Write a program takes in a binary value, protected by a try/catch block, from the console and displays back to the user the decimal value and exits. You can use Integer.parseInt(bin, 2). If input is not base 2, parseInt will throw an error, which the try block will catch and add display a meaningful message to user, then prompt for a binary number again. 

 

Here is the code I have. However it is not displaying the decimal value.

 

import java.util.Scanner;

 

public class Decimal {

 

  public static void main(String[] args) {

    Scanner input = new Scanner(System.in);

    String binary;

    int decimal;

 

    while (true) {

      decimal = 0;

 

      // prompt and read the binary value

      System.out.println("Enter a binary string:");

      binary = input.nextLine();

 

      try {

 

        // iterate through character by character and convert to decimal

        for (int i = 0; i < binary.length(); i++) {

          if (binary.charAt(i) != '0' && binary.charAt(i) != '1') {

            throw new IllegalArgumentException("That is not a binary string");

          }

 

          if (binary.charAt(i) == '1') {

            decimal += Math.pow(2, (binary.length() - i - 1));

          }

        }

 

        // display the decimal value

        System.out.println("Binary: " + binary);

        System.out.println("Decimal: " + decimal);

        break;

 

      } catch (IllegalArgumentException ex) {

        System.out.println(ex.getMessage());

        System.out.println("Press enter to try again.");

        input.nextLine();

      }

    }

  }

 

}