[RESOLVED] Read DWORD & extract BITs

[some1uk03]

Hello,

I'm not a bits & bytes person, but i need to figure this out:

So, we have a file loaded in our buffer and the GUIDE says:

at DWORD @ POS xx

so: we read 2 BYTES
convert to an INTEGER
then do a DecToBin conversion?

Whats the theory?

[LaVolpe]

DWORD is 4 bytes, not 2.

By bits, do you mean you have to display each bit in the 32bit Long (DWORD)?

More info maybe.

[some1uk03]

DWORD is 4 bytes, not 2.

By bits, do you mean you have to display each bit in the 32bit Long (DWORD)?

More info maybe.

SORRY I meant WORD not DWORD,

yes:
WORD = 2
DWORD = 4


Yes, I need to display each bit in the result! I'm looking for 16, 0 & 1's

I then need to be able to modify/change each bit and write back.

[LaVolpe]

For an integer or long, you'll need to handle the high bit separately.

Integer high bit is &H8000. If it's set then your integer will be negative or you can test for it.
Example: (-1234 And &H8000)<>0 also -1234 is obviously negative, so bit 16 is a 1
For the remainder, you can use a loop

Dim iMask As Integer, myInt As Integer
iMask = &H4000 ' equivalent of bit 15 (2^(15-1))
myInt = CInt(Rnd * &H7FFF)
If (myInt And 1) Then myInt = -myInt ' toggle to negative for testing

Debug.Print "Value: "; myInt
If myInt < 0 Then Debug.Print "1"; Else Debug.Print "0";
Do
     If (myInt And iMask) = 0 Then Debug.Print "0"; Else Debug.Print "1";
     iMask = iMask \ 2
Loop Until iMask = 0
Debug.Print

Bits are listed from highest to lowest, left to right

[LaVolpe]

Hmmm. I didn't offer a clear way of modifying them, did I?

To modify the 16th bit of an Integer: Toggle the high bit
If you want the high bit set: myInt = myInt Or &H8000
Strip the high bit if not wanted: myInt = (myInt And &H7FFF). Remember high bit is &H8000
If you want to toggle it regardless of what it is set at: myInt = myInt Xor &H8000

For all other bits (15-1), a bit different (pun intended)
To include a bit (ensure it is 1): myInt = myInt Or (2^(bit# - 1))
To remove a bit (ensure it is 0): myInt = myInt And Not (2^(bit# - 1))
To toggle a bit regardless of what it is set at: myInt = myInt Xor (myInt And (2^(bit# -1))

In above samples, bit# is a value between 1 and 15
Wikipedia has a somewhat easy to understand description of bitwise operations

[some1uk03]

Hey.. sorry I've been away on this for a while.. and I'm totally lost with these bits (

Can we put this in a working Sub/function please?
i.e

Private theBits(15) as integer/long   'Do we handle as integers or longs?
''So the theBits() array can display each bit; i.e  1000001011010110

Private Sub GetBits(byval WORD_Data as string)
'''so we do a loop which adds each of the bits to the   theBits()  array
''so that they can easily be manipulated
End Sub


Private Sub WriteBits(theArray() as integer/long) as String
'We must write the array back to a WORD string  ; {2 chars}
End Sub

[Milk]

DWORD_Data as string? 2 chars? theBits(15)?

How many bytes are encoded into your 2 character string? 2 or 4

Why string?

Just want to get a grasp of the bigger picture.

Edit: some1uk03, I'm being a little pedantic with the above I'm guessing you mean Word_Data and there is 1 byte represented by each 2 byte character. theBits(15) could also be defined as Boolean.

[some1uk03]

Sorry that's a typo... meant to be WORD not DWORD

So the idea or the Bigger picture is this:
I have specific file with CHUNKS etc.. assuming:

FilePos:       Data Type
xPOS  30     Byte
xPOS  31     Long
xPOS  35     WORD
'etc..
'etc...

So I need to read pos 35. This being a WORD type, will be 35&36.

Then I need to convert/get the bits out of these!! presumably pos35=1Byte/8 bits & pos36=1Byte/8 bits ; in total 16

Make sense?

[Milk]

It does. Why string though? Are you reading the file into one?

[some1uk03]

Yes the whole file is read to a string.
And I normally do a ToInteger / ToLong conversions.

I could have that as byval WORD_Data as integer/long no prob.

