Convert Decimal to Hex using Recursive method Java

Question

I need to make a recursive method that converts a decimal into hexadecimal. I can't use Integer.toHexString. EDIT:I tried this code but it doesn't work properly

public static String Hexa(String s) {
    String result = "";
    int n = Integer.parseInt(s);
    int remainder = n % 16;

    if (n == 0) {
        return Integer.toString(0);
    } else {
        switch (remainder) {
            case 10:
                result = "A" + result;
                break;
            case 11:
                result = "B" + result;
                break;
            case 12:
                result = "C" + result;
                break;
            case 13:
                result = "D" + result;
                break;
            case 14:
                result = "E" + result;
                break;
            case 15:
                result = "F" + result;
                break;
            default: result = Integer.toString(n/16) + result; break;
        }
        System.out.println(result);
        return Hexa(Integer.toString(n/16)) + result;
    }
}

Edit: Changed the default case and the if (n == 0) loop return statement and it works beautifully now.

new code:

 public static String Hexa(String s) {
        String result = "";
        int n = Integer.parseInt(s);
        int remainder = n % 16;

        if (n == 0) {
            return "";
        } else {
            switch (remainder) {
                case 10:
                    result = "A";
                    break;
                case 11:
                    result = "B";
                    break;
                case 12:
                    result = "C";
                    break;
                case 13:
                    result = "D";
                    break;
                case 14:
                    result = "E";
                    break;
                case 15:
                    result = "F";
                    break;
                default:
                    result = remainder + result;
                    break;
            }
            return Hexa(Integer.toString(n / 16)) + result;
        }
    }

`Integer.toHexString` doesn't convert decimal to hexadecimal, anyway :) — Marko Topolnik, Nov 22 '13 at 17:29
You're concatenating twice, that's why it doesn't work properly. You are mixing up iteration with recursion. — Marko Topolnik, Nov 22 '13 at 17:38
So do a Search for: `" + result"` Replace with: `` and apply that to all but the last match. — Marko Topolnik, Nov 22 '13 at 17:39
Do you notice in your recursive call you are converting your Integer to a String, just to convert it back at the beginning of your function? — Daniel, Nov 22 '13 at 17:41
I could be that i'm mixing stuff up, i just started learning java for college and this is a assignment i have to do. i have a working thing but that's not recursive — Liam de Haas, Nov 22 '13 at 17:43
@DWilches when i put in 1234 it should return 4D2 but instead returns 00D77 — Liam de Haas, Nov 22 '13 at 17:44
The default case for the switch doesn't seem right to me...try `default: result = remainder + result`? — Jonathan Apodaca, Nov 22 '13 at 17:47
For your base case, you are returning leading zeros as `"0"`, whereas I think you just want to return nothing (i.e., the empty string) — Jonathan Apodaca, Nov 22 '13 at 17:51

Daniel · Accepted Answer · 2013-11-22T18:02:14.250

3

The problem is in your default clause:

default: result = Integer.toString(n/16) + result; break;

it should read:

default: result = Integer.toString(remainder) + result; break;

That will make your program return "04D2".

But there are several other corrections you can make:

Stop converting back and forth to String. For example that same line can be just:

default: result = remainder + result; break;

Also, change your parameters time to int. If you do need to receive a String, then make this an auxiliary function and make your main function receive a String.
You don't need that breakat the end of your default
You don't need a switch. Isn't 'F' = 'A' + (15 - 10) ? You can figure out how to make a formula that translates any number in the range [10,15] to its corresponding letter.
Instead of Integer.toString(0) you can use "0" ... but that isn't even necessary, you can use "" to avoid that leading 0 in your output. If your are worried for handling the special case where the whole number is "0" add a special clause.

