Modify a Bit at Given Position

/* We are given an integer number n, a bit value v (v=0 or 1) and a position p. Write a sequence of operators
 * that modifies n to hold the value v at the position p from the binary representation of n while preserving all other bits in n. */

import java.util.Scanner;

public class SetBitAtGivenPositionInGivenIntegerToGivenBitValue {

    public static void main(String[] args) {
        // TODO Auto-generated method stub
        Scanner scan = new Scanner(System.in);
        System.out.print("Enter a Integer number: ");
        long number = scan.nextLong();
        System.out.print("Enter whole number for Bit position: ");
        int bitP = scan.nextInt();
        System.out.print("Enter value to set Bit '0' or '1' : ");
        int givenBitValue = scan.nextInt();
        scan.close();

        if (bitP >= 0 && bitP <= 31 && (givenBitValue == 0 || givenBitValue == 1)) {
            long bitMask = (long)1 << bitP;
            int currentBitValue = ((number & bitMask) == 0) ? 0 : 1;
            if (currentBitValue != givenBitValue) {
                number ^= bitMask;
            }

            System.out.printf("The Number after Bit Modification is: %d !\n", number);

        } else {
            System.out.println("Error! - One or more of Input arguments is Out of Range!!!");
        }
    }

}