Files
myAut2Exe/!SourceCode/SRC/ModCRC32.bas
2015-11-26 14:28:00 +01:00

268 lines
11 KiB
QBasic

Attribute VB_Name = "ModCRC32"
Option Explicit
' Src based on ... http://vb-tec.de/crc.htm
Private pInititialized As Boolean
Private pTable(0 To 255) As Long
Const Reserved& = 0
Private Declare Function Mul Lib "MSVBVM60.DLL" Alias "_allmul" (ByVal dw1 As Long, ByVal Reserved As Long, ByVal dw3 As Long, ByVal Reserved As Long) As Long
Private m_l2Power(0 To 30) As Long
Private m_lOnBits(0 To 30) As Long
Public Sub CRCInit(Optional ByVal Poly As Long = &HEDB88320)
' init for Lshift...
Class_Initialize
Dim CRC As Long
Dim int8 As Integer
Dim Bits As Integer
For int8 = 0 To 255
CRC = Mul(int8, 0, &H1000000, 0) '* (2 ^ &H18)
For Bits = 7 To 0 Step -1
'crc32 & 0x80000000) ?
If CRC < 0 Then
'CRC = (CRC << 1) ^ Poly
CRC = Mul(CRC, 0, 2, 0) Xor Poly
Else
'CRC = (CRC << 1)
CRC = Mul(CRC, 0, 2, 0)
End If
Next Bits
pTable(int8) = CRC
Next int8
pInititialized = True
End Sub
Public Function CRC32(ByRef Bytes() As Byte) As Long
If Not pInititialized Then CRCInit
' BenchStart
' CRC berechnen:
CRC32 = &HFFFFFFFF
Dim i As Long
For i = LBound(Bytes) To UBound(Bytes)
'CRC = (CRC << 0x18) ^ pTable[Bytes[i] ^ (CRC >> 0x08)]
' CRC32 = LShift(CRC32, &H8) Xor pTable(Bytes(i) Xor RShift(CRC32, &H18) And &HFF&)
DoEventsVerySeldom
CRC32 = Mul(CRC32, 0, &H100, 0) Xor pTable((Bytes(i) Xor _
RShift(CRC32, &H18) And &HFF&))
Next i
'BenchEnd
End Function
Private Sub Class_Initialize()
' Could have done this with a loop calculating each value, but simply
' assigning the values is quicker - BITS SET FROM RIGHT
m_lOnBits(0) = 1 ' 00000000000000000000000000000001
m_lOnBits(1) = 3 ' 00000000000000000000000000000011
m_lOnBits(2) = 7 ' 00000000000000000000000000000111
m_lOnBits(3) = 15 ' 00000000000000000000000000001111
m_lOnBits(4) = 31 ' 00000000000000000000000000011111
m_lOnBits(5) = 63 ' 00000000000000000000000000111111
m_lOnBits(6) = 127 ' 00000000000000000000000001111111
m_lOnBits(7) = 255 ' 00000000000000000000000011111111
m_lOnBits(8) = 511 ' 00000000000000000000000111111111
m_lOnBits(9) = 1023 ' 00000000000000000000001111111111
m_lOnBits(10) = 2047 ' 00000000000000000000011111111111
m_lOnBits(11) = 4095 ' 00000000000000000000111111111111
m_lOnBits(12) = 8191 ' 00000000000000000001111111111111
m_lOnBits(13) = 16383 ' 00000000000000000011111111111111
m_lOnBits(14) = 32767 ' 00000000000000000111111111111111
m_lOnBits(15) = 65535 ' 00000000000000001111111111111111
m_lOnBits(16) = 131071 ' 00000000000000011111111111111111
m_lOnBits(17) = 262143 ' 00000000000000111111111111111111
m_lOnBits(18) = 524287 ' 00000000000001111111111111111111
m_lOnBits(19) = 1048575 ' 00000000000011111111111111111111
m_lOnBits(20) = 2097151 ' 00000000000111111111111111111111
m_lOnBits(21) = 4194303 ' 00000000001111111111111111111111
m_lOnBits(22) = 8388607 ' 00000000011111111111111111111111
m_lOnBits(23) = 16777215 ' 00000000111111111111111111111111
m_lOnBits(24) = 33554431 ' 00000001111111111111111111111111
m_lOnBits(25) = 67108863 ' 00000011111111111111111111111111
m_lOnBits(26) = 134217727 ' 00000111111111111111111111111111
m_lOnBits(27) = 268435455 ' 00001111111111111111111111111111
m_lOnBits(28) = 536870911 ' 00011111111111111111111111111111
m_lOnBits(29) = 1073741823 ' 00111111111111111111111111111111
m_lOnBits(30) = 2147483647 ' 01111111111111111111111111111111
' Could have done this with a loop calculating each value, but simply
' assigning the values is quicker - POWERS OF 2
m_l2Power(0) = 1 ' 00000000000000000000000000000001
m_l2Power(1) = 2 ' 00000000000000000000000000000010
m_l2Power(2) = 4 ' 00000000000000000000000000000100
m_l2Power(3) = 8 ' 00000000000000000000000000001000
m_l2Power(4) = 16 ' 00000000000000000000000000010000
m_l2Power(5) = 32 ' 00000000000000000000000000100000
m_l2Power(6) = 64 ' 00000000000000000000000001000000
m_l2Power(7) = 128 ' 00000000000000000000000010000000
m_l2Power(8) = 256 ' 00000000000000000000000100000000
m_l2Power(9) = 512 ' 00000000000000000000001000000000
m_l2Power(10) = 1024 ' 00000000000000000000010000000000
m_l2Power(11) = 2048 ' 00000000000000000000100000000000
m_l2Power(12) = 4096 ' 00000000000000000001000000000000
m_l2Power(13) = 8192 ' 00000000000000000010000000000000
m_l2Power(14) = 16384 ' 00000000000000000100000000000000
m_l2Power(15) = 32768 ' 00000000000000001000000000000000
m_l2Power(16) = 65536 ' 00000000000000010000000000000000
m_l2Power(17) = 131072 ' 00000000000000100000000000000000
m_l2Power(18) = 262144 ' 00000000000001000000000000000000
m_l2Power(19) = 524288 ' 00000000000010000000000000000000
m_l2Power(20) = 1048576 ' 00000000000100000000000000000000
m_l2Power(21) = 2097152 ' 00000000001000000000000000000000
m_l2Power(22) = 4194304 ' 00000000010000000000000000000000
m_l2Power(23) = 8388608 ' 00000000100000000000000000000000
m_l2Power(24) = 16777216 ' 00000001000000000000000000000000
