Java.lang.Long.highestOneBit() Method


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Description

The java.lang.Long.highestOneBit() method returns a long value with at most a single one-bit, in the position of the highest-order ("leftmost") one-bit in the specified long value. It returns zero if the specified value has no one-bits in its two's complement binary representation, that is, if it is equal to zero.

Declaration

Following is the declaration for java.lang.Long.highestOneBit() method

public static long highestOneBit(long i)

Parameters

i − This is the long value.

Return Value

This method returns a long value with a single one-bit, in the position of the highest-order one-bit in the specified value, or zero if the specified value is itself equal to zero.

Exception

NA

Example

The following example shows the usage of java.lang.Long.highestOneBit() method.

Live Demo
package com.tutorialspoint;

import java.lang.*;

public class LongDemo {

   public static void main(String[] args) {

      long l = 220;
      System.out.println("Number = " + l);

      /* returns the string representation of the unsigned long value 
         represented by the argument in binary (base 2) */
      System.out.println("Binary = " + Long.toBinaryString(l));

      // returns the number of one-bits 
      System.out.println("Number of one bits = " + Long.bitCount(l));

      /* returns a long value with at most a single one-bit, in the position 
         of the highest-order ("leftmost") one-bit in the specified int value */ 
      System.out.println("Highest one bit = " + Long.highestOneBit(l));
   }
}  

Let us compile and run the above program, this will produce the following result −

Number = 220
Binary = 11011100
Number of one bits = 5
Highest one bit = 128

java_lang_long.htm

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