Find the position of rightmost set bit of number in swift

Swift program for Find the position of rightmost set bit of number. Here problem description and explanation.

import Foundation
/*
  Swift 4 Program for
  Find position of rightmost set bit
*/
class BitPosition
{
	// Finding the right most active bit position
	func rightmostActiveBit(_ n: Int)
	{
		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
		let result: Int = Int(log(Double(n & -n)) / log(2.0))+1
		print(" Number :",n,"Result :",result);
	}
	static func main()
	{
		let task: BitPosition = 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);
	}
}
BitPosition.main();

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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