edited Nov 22 '13 at 18:02

answered Nov 22 '13 at 17:53

Daniel

21,933
14
72
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1

i did what you said just before you posated it, see my edit. thank you nonetheless – Liam de Haas Nov 22 '13 at 18:00
In the default clause, you are doing: `result = remainder + result`. Is that only to change the int to String ? You can do this clearer: `result = "" + remainder`. Or in this case with your version of `Integer.toString` – Daniel Nov 22 '13 at 18:07
how would i make it recursive if it returns a string and the parameter is an int? – Liam de Haas Nov 22 '13 at 18:09
Look closely at your function, the type of the returned value is not related to the type of the input parameter. – Daniel Nov 22 '13 at 18:13
I don't get it. i need the return type to be string otherwise a can't print the hex vaue an in order for the function to be recursive i need to let function call itself. it returns a string so how can the input parameter type differ fromn the return type? – Liam de Haas Nov 22 '13 at 18:17
No problem. Look, if you change your input parameter to int, and your recursive call is like this, would it work? `return Hexa(n/16) + result;` – Daniel Nov 22 '13 at 18:19
yes it does, guess that's because result is a string. now the whole `'F' = 'A' + (15 - 10)` thing, how does that work? – Liam de Haas Nov 22 '13 at 18:21
It's a way to turn numbers into letters by using their ASCII values. As it is out scope, check other posts about it [here is one](http://stackoverflow.com/questions/10813154/converting-number-to-letter) – Daniel Nov 22 '13 at 22:46

score 1 · Answer 2 · edited Apr 04 '17 at 07:37

1

The code below may help you to solve your problem:

public static String decimalToAnyBase(int num,int base) {       
    if(num<base) {
        return num+"";
    }
    String result=null;
    int rem=num%base;
    String str=decimalToAnyBase(num/base, base);
    result=str+((rem>=10)? (char)(rem-10+'A')+"":rem);
    return result;
}

edited Apr 04 '17 at 07:37

boutta

24,189
7
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answered Apr 04 '17 at 07:21

user7812679

11
1

score 0 · Answer 3 · edited May 02 '22 at 07:24

    import java.util.Scanner;

    public class Assign01_05 
    {
        static String res;
        public static void hex(int num) //125
        {
            if(num>=0 && num<10)
                System.out.print(num);
            else if(num>=10 && num<=15)
            {
                switch(num)
                {
                case 10:
                    System.out.print('A');
                    break;
                case 11:
                    System.out.print('B');
                    break;
                case 12:
                    System.out.print('C');
                    break;
                case 13:
                    System.out.print('D');
                    break;
                case 14:
                    System.out.print('E');
                    break;
                case 15:
                    System.out.println('F');
                    break;
                }
            }
            else
            {
                hex(num/16);
                hex(num%16);
            }
        }
        public static void main(String[] args) 
        {
            Scanner sc = new Scanner(System.in);
            System.out.println("Enter the Demical number :");
            int num=sc.nextInt();
            hex(num);
        }
    }

import java.util.Scanner;

public class Assign01_05 
{
    static String res;
    public static void hex(int num) //125
    {
        if(num>=0 && num<10)
            System.out.print(num);
        else if(num>=10 && num<=15)
        {
            switch(num)
            {
            case 10:
                System.out.print('A');
                break;
            case 11:
                System.out.print('B');
                break;
            case 12:
                System.out.print('C');
                break;
            case 13:
                System.out.print('D');
                break;
            case 14:
                System.out.print('E');
                break;
            case 15:
                System.out.println('F');
                break;
            }
        }
        else
        {
            hex(num/16);
            hex(num%16);
        }
    }
    public static void main(String[] args) 
    {
        Scanner sc = new Scanner(System.in);
        System.out.println("Enter the Demical number :");
        int num=sc.nextInt();
        hex(num);
    }
}

score 0 · Answer 4 · edited May 03 '22 at 05:38

import java.util.Scanner;

public class Ques5 {
    
    public static void hex(int n) {
        
        if(n>9&&n<=15) {
            System.out.printf("%c",'A'+(n-10));
        }
        else if(n>=0&&n<=9){
            System.out.printf("%d",n);
        }
        else {
            hex(n/16);
            hex(n%16);
        }       
    }
    
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.println("Enter the decimal number : ");
        int i = sc.nextInt();
        System.out.println("The hexadecimal number is : ");
        hex(i);
        sc.close();
    }

}

score 0 · Answer 5 · answered Sep 06 '22 at 12:09

0

const char hex_string[17] = "0123456789ABCDEF";
void dec_to_hex(long long x){
    if(x == 0) return;
    dec_to_hex(x/16);
    printf("%c",hex_string[x%16]);
}

Same in Java

answered Sep 06 '22 at 12:09

Thanh Sơn Đàm

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Convert Decimal to Hex using Recursive method Java

5 Answers5