# Position of rightmost set bit in java

Java program for Position of rightmost set bit. Here problem description and other solutions.

```
/*
Java Program for
Find position of rightmost set bit
*/
class BitPosition
{
// Finding the right most active bit position
public 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)(Math.log(n & -n) / Math.log(2) + 1);
System.out.println(" Number : " + n + " Result : " + result);
}
public static void main(String[] args)
{
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);
}
}
```

**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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