Attribute VB_Name = "HexToASCII"
Option Explicit

' I produced this when I started doing some machine code and assembaly language
' programming, when knowing hex was useful.
' Like the about box says i started work on a binary calculator to deal with
' 2s complement, binary fractions, binary exponential etc but unfortunatly that
' has been put on hold due to a general excess of work.

' If you use this code for anything than e-mail me; grredo@hotmail.com

Private integer1 As Integer
Private integer2 As Integer
Private integer3 As Integer
Private string1 As String
Private string2 As String
Private string3 As String

Private Function pad(ByVal binstring As String) As String

' the variables to be used are reset

integer1 = 0
integer2 = 0

' this function will pad out a string with 0s to make it into a byte

integer2 = 8 - Len(binstring)  'the eight in this line is seting the length of the string to be produced
For integer1 = 1 To integer2
    binstring = "0" & binstring     '0 is added to the front of the string
Next integer1
pad = binstring
End Function

Public Function calcden2bin(ByVal den As String) As String

' the two variables to be used in this function are reset

string1 = ""
integer1 = 0

' integer1 = mod 2 of den
' den = integer division of itself
' string1 = integer1 and itself

Do Until den = 0
    integer1 = den Mod 2
    den = den \ 2
    string1 = integer1 & string1
Loop

'the string is then padded out using the pad function then returned

string1 = pad(string1)
calcden2bin = string1
End Function

Public Function calcbin2den(ByVal bin As String) As String

' the variables to be used in this function are set to their starting values

integer1 = 0
string1 = ""
integer3 = 0
integer2 = 1


Do While integer3 < Len(bin) 'do while the digit number being checked is 1 less than the total length of the code to be checked
    integer3 = integer3 + 1
    If integer3 > 1 Then integer2 = integer2 * 2 'increase the digno (used for digit ident + weighting) by one
    string1 = Left((Right(bin, integer3)), 1) 'set currentdigit to = the
    integer1 = integer1 + (integer2 * string1) 'total = total + the digno(weighting) * the current digit
Loop

' the end value is returned

calcbin2den = integer1
End Function

Public Function calcden2hex(ByVal den As String) As String

' all the variables are reset

string1 = ""
string2 = ""
string3 = ""

' string1 = the integer division of the input number

string1 = (den \ 16)

' if string 1 is greater than nine then its charecter is calculated using the lcalc function

If string1 > 9 Then string1 = lcalc(string1)
string2 = den - ((den \ 16) * 16)

' if the string2 is greater than 9 then its charecter is calculated using the lcalc function
' if not then string3 = string 2
If string2 > 9 Then
    string3 = lcalc(string2)
Else
    string3 = string2
End If

' the value bearing strings are combined and returned

string2 = string1 & string3
calcden2hex = string2
End Function

Public Function calchex2den(ByVal hex As String) As String

' all the variables to be used in this function are reset

string1 = "0"
string2 = "0"
integer1 = 0
integer2 = 0
integer3 = 0

' string1 = ther rightermost digit of the input hex value

string1 = Right(hex, 1)

' if string1 isn't numeric then integer1 = l2calc of the charecter string1
' if string1 is numeric then integer1 = string1

If IsNumeric(string1) = False Then
    integer1 = l2calc(string1)
    Else:
    integer1 = Val(string1)
End If

' if the length of the input string is 2 then
' string2 = leftermost digit of the input string
' if string2 isnt numeric then string 2 = l2calc of string2
' integer2 = the value held by string2

If Len(hex) = 2 Then
    string2 = Left(hex, 1)
        If IsNumeric(string2) = False Then
            integer2 = l2calc(string2)
        Else:
            integer2 = Val(string2)
        End If
End If

' integer3 = integer1 + (integer2 * base value)
' integer3 is then returned

integer3 = integer1 + (integer2 * 16)
calchex2den = integer3
End Function

Public Function calcden2asci(ByVal den As Integer) As String

' to keep in line with the rest of the program i have set the simple function chr

calcden2asci = Chr(den)
End Function

Public Function calcasci2den(ByVal asci As String) As Integer

' to keep in line with the rest of the program i have set the simple function asc

calcasci2den = Asc(asci)
End Function

Public Function lcalc(ByVal remainder As String) As String

' using select case the letter associated with the value input is calculated

Select Case remainder
    Case 10
        remainder = "a"
    Case 11
        remainder = "b"
    Case 12
        remainder = "c"
    Case 13
        remainder = "d"
    Case 14
        remainder = "e"
    Case 15
        remainder = "f"
End Select

' the letter for the number is returned

lcalc = remainder
End Function

Public Function l2calc(ByVal hexletter As String) As Integer

' using select case the value associated with the letter input is calculated

Select Case hexletter
    Case "a"
        hexletter = 10
    Case "b"
        hexletter = 11
    Case "c"
        hexletter = 12
    Case "d"
        hexletter = 13
    Case "e"
        hexletter = 14
    Case "f"
        hexletter = 15
End Select

' the number for the letter is returned

l2calc = hexletter
End Function