I could even have them as 2 separate inputs? byval part1 as integer/long, byval part2 as integer/long

It's extracting the bits I'm more worried about, rather than the input.

[Milk]

Sorry I'm getting off track, the following is for an Integer rather than string and it assumes theBits() is boolean.

Private Sub GetBits(byval Word as Integer)
   Dim i as Long, bit as Long
   bit = 1
   For i = 0 to 14
      theBits(i) = cbool(Word And bit) 'use this if theBits is boolean
      'If (Word And bit) <> 0 Then theBits(i) = 1 Else theBits(i) = 0 'use this is if theBits is byte, integer, long
      'theBits(i) = -cBool(Word And bit) 'same as above line but not very clear
      bit = bit * 2
   Next
   theBits(15) = cbool(Word And &h8000) 'handle the high bit
End Sub

Private Sub WriteBits() as Integer
   Dim i as Long, bit as Long, wordOut as Integer
   bit = 1
   For i = 0 to 14
      If theBits(i) Then wordOut = wordOut Or bit
      bit = bit * 2
   Next
   If theBits(15) Then wordOut = wordOut Or &h8000
   WriteBits = wordOut
End Sub

Not tested

[Logophobic]

If the file contains binary data, then reading it as a string is probably not a good idea. Reading a Word (Integer) or DWord (Long) value from a specific location in the file is simple: Get #FileNumber, Pos, Value where Value is a variable of the appropriate type.


Anyway, here is something that may help with bit-wise operations:

Private IntegerBit(15) As Integer
Private LongBit(31) As Long

Private Sub Form_Load()
  Dim i As Long, n As Long

  n=1

  For i = 0 To 14
    IntegerBit(i) = n
    LongBit(i) = n
    n=n+n
  Next i
  IntegerBit(15) = &H8000

  For i = 15 To 29
    LongBit(i) = n
    n=n+n
  Next i
  LongBit(30) = n
  LongBit(31) = &H80000000
End Sub

You can use it like this:

value = value Or bit(x) ' make the bit '1'
value = value And Not bit(x) ' make the bit '0'
value = value XOr bit(x) ' change the bit

The arrays also make it easy to 'extract' the bits into a string of 1s and 0s. Start with a string of 0s and use Mid() to put 1s in where needed.

[Doogle]

You could go completely over the top and treat Bytes as Bytes....(Saves having to mess about with sign bits etc.)

Option Explicit

Private Sub Command_Click()
Dim strData As String
Dim bytData(1) As Byte
Dim intI As Integer
strData = "AB"
'
' Stick the data into a byte array
'
bytData(0) = CByte(Asc(Mid$(strData, 1, 1)))
bytData(1) = CByte(Asc(Mid$(strData, 2, 1)))
Debug.Print Hex(bytData(0)); Hex(bytData(1))
'
' Display the bits
' bit 0 = MSb, bit 15=LSb
'
For intI = 0 To 15
    Debug.Print CStr(GetBit(bytData, intI));
Next intI
Debug.Print
'
' Set bit 1 to zero
'
Call ChangeBit(bytData, 1, 0)
Debug.Print Hex(bytData(0)); Hex(bytData(1))
'
' Set bit 15 to 1
'
Call ChangeBit(bytData, 15, 1)
Debug.Print Hex(bytData(0)); Hex(bytData(1))
End Sub

Private Function GetBit(bytD() As Byte, intBit As Integer) As Integer
Dim intElement As Integer
Dim intElementBit As Integer
If intBit >= 0 And intBit <= 15 Then
    intElement = intBit \ 8
    intElementBit = 7 - (intBit - (intElement * 8))
    If (bytD(intElement) And (2 ^ intElementBit)) = 0 Then
        GetBit = 0
    Else
        GetBit = 1
    End If
End If
End Function

Private Sub ChangeBit(bytD() As Byte, intBit As Integer, intV As Integer)
Dim intElement As Integer
Dim intElementBit As Integer
If intBit >= 0 And intBit <= 15 And (intV = 0 Or intV = 1) Then
    intElement = intBit \ 8
    intElementBit = 7 - (intBit - (intElement * 8))
    If intV = 1 Then
        bytD(intElement) = bytD(intElement) Or 2 ^ intElementBit
    Else
        bytD(intElement) = bytD(intElement) And Not (2 ^ intElementBit)
    End If
End If
End Sub

(Just stick the data back into the string when you've finished messing about with the bits)

[some1uk03]

and it works... ) thanks guys..