'
' This will encrypt any string that you can type on the
' (English) keyboard. Extended ASCII is not supported in
' either string to be encrypted or the encryption key.
Public Function EncryptString(ByVal InString As String, ByVal EncryptKey As String) As String
Dim TempKey, OutString As String
Dim OldChar, NewChar, CryptChar As Long
Dim i As Integer
' Initilize i and make sure the EncryptKey is long enough
i = 0
Do
TempKey = TempKey + EncryptKey
Loop While Len(TempKey) < Len(InString)
' Loop through the string to encrypt each character.
Do
i = i + 1
OldChar = Asc(Mid(InString, i, 1))
CryptChar = Asc(Mid(TempKey, i, 1))
' If it's an even character, add the ASCII value of the
' appropriate character in the Key, otherwise, subract it.
' Also, make sure the value is between 0 and 127.
Select Case i Mod 2
Case 0 'Even Character
NewChar = OldChar + CryptChar
If NewChar > 127 Then NewChar = NewChar - 127
Case Else 'Odd Character
NewChar = OldChar - CryptChar
If NewChar < 0 Then NewChar = NewChar + 127
End Select
' If the value is less than 35, add 40 to it (to make sure
' it's in the display range) and put it in an escape
' sequence (using ! [ASCII Value 33] as the escape char)
If NewChar < 35 Then
OutString = OutString + "!" + Chr(NewChar + 40)
Else
OutString = OutString + Chr(NewChar)
End If
Loop Until i = Len(InString)
EncryptString = OutString
End Function
Public Function DecryptString(ByVal InString As String, ByVal EncryptKey As String) As String
Dim TempKey, OutString As String
Dim OldChar, NewChar, CryptChar As Long
Dim i, c As Integer
' Initialize c and i (loop variables)
c = 0 ' c is used for InString
i = 0 ' i is used for EncryptKey
' Make sure the EncryptKey is long enough
Do
TempKey = TempKey + EncryptKey
Loop While Len(TempKey) < Len(InString)
Do
' In the decrypt function, two integers are need keeping
' track of location (becuase the escape sequence it two
' chars long, but only has one placeholder in the key)
i = i + 1
c = c + 1
OldChar = Asc(Mid(InString, c, 1))
' If this is an escape sequence, get the next character and
' subtract 40 from it's value.
If OldChar = 33 Then
c = c + 1
OldChar = Asc(Mid(InString, c, 1))
OldChar = OldChar - 40
End If
CryptChar = Asc(Mid(TempKey, i, 1))
' If it's an even character, subract the appropriate key
' value... also, if it's out of range, bring it back in.
Select Case i Mod 2
Case 0 'Even Character
NewChar = OldChar - CryptChar
If NewChar < 0 Then NewChar = NewChar + 127
Case Else 'Odd Character
NewChar = OldChar + CryptChar
If NewChar > 127 Then NewChar = NewChar - 127
End Select
OutString = OutString + Chr(NewChar)
Loop Until c = Len(InString)
DecryptString = OutString
End Function
