Java, converting from hexadecimal to decimal using casting

Question

I found an exercise to convert from hex to decimal using casting, but I did not understand how that decimal + 'A' - 10 worked. Can anybody explain me?

code is here:

 import java.util.Scanner;
 public class Test {
     public static void main(String[] args) {
         Scanner input = new Scanner(System.in);

         System.out.println("Please enter a decimal value (0-15):");
         int decimal = input.nextInt();

         if (decimal <= 9 && decimal >= 0) {
             System.out.println("The hex value is: " + decimal);

         }

         else if (decimal >= 10 && decimal <= 15) {
             System.out.println("The hex value is " + (char)(decimal + 'A' - 10));

         }

         else {
             System.out.println("It's an invalid input.");
         }
     }
 }

Thanks.

Answer 1

Each char has integer value associated with it. Hence casting int to (char) will yield to char value.

Answer 2

The alphabetical characters in Unicode have sequential codes:

A = 65
B = 66
C = 67
D = 68
...

So, the value of (char)('A'+1) is the same as 'B' , as the char value 'A' is interpreted as 65 when used in an arithmetic expression, and then the +1 makes it 66, and then you cast it as char again, it's 'B' .

So what you have there is the value of 'A' plus the difference between the decimal and 10 (which is 0 through 5). This will give you the values 'A' + 0 , 'A' + 1 , 'A' + 2 etc, which means 'A' , 'B' , 'C' respectively when you calculate it.