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)
task.rightmostActiveBit(320)
' (1000) = Binary(1111101000)
task.rightmostActiveBit(1000)
' (153) = Binary(10011001)
task.rightmostActiveBit(153)
' (354) = Binary(101100010)
task.rightmostActiveBit(354)
' 160 = Binary(10100000)
task.rightmostActiveBit(160)
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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