Java: Checking if a bit is 0 or 1 in a long

Java: Checking if a bit is 0 or 1 in a long

To check if a specific bit is 0 or 1 in a long value in Java, you can use bitwise operations, specifically the bitwise AND (&) operator. Here's how you can do it:

Suppose you have a long value number and you want to check the nth bit (0-based index) from the right:

public class BitCheckExample { public static void main(String[] args) { long number = 12345L; // Your long value int bitIndex = 3; // The bit index to check (0-based index) // Create a mask with the bit of interest set to 1 and all other bits set to 0 long mask = 1L << bitIndex; // Use the bitwise AND operator to check if the bit is 0 or 1 boolean isBitSet = (number & mask) != 0; // Print the result if (isBitSet) { System.out.println("Bit at index " + bitIndex + " is 1."); } else { System.out.println("Bit at index " + bitIndex + " is 0."); } } }

In this example:

  1. We have a long value called number.

  2. We specify the index of the bit we want to check using the bitIndex variable (0-based index).

  3. We create a bitmask (mask) by left-shifting 1 by bitIndex positions. This sets the bit of interest to 1 and leaves all other bits as 0.

  4. We use the bitwise AND (&) operator to perform a bitwise AND operation between number and mask. If the result is non-zero, it means that the bit at the specified index is 1; otherwise, it is 0.

  5. We print the result based on whether the bit is set or not.

You can adjust the bitIndex variable to check different bits within the number value.

More Tags

core mediarecorder new-operator unique-values .htaccess threadpoolexecutor count-unique homekit nginfinitescroll privileges

More Java Questions

More Fitness-Health Calculators

More Internet Calculators

More Animal pregnancy Calculators

More Fitness Calculators