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.

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



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