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Integer as Unsigned Integer

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Submitted on: 9/17/2012 3:10:15 PM
By: Je. 
Level: Beginner
User Rating: By 3 Users
Compatibility: VB 5.0, VB 6.0
Views: 1441
 
     Use an integer as an unsigned integer.
 
code:
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'**************************************
' Name: Integer as Unsigned Integer
' Description:Use an integer as an unsigned integer.
' By: Je.
'
'This code is copyrighted and has' limited warranties.Please see http://www.Planet-Source-Code.com/vb/scripts/ShowCode.asp?txtCodeId=74553&lngWId=1'for details.'**************************************

'in a form
Dim ii(0 To &HFFFF&) As Integer
Private Sub Form_Load()
Dim i As Long
For i = 0 To &HFFFF&
ii(i) = LtoU(i)
Next
Me.Caption = UtoL(ii(55961))
End Sub
Function LtoU(ByVal l As Long) As Integer
If (l And &H8000&) = &H8000& Then LtoU = (l Or &H8000) Else LtoU = l
End Function
Function UtoL(ByVal u As Integer) As Long
UtoL = u And &HFFFF&
End Function
Function AU(u As Integer, Nadd As Integer) As Integer
Dim ul As Long
ul = UtoL(u) + UtoL(Nadd)
If ul > &HFFFF& Then Error 6 'out of bounds
AU = LtoU(ul)
End Function
Function SU(u As Integer, nSubtract As Integer) As Integer
Dim ul As Long
ul = UtoL(u) - UtoL(nSubtract)
If ul < 0 Then Error 6 'out of bounds
SU = LtoU(ul)
End Function
Function MU(u As Integer, nMultiply As Integer) As Integer
Dim ul As Long
ul = UtoL(u) * UtoL(nMultiply)
If ul > &HFFFF& Then Error 6 'out of bounds
MU = LtoU(ul)
End Function
Function DU(u As Integer, nDivide As Integer) As Integer
Dim ul As Long
ul = UtoL(u) \ UtoL(nDivide)
If ul < 0 Then Error 6 'out of bounds
DU = LtoU(ul)
End Function
Function UMod(u As Integer, nMod As Integer) As Integer
Dim ul As Long
ul = UtoL(u) Mod UtoL(nMod)
UMod = LtoU(ul)
End Function
Function UPow(u As Integer, nPow As Integer) As Integer
Dim ul As Long
ul = UtoL(u) ^ UtoL(nPow)
If ul > &HFFFF& Then Error 6 'out of bounds
UPow = LtoU(ul)
End Function
Sub incU(u As Integer, NIncrement As Integer)
Dim ul As Long
ul = UtoL(u) + UtoL(NIncrement)
If ul > &HFFFF& Then Error 6 'out of bounds
u = LtoU(ul)
End Sub


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