Java Exercises: Convert a hexadecimal to a octal number

Java Basic: Exercise-30 with Solution

Write a Java program to convert a hexadecimal to a octal number.

Hexadecimal number: This is a positional numeral system with a radix, or base, of 16. Hexadecimal uses sixteen distinct symbols, most often the symbols 0-9 to represent values zero to nine, and A, B, C, D, E, F (or alternatively a, b, c, d, e, f) to represent values ten to fifteen.

Octal number: The octal numeral system is the base-8 number system, and uses the digits 0 to 7.

Test Data:
Input any hexadecimal number: 40

Pictorial Presentation: Hexadecimal to Octal number

Java: Convert a hexadecimal to a octal number

Sample Solution:

Java Code:

import java.util.Scanner;

public class Exercise30 {
 public static int hex_to_decimal(String s)
             String digits = "0123456789ABCDEF";
             s = s.toUpperCase();
             int val = 0;
             for (int i = 0; i < s.length(); i++)
                 char c = s.charAt(i);
                 int d = digits.indexOf(c);
                 val = 16*val + d;
             return val;
    public static void main(String args[])
        String hexdec_num;
        int dec_num, i=1, j;
        int octal_num[] = new int[100];
        Scanner in = new Scanner(System.in);
        System.out.print("Input a hexadecimal number: ");
        hexdec_num = in.nextLine();
        // Convert hexadecimal to decimal
        dec_num = hex_to_decimal(hexdec_num);
        //Convert decimal to octal
        while(dec_num != 0)
            octal_num[i++] = dec_num%8;
            dec_num = dec_num/8;
        System.out.print("Equivalent of octal number is: ");
        for(j=i-1; j>0; j--)

Sample Output:

Input a hexadecimal number: 40                                                                                
Equivalent of octal number is: 100


Flowchart: Java exercises: Convert a hexadecimal to a octal number

Java Code Editor:

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Java: Tips of the Day

Try and catch:

Java is excellent at catching errors, but it can only recover gracefully if you tell it what to do. The cascading hierarchy of attempting to perform an action in Java starts with try, falls back to catch, and ends with finally. Should the try clause fail, then catch is invoked, and in the end, there's always finally to perform some sensible action regardless of the results. Here's an example:

try {
        cmd = parser.parse(opt, args); 
        if(cmd.hasOption("help")) {
                HelpFormatter helper = new HelpFormatter();
                helper.printHelp("Hello ", opt);
        else {
                if(cmd.hasOption("shell") || cmd.hasOption("s")) {
                String target = cmd.getOptionValue("tgt");
                } // else
        } // fi
} catch (ParseException err) {
        } //catch
        finally {
                new Hello().helloWorld(opt);
        } //finally
} //try

It's a robust system that attempts to avoid irrecoverable errors or, at least, to provide you with the option to give useful feedback to the user. Use it often, and your users will thank you!

Ref: https://red.ht/3EZc9OC