# Find the position of rightmost set bit of integer in vb.net

Vb program for Find the position of rightmost set bit of integer. Here problem description and explanation.

``````' Include namespace system
Imports System
'  Vb.net Program for
'  Find position of rightmost set bit

Public Class BitPosition
'  Finding the right most active bit position
Public Sub rightmostActiveBit(ByVal n As Integer)
if (n <= 0) Then
Return
End If
'    Example
'    ———————
'    n    = 320
'    n    = (00101000000)   Binary
'    -n   = (11011000000)   (2s)
'    Calculate log2
'    ——————————————
'    Formula : log(n & -n) / log(2) + 1
'    -----------------------------------
'    log(320 & -320) / log(2) + 1)
'    Here : log(320 & -320) = log(64) = 4.158883
'           log(2)  = 0.693147
'    (4.158883 / 0.693147) + 1 = (7) position
'    ————————————————————————————————————————
'  Calculate rightmost active bits
Dim result As Integer = CInt((Math.Log(n And -n) / Math.Log(2) + 1))
Console.WriteLine(" Number : " +
n.ToString() + " Result : " +
result.ToString())
End Sub
Public Shared Sub Main(ByVal args As String())
Dim task As BitPosition = New BitPosition()
'  Test Cases
'  320 = Binary(101000000)
'  (1000) = Binary(1111101000)
'  (153) = Binary(10011001)
'  (354) = Binary(101100010)
'  160 = Binary(10100000)
End Sub
End Class

``````

Output

`````` Number : 320 Result : 7
Number : 1000 Result : 4
Number : 153 Result : 1
Number : 354 Result : 2
Number : 160 Result : 6``````

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