Position of rightmost set bit in c++

C++ program for Position of rightmost set bit. Here problem description and explanation.

// Include header file
#include <iostream>
#include <math.h>

using namespace std;
/*
  C++ Program for
  Find position of rightmost set bit
*/
class BitPosition
{
	public:
		// Finding the right most active bit position
		void rightmostActiveBit(int n)
		{
			if (n <= 0)
			{
				return;
			}
			/*
			    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
			int result = (int)(log(n &-n) / log(2) + 1);
			cout << " Number : " << n 
                 << " Result : " << result << endl;
		}
};
int main()
{
	BitPosition *task = 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);
	return 0;
}

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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