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Old March 17th, 2004, 07:39 AM
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Default integer overflow

Hello!

I am trying to modify the bits of an integer (16 bits). What I want to do is to change the 16th bit of that number. So I am doing the following:

MyNumber = MyNumber Or LShift(1, 15) where LShift is the following function:

Public Function LShift(ByVal Value As Integer, _
                                 Count As Integer) As Integer
Dim i As Integer

  If Count = 0 Then
    LShift = Value
    Exit Function
  End If

  LShift = Value

  For i = 1 To Count
    LShift = LShift * 2
  Next

End Function

The problem is that as soon as the counter gets to 15, the function returns an overflow. So the maximun number I get is 16384. Is there another way to do this?

Thanks,

Paul
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Old March 17th, 2004, 08:39 AM
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The usual way to set the nth bit is just to 'Or' the current value with 2^(n-1), but VB's 16-bit integers are signed so the hi-bit (16th bit) denotes the sign. The easiest way to change the sign of an Int is to * -1.
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Old March 17th, 2004, 09:01 AM
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Thanks mate!


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