Question

Hi everyone I have implemented a solution to convert a value from binary to hexidecimal. So I was wondering if there could be more elegant solution(pretty sure there is) than mine. I have tested the program and it is working. Here is the code:

public class BinaryToHex
{
    public static void main(String[] args) 
    {
        String binary = "1110001101";
        binaryToHex(binary);
        //38d
    }

    public static void binaryToHex(String binaryValue)
    {
        StringBuilder sb = new StringBuilder(binaryValue);

        System.out.println("Original StringBuilder: " + sb);
        sb.reverse();

        System.out.println("reversed StringBuilder: " + sb);

        int convert = binaryValue.length();
        System.out.println("Legth of the binary: " +  convert);

        if(convert % 4 != 0)
        {
            while(convert % 4 != 0)
            {


                sb.append(0);
                convert ++;

                System.out.println("StringBuilder in loop: " + sb);
                System.out.println("Convert in loop: " + convert);
            }
        }
        sb.reverse();
        System.out.println("Ready StringBuilder for use? " + sb);
        String test = null;
//      String test = sb.toString();
        for(int i=0; i<sb.length(); i ++)
        {
            if(i % 4 == 0)
            {
                test = sb.substring(i, (i+4));
                System.out.print(getChar(test));
            }
        }
    }

    public static String getChar(String num)
    {
        String number = "";
        switch(num)
        {
            case "0000" : number = "0"; break;
            case "0001" : number = "1"; break;
            case "0010" : number = "2"; break;
            case "0011" : number = "3"; break;
            case "0100" : number = "4"; break;
            case "0101" : number = "5"; break;
            case "0110" : number = "6"; break;
            case "0111" : number = "7"; break;
            case "1000" : number = "8"; break;
            case "1001" : number = "9"; break;
            case "1010" : number = "A"; break;
            case "1011" : number = "B"; break;
            case "1100" : number = "C"; break;
            case "1101" : number = "D"; break;
            case "1110" : number = "E"; break;
            case "1111" : number = "F"; break;

        }
        return number;
    }

}

Answer 1

It was an funny exercise to do things we take for granted in a library implementation.
You can improve you algo a lot by using counters to track where you are and prevent the reverse operations. Also the switch block is rather un-elegant.

Here’s my go at it. Not claiming it’s highly performant, but I think it’s an improvement on your proposal.

public class BinToHex {

    static char[] HEX = {'0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F'};
    static int[] POW_2 = {1,2,4,8};
    public static String toHex(String bin) {
        int len = bin.length();
        char[] result = new char[(int)Math.ceil(len*1D/4)];
        int pos = result.length-1;
        int hexval = 0;
        for (int i=0; i<len;i++ ){
            hexval += (bin.charAt(len-i-1)-HEX[0]) * POW_2[i%4];
            if ((i+1)%4==0) {
                result[pos--] = HEX[hexval];
                hexval = 0;
            }
        }
        if (pos==0) {
            result[0] = HEX[hexval];
        }
        return new String(result);
    }

    public static void main(String [] param) {
        System.out.println(BinToHex.toHex("1")); // border case 1 char => 0x1
        System.out.println(BinToHex.toHex("101")); // testcase less than 4 chars => 0x5
        System.out.println(BinToHex.toHex("1011")); // testcase eq 4 chars => 0xB
        System.out.println(BinToHex.toHex("101101011011")); // testcase lenght % 4  = 0
        System.out.println(BinToHex.toHex("11101101011011")); // testcase lenght % 4  != 0
        System.out.println(BinToHex.toHex("000101101011011")); // testcase leading 0
    }

}