How can I convert a hexadecimal number into binary without using parse or automatic conversion

Question

Below I have a method named 'hextoBinary' that returns a hexadecimal to binary conversion through type void.

In order for me to continue with my program I need a conversion from hex to binary method that returns and int so I can convert that binary int into a decimal with my 'hextoDecimal' method.

Can anybody help me or guide me on what approach to take, i've been stuck on this for a while now. i am limited to doing this manually instead of using parse or java automatic conversions.

import java.io.*;
import java.util.Scanner;
import java.util.ArrayList;

public class Main
{
    static void hexToBinary(char hexdec[])
    {
        for (char c: hexdec)
        {
            switch (c)
            {
                case '0':
                    System.out.print("0000");
                    break;
                case '1':
                    System.out.print("0001");
                    break;
                case '2':
                    System.out.print("0010");
                    break;
                case '3':
                    System.out.print("0011");
                    break;
                case '4':
                    System.out.print("0100");
                    break;
                case '5':
                    System.out.print("0101");
                    break;
                case '6':
                    System.out.print("0110");
                    break;
                case '7':
                    System.out.print("0111");
                    break;
                case '8':
                    System.out.print("1000");
                    break;
                case '9':
                    System.out.print("1001");
                    break;
                case 'A':
                    System.out.print("1010");
                    break;
                case 'B':
                    System.out.print("1011");
                    break;
                case 'C':
                    System.out.print("1100");
                    break;
                case 'D':
                    System.out.print("1101");
                    break;
                case 'E':
                    System.out.print("1110");
                    break;
                case 'F':
                    System.out.print("1111");
                    break;
                default:
                    System.out.print("\nInvalid hexadecimal digit " + hexdec[c]);
            }
        }
    }

    public static int hextoDecimal(int n)
    {
        int decimal = 0, p = 0;
        while(n != 0)
        {
            decimal += ((n % 10) * Math.pow(2,p));
            n = n / 10;
            p++;
        }
        return decimal;
    }

    public static void main(String[] args) throws IOException
    {
        Scanner sc = new Scanner(new File("RAMerrors8x4c"));
        ArrayList<String> hexValues = new ArrayList<>();

        while(sc.hasNext())
        {
            hexValues.add(sc.nextLine());
        }

        hexToBinary(hexValues.get(0).toCharArray());
    }
}

Answer 1

I modified your code a little.

a. In your code only the first hex was printed.

Change:

• call hexToBinary for every hex String.

b. the binary value was discarded after printing, so it couldn't be reused.

Change:

• Changed returntype of hexToBinary from void to String and returned the binary value calculated.

• To be able to return a String I add the peaces(nibbles) of the hex/binary to a String in every switch(case) clause.(a Stringbuilder might be better than a String - you can additionally improve that)

• in the main: additionally collect all the returned binary values in a arraylist called "binaryValues" in order to have them for the next step.

With the above (little) changes I now have all the binary values that had already been calculated.

So I am able to simply use them in a binaryToDecimal method which just sums up the binary values weighted by their position.

Why not do it again? Because youd need to convert the A-F to numbers what your hexToBinary already did. So storing the values saves you doing that step again. I have a feeling that is what your teacher had in mind when he/she combined the tasks like this.

The resulting code is:

import java.io.*;
import java.util.Scanner;
import java.util.ArrayList;

public class Main
{
     static String hexToBinary(char hexdec[]) {
            String hex = "";
            for (char c : hexdec) {
                switch (c) {
                    case '0':
                        System.out.print("0000");
                        hex += "0000";
                        break;
                    case '1':
                        System.out.print("0001");
                        hex += "0001";
                        break;
                    case '2':
                        System.out.print("0010");
                        hex += "0010";
                        break;
                    case '3':
                        System.out.print("0011");
                        hex += "0011";
                        break;
                    case '4':
                        System.out.print("0100");
                        hex += "0100";
                        break;
                    case '5':
                        System.out.print("0101");
                        hex += "0101";
                        break;
                    case '6':
                        System.out.print("0110");
                        hex += "0110";
                        break;
                    case '7':
                        System.out.print("0111");
                        hex += "0111";
                        break;
                    case '8':
                        System.out.print("1000");
                        hex += "1000";
                        break;
                    case '9':
                        System.out.print("1001");
                        hex += "1001";
                        break;
                    case 'A':
                        System.out.print("1010");
                        hex += "1110";
                        break;
                    case 'B':
                        System.out.print("1011");
                        hex += "1111";
                        break;
                    case 'C':
                        System.out.print("1100");
                        hex += "1100";
                        break;
                    case 'D':
                        System.out.print("1101");
                        hex += "1110";
                        break;
                    case 'E':
                        System.out.print("1110");
                        hex += "1110";
                        break;
                    case 'F':
                        hex += "1111";
                        System.out.print("1111");
                        break;
                    default:
                        System.out.print("\nInvalid hexadecimal digit " + hexdec[c]);
                }
            }
            System.out.println();
            return hex;
        }

        public static int binaryToDecimal(String binary) {
            int decimal = 0;
            for (int i = 1; i < binary.length()-1; i++) {
                    decimal += Math.pow(2, i-1) * (binary.charAt(binary.length()-i) - '0');
            }
            return decimal;
        }

        public static void main(String[] args) throws IOException {
            Scanner sc = new Scanner(new File("RAMerrors8x4c"));
            ArrayList<String> hexValues = new ArrayList<>();
            ArrayList<String> binaryValues = new ArrayList<>();
            while (sc.hasNext()) {
                hexValues.add(sc.nextLine());
            }
            for (String hex : hexValues) {
                String binary = hexToBinary(hex.toCharArray());
                binaryValues.add(binary);
                System.out.println(binary);
            }
            for (String binary : binaryValues) {
                int decimal = binaryToDecimal(binary);
                System.out.println(decimal);
            }
        }
    }
}

Besides using a Stringbuilder another idea could be to do all the printing of the binary values in the main. The hexToBinary returns the String - so you can print it in the loop - if you want.

Answer 2

This code is based on some that came from here but that link no longer seems to be active. Anyway, from a hex string you can get an int like this:

int hexToDecimal(String s){
  int result = 0;
  int digit = 0;
  for (int i = 0; i < s.length(); i++) {
    char c = s.charAt(i);
    if (c >= '0' && c <= '9')
       digit = c - '0';
    else 
      if (c >= 'A' && c <= 'F')
        digit = 10 + c - 'A';
      else 
        inputError(s);
    result = 16 * result + digit;
  }
  return result
}