[Gambas-user] perl unpack

[Ron]

Great work!

I had a few vb code to start from, but the all where slightly
different, so where the results.
This shows it's sometimes better to just start from the basic info and
work from there line by line...

Thanks alot!
Going to put this in my project...

Regards,
Ron_2nd.


2011/9/20 Caveat <Gambas at ...1950...>:
> Don't stress too much over the `, it's just a kind of non-standard
> padding character.  The % at the beginning of the string says we only
> have 5 characters to decode so we shouldn't worry...we SHOULD always
> have an exact multiple of 4 characters after the first length byte...
> but some of them may not matter...
>
> This should do it:
>
> ======================================================================
> Private Function decodeUU(codedStr As String) As Byte[]
>
>  Dim idx, idy, ptr As Integer
>  Dim result As Byte[]
>  Dim lengthUU, ascAChar, ascMin32 As Integer
>  Dim binFour As String = ""
>  ' First character's ascii code - 32 is the length
>  ascAChar = Asc(Left$(codedStr, 1))
>  ascMin32 = ascAChar - 32
>  lengthUU = ascMin32
>  Print "Expecting a length of: " & lengthUU
>  ' Set the size of the result array
>  result = New Byte[lengthUU]
>  ' Initialise pointer into the result array
>  ptr = 0
>  ' Step through the uuencoded string character by character starting at
> the 2nd character (1st is the length)
>  For idx = 2 To Len(codedStr)
>    ascAChar = Asc(Mid$(codedStr, idx, 1))
>    ' Only include what is not whitespace
>    If ascAChar > 31 And ascAChar < 97 Then
>      ' Subtract 32 from the ascii code of the character
>      ascMin32 = ascAChar - 32
>      ' Assemble a block of four 6-bit values
>      binFour = binFour & Right$("000000" & intToBase(ascMin32,
> BASE_BINARY), 6)
>      ' Once we have 4 binary 6-bit 'characters' in our string
>      If Len(binFour) = 24 Then
>        ' Treat the 4 6-bit characters as 3 8-bit characters
>        For idy = 1 To 3
>          ' Make sure we don't go trying to convert more than the length
> says we have to
>          If ptr < result.Length
>            Print "Bin to convert: " & Mid$(binFour, 1 + ((idy - 1) *
> 8), 8)
>            ' Converts each block of 8 bits to its decimal value and
> assigns to the output byte array
>            result[ptr] = toInt(Mid$(binFour, 1 + ((idy - 1) * 8), 8),
> BASE_BINARY)
>            Inc ptr
>          End If
>        Next
>        ' Be sure to clear out binFour for the next unit of UUencoding
>        binFour = ""
>      End If
>    End If
>  Next
>  Return result
>
> End
> ======================================================================
>
> You probably need the routines to convert between bases too:
>
> ======================================================================
> Private Function convertBase(numberIn As String, fromBase As Integer,
> toBase As Integer) As String
>
>  Dim value As Integer
>  value = toInt(numberIn, fromBase)
>  Return intToBase(value, toBase)
>
> End
>
> Private Function intToBase(numberIn As Integer, base As Integer) As
> String
>
>  Dim remain, numToDivide As Integer
>  Dim result As String = ""
>
>  numToDivide = numberIn
>  Do While numToDivide / base > 0
>    remain = numToDivide Mod base
>    numToDivide = (Int)(numToDivide / base)
>    result = DIGITS[remain] & result
>  Loop
>
>  Return result
>
> End
>
> Private Function toInt(inputStr As String, base As Integer) As Integer
>
>  Dim idx, mult, result, value As Integer
>  mult = 1
>  For idx = Len(inputStr) To 1 Step -1
>    ' If we're in a base with digits bigger than 9
>    ' we need the Find to return 10 for A, 11 for B, 12 for C etc.
>    value = DIGITS.Find(UCase(Mid$(inputStr, idx, 1))) * mult
>    result = result + value
>    mult = mult * base
>  Next
>  Return result
>
> End
> ======================================================================
>
> And don't forget a few Consts for convenience:
>
> ======================================================================
> Private Const TEST_STR As String = "%<V5N9#H`"
> Private Const BASE_BINARY As Integer = 2
> Private Const BASE_OCTAL As Integer = 8
> Private Const BASE_DENARY As Integer = 10
> Private Const BASE_HEX As Integer = 16
> Private DIGITS As String[] = ["0", "1", "2", "3", "4", "5", "6", "7",
> "8", "9", "A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K", "L",
> "M", "N", "O", "P", "Q", "R", "S", "T", "U", "V", "W", "X", "Y", "Z"]
> ======================================================================
>
> Oh and just for fun, here's an encode function too... you will notice
> that I encode "send:" CORRECTLY... LOL!
>
> ======================================================================
> Private Function encodeUU(source As Byte[]) As String
>
>  Dim idx, idy, idxThree As Integer
>  Dim result As String
>  Dim aByte As Byte
>  Dim aBinChar, binCharGroup As String
>  Dim threeChars As Byte[]
>  binCharGroup = ""
>  result = result & Chr$(source.Count + 32)
>  For idx = 0 To source.Max
>    aByte = source[idx]
>    ' Convert the byte to exactly 8 digits of binary
>    ' so for e.g. pad 1 to become 00000001
>    aBinChar = Right$("00000000" & intToBase(aByte, BASE_BINARY), 8)
>    Print "aByte: " & aByte & " abinChar: " & aBinChar
>    ' Add bytes together to make blocks of 3 8-bit characters
>    binCharGroup = binCharGroup & aBinChar
>    ' Pad if we're at the end of the string and don't have a full 3-char
> block
>    If idx = source.Max Then
>      binCharGroup = Left$(binCharGroup & "000000000000000000000000",
> 24)
>    Endif
>    If Len(binCharGroup) = 24 Then
>      Print binCharGroup
>      ' Now treat the 3 blocks of 8 bits like 4 blocks of 6 bits....
>      For idy = 1 To 4
>        Print "char: " & idy & " has value: " & (toInt(Mid
> $(binCharGroup, 1 + ((idy - 1) * 6), 6), BASE_BINARY) + 32)
>        ' Append the Chr$ of the value of the 6-bit byte + 32 to our
> result
>        result = result & Chr$(toInt(Mid$(binCharGroup, 1 + ((idy - 1) *
> 6), 6), BASE_BINARY) + 32)
>      Next
>      binCharGroup = ""
>    Endif
>  Next
>  Return result
>
> End
> ======================================================================
>
> Kind regards,
> Caveat
>
> On Mon, 2011-09-19 at 13:24 +0200, Ron wrote:
>> I'm trying to decode this with gambas, no luck, anyone has an idea?
>>
>> #!/usr/bin/perl
>> print pack('u', "send:");
>>
>> %<V5N9#H`
>>
>> So decoding %<V5N9#H` should result in 'send:'
>>
>> The pack 'u' function does uuencoding  but all vb alike code doesn't
>> reproduce the correct result, or struggles with the `...
>>
>>
>> Thanks in advance!!
>>
>> Regards,
>> Ron_2nd.
>>
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