Java int type literals convert Hexadecimal Digit to Decimal Value

Question

The hexadecimal number system has 16 digits: 0-9, A-F.

The letters A, B, C, D, E, and F correspond to the decimal numbers 10, 11, 12, 13, 14, and 15.

Write a program that prompts the user to enter a hex digit and display its corresponding decimal value. 

import java.util.Scanner;

public class Main {
   public static void main(String[] args) {
      Scanner input = new Scanner(System.in);
      System.out.print("Enter a hex digit: ");
      String hexString = input.nextLine();

     //your code here
   }//  ww w  .j  av a 2 s .  com
}




import java.util.Scanner;

public class Main {
   public static void main(String[] args) {
      Scanner input = new Scanner(System.in);
      System.out.print("Enter a hex digit: ");
      String hexString = input.nextLine();

      // Check if the hex string has exactly one character
      if (hexString.length() != 1) {
         System.out.println("You must enter exactly one character");
         System.exit(1);
      }

      // Display decimal value for the hex digit
      char ch = Character.toUpperCase(hexString.charAt(0));
      if (ch <= 'F' && ch >= 'A') {
         int value = ch - 'A' + 10;
         System.out.println("The decimal value for hex digit " + ch + " is " + value);
      } else if (Character.isDigit(ch)) {
         System.out.println("The decimal value for hex digit " + ch + " is " + ch);
      } else {
         System.out.println(ch + " is an invalid input");
      }
   }
}



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