m_l2Power(25) = 33554432 ' 00000010000000000000000000000000
m_l2Power(26) = 67108864 ' 00000100000000000000000000000000
m_l2Power(27) = 134217728 ' 00001000000000000000000000000000
m_l2Power(28) = 268435456 ' 00010000000000000000000000000000
m_l2Power(29) = 536870912 ' 00100000000000000000000000000000
m_l2Power(30) = 1073741824 ' 01000000000000000000000000000000
End Sub
'*******************************************************************************
' LShift (FUNCTION)
'
' PARAMETERS:
' (In) - lValue - Long - The value to be shifted
' (In) - iShiftBits - Integer - The number of bits to shift the value by
'
' RETURN VALUE:
' Long - The shifted long integer
'
' DESCRIPTION:
' A left shift takes all the set binary bits and moves them left, in-filling
' with zeros in the vacated bits on the right. This function is equivalent to
' the << operator in Java and C++
'*******************************************************************************
Public Function LShift(ByVal lValue As Long, _
ByVal iShiftBits As Integer) As Long
' NOTE: If you can guarantee that the Shift parameter will be in the
' range 1 to 30 you can safely strip of this first nested if structure for
' speed.
'
' A shift of zero is no shift at all.
If iShiftBits = 0 Then
LShift = lValue
Exit Function
' A shift of 31 will result in the right most bit becoming the left most
' bit and all other bits being cleared
ElseIf iShiftBits = 31 Then
If lValue And 1 Then
LShift = &H80000000
Else
LShift = 0
End If
Exit Function
' A shift of less than zero or more than 31 is undefined
ElseIf iShiftBits < 0 Or iShiftBits > 31 Then
Err.Raise 6
End If
' If the left most bit that remains will end up in the negative bit
' position (&H80000000) we would end up with an overflow if we took the
' standard route. We need to strip the left most bit and add it back
' afterwards.
If (lValue And m_l2Power(31 - iShiftBits)) Then
' (Value And OnBits(31 - (Shift + 1))) chops off the left most bits that
' we are shifting into, but also the left most bit we still want as this
' is going to end up in the negative bit marker position (&H80000000).
' After the multiplication/shift we Or the result with &H80000000 to
' turn the negative bit on.
LShift = ((lValue And m_lOnBits(31 - (iShiftBits + 1))) * _
m_l2Power(iShiftBits)) Or &H80000000
Else
' (Value And OnBits(31-Shift)) chops off the left most bits that we are
' shifting into so we do not get an overflow error when we do the
' multiplication/shift
LShift = ((lValue And m_lOnBits(31 - iShiftBits)) * _
m_l2Power(iShiftBits))
End If
End Function
'*******************************************************************************
' RShift (FUNCTION)
'
' PARAMETERS:
' (In) - lValue - Long - The value to be shifted
' (In) - iShiftBits - Integer - The number of bits to shift the value by
'
' RETURN VALUE:
' Long - The shifted long integer
'
' DESCRIPTION:
' The right shift of an unsigned long integer involves shifting all the set bits
' to the right and in-filling on the left with zeros. This function is
' equivalent to the >>> operator in Java or the >> operator in C++ when used on
' an unsigned long.
'*******************************************************************************
Public Function RShift(ByVal lValue As Long, _
ByVal iShiftBits As Integer) As Long
' NOTE: If you can guarantee that the Shift parameter will be in the
' range 1 to 30 you can safely strip of this first nested if structure for
' speed.
'
' A shift of zero is no shift at all
If iShiftBits = 0 Then
RShift = lValue
Exit Function
' A shift of 31 will clear all bits and move the left most bit to the right
' most bit position
ElseIf iShiftBits = 31 Then
If lValue And &H80000000 Then
RShift = 1
Else
RShift = 0
End If
Exit Function
' A shift of less than zero or more than 31 is undefined
ElseIf iShiftBits < 0 Or iShiftBits > 31 Then
Err.Raise 6
End If
' We do not care about the top most bit or the final bit, the top most bit
' will be taken into account in the next stage, the final bit (whether it
' is an odd number or not) is being shifted into, so we do not give a jot
' about it
RShift = (lValue And &H7FFFFFFE) \ m_l2Power(iShiftBits)
' If the top most bit (&H80000000) was set we need to do things differently
' as in a normal VB signed long integer the top most bit is used to indicate
' the sign of the number, when it is set it is a negative number, so just
' deviding by a factor of 2 as above would not work.
' NOTE: (lValue And &H80000000) is equivalent to (lValue < 0), you could
' get a very marginal speed improvement by changing the test to (lValue < 0)
If (lValue And &H80000000) Then
' We take the value computed so far, and then add the left most negative
' bit after it has been shifted to the right the appropriate number of
' places
RShift = (RShift Or (&H40000000 \ m_l2Power(iShiftBits - 1)))
End If
End Function