Java Program to Check if Bit Position is Set to One or not

This is a Java Program to Check if a given Bit Position is set to One or not.

Enter any decimal number as an input. Also mention the bit position at which you want to check whether one is present or not. After that we first convert the given decimal number into binary number and then check the element at that given position and generate output.

Here is the source code of the Java Program to Check if a given Bit Position is set to One or not. The Java program is successfully compiled and run on a Windows system. The program output is also shown below.

  1. import java.util.Scanner;
  2. public class Bit_Postion 
  3. {
  4.     public static void main(String[] args) 
  5.     {
  6.         int n, m;
  7.         String x = "";
  8.         Scanner s = new Scanner(System.in);
  9.         System.out.print("Enter any Decimal Number:");
  10.         n = s.nextInt();
  11.         while(n > 0)
  12.         {
  13.             int a = n % 2;
  14.             x = a + x;
  15.             n = n / 2;
  16.         }
  17.         System.out.print("Enter the position where you want to check:");
  18.         int l = x.length();
  19.         m = s.nextInt();
  20.         if((l - m) >= 0 && (x.charAt(l - m) == '1'))
  21.         {
  22.             System.out.println("1 is present at given bit position");
  23.         }
  24.         else
  25.         {
  26.             System.out.println("0 is present at given bit position");
  27.         }
  28.     }
  29. }

Output:

$ javac Bit_Postion.java
$ java Bit_Postion
 
Enter any Decimal Number:6
Enter the position where you want to check:2
1 is present at given bit position

Sanfoundry Global Education & Learning Series -– 1000 Java Programs.

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Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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