When translating C code to VB, often you come across the Right Shift and Left Shift operators ( << and >> respectively ). There is no inbuilt support for Right and Left Shift in VB, so you have to code your own. One of the main problems with replacing these functions is that C code often uses them in unsigned operations. VB doesn't support unsigned long integers either, so code can often fail if the highest bit of the long value is involved.
This tip demonstrates how to implement unsigned Right Shift and Left Shift operators without overflow.
Start a new project then add a module. Add the following code to the module:
Option Explicit
Private m_lPower2(0 To 31) As Long
Public Function RShift(ByVal lThis As Long, ByVal lBits As Long) As Long
If (lBits <= 0) Then
RShift = lThis
ElseIf (lBits > 63) Then
' .. error ...
ElseIf (lBits > 31) Then
RShift = 0
Else
If (lThis And m_lPower2(31 - lBits)) = m_lPower2(31 - lBits) Then
RShift = (lThis And (m_lPower2(31 - lBits) - 1)) * m_lPower2(lBits) Or m_lPower2(31)
Else
RShift = (lThis And (m_lPower2(31 - lBits) - 1)) * m_lPower2(lBits)
End If
End If
End Function
Public Function LShift(ByVal lThis As Long, ByVal lBits As Long) As Long
If (lBits <= 0) Then
LShift = lThis
ElseIf (lBits > 63) Then
' ... error ...
ElseIf (lBits > 31) Then
LShift = 0
Else
If (lThis And m_lPower2(31)) = m_lPower2(31) Then
LShift = (lThis And &H7FFFFFFF) \ m_lPower2(lBits) Or m_lPower2(31 - lBits)
Else
LShift = lThis \ m_lPower2(lBits)
End If
End If
End Function
Public Sub Init()
m_lPower2(0) = &H1&
m_lPower2(1) = &H2&
m_lPower2(2) = &H4&
m_lPower2(3) = &H8&
m_lPower2(4) = &H10&
m_lPower2(5) = &H20&
m_lPower2(6) = &H40&
m_lPower2(7) = &H80&
m_lPower2(8) = &H100&
m_lPower2(9) = &H200&
m_lPower2(10) = &H400&
m_lPower2(11) = &H800&
m_lPower2(12) = &H1000&
m_lPower2(13) = &H2000&
m_lPower2(14) = &H4000&
m_lPower2(15) = &H8000&
m_lPower2(16) = &H10000
m_lPower2(17) = &H20000
m_lPower2(18) = &H40000
m_lPower2(19) = &H80000
m_lPower2(20) = &H100000
m_lPower2(21) = &H200000
m_lPower2(22) = &H400000
m_lPower2(23) = &H800000
m_lPower2(24) = &H1000000
m_lPower2(25) = &H2000000
m_lPower2(26) = &H4000000
m_lPower2(27) = &H8000000
m_lPower2(28) = &H10000000
m_lPower2(29) = &H20000000
m_lPower2(30) = &H40000000
m_lPower2(31) = &H80000000
End Sub
To use the code, you initially call Init to set up the precompiled array of powers of 2. This is implemented to help speed the operation of the RShift and LShift functions. Then you use the functions like this:
Dim lValue As Long lValue = &HFFFFFFFF& ' VB thinks this is -1 Debug.Print RShift(lValue,4) ' Shift value right by 4 bits ' Answer is &HFFFFFFF0.
I am trying to reproduce this c line in vb6
ReplyDeletechecksum = (checksum<<31) | (checksum>> 1) + data[i];
The problem is when I hit a string of data[i]=0 (all 0's) in a data stream, the checksum I end up with is either all FFFFFFFF's (if rotate shift is used) or all 00000000's if non-rotate shift is used.
I can't see how this algorithm works to produce a complex checksum???
Any ideas?
Hello Steve,
ReplyDeletedo you mean the algorithm in the snippet isn't working? Or is it just that one line of c you posted isn't working?
Steve