Clearing the right most set bit in a binary number

Given an integer, write a program to clear the right most set bit in it’s binary representation.

For example, consider a number 54. It is represented in binary as

1 1 0 1 1 0.

The right most set bit is marked in bold. We need to convert this to

1 1 0 1 0 0

Let us look at the approach step by step.

If you negate the number to -54, it is represented in computer’s memory in 2’s compliment.

0 0 1 0 1 0

Let us perform a bit-wise AND operation between these two.

1 1 0 1 1 0
0 0 1 0 1 0
————&
0 0 0 0 1 0

Subtract this from the original number to get the result. So x – ( x & -x) gives the required answer.

Here is the simple program to test the above.

	/**
	* Created by Ravi on 18-Nov-14.
	*/
	public class ClearRightmostSetBit
	{
	public static void main(String []args)
	{
	Test();
	}

	private static void Test()
	{
	int x = 54;
	System.out.println("Input: " + Integer.toBinaryString(x));
	x = clearRightmostSetBit(x);
	System.out.println("Output: " + Integer.toBinaryString(x));

	x = 1;
	System.out.println("Input: " + Integer.toBinaryString(x));
	x = clearRightmostSetBit(x);
	System.out.println("Output: " + Integer.toBinaryString(x));

	x = 8;
	System.out.println("Input: " + Integer.toBinaryString(x));
	x = clearRightmostSetBit(x);
	System.out.println("Output: " + Integer.toBinaryString(x));

	x = 15;
	System.out.println("Input: " + Integer.toBinaryString(x));
	x = clearRightmostSetBit(x);
	System.out.println("Output: " + Integer.toBinaryString(x));
	}

	public static int clearRightmostSetBit(int num)
	{
	if( num != 0 )
	return num - (num & -num);
	return num;
	}
	}

by GitHub