CODEWITHSUNDEEP
javabinaryhex
Mark
Mark

Reputation: 489

Convert hexadecimal string (hex) to a binary string

I found the following way hex to binary conversion:

String binAddr = Integer.toBinaryString(Integer.parseInt(hexAddr, 16));

While this approach works for small hex numbers, a hex number such as the following

A14AA1DBDB818F9759

Throws a NumberFormatException.

I therefore wrote the following method that seems to work:

private String hexToBin(String hex){
    String bin = "";
    String binFragment = "";
    int iHex;
    hex = hex.trim();
    hex = hex.replaceFirst("0x", "");

    for(int i = 0; i < hex.length(); i++){
        iHex = Integer.parseInt(""+hex.charAt(i),16);
        binFragment = Integer.toBinaryString(iHex);

        while(binFragment.length() < 4){
            binFragment = "0" + binFragment;
        }
        bin += binFragment;
    }
    return bin;
}

The above method basically takes each character in the Hex string and converts it to its binary equivalent pads it with zeros if necessary then joins it to the return value. Is this a proper way of performing a conversion? Or am I overlooking something that may cause my approach to fail?

Thanks in advance for any assistance.

Upvotes: 33

Views: 127515

Answers (7)

Aleksander Wostal
Aleksander Wostal

Reputation: 141

Fast, and works for large strings:

    private String hexToBin(String hex){
        hex = hex.replaceAll("0", "0000");
        hex = hex.replaceAll("1", "0001");
        hex = hex.replaceAll("2", "0010");
        hex = hex.replaceAll("3", "0011");
        hex = hex.replaceAll("4", "0100");
        hex = hex.replaceAll("5", "0101");
        hex = hex.replaceAll("6", "0110");
        hex = hex.replaceAll("7", "0111");
        hex = hex.replaceAll("8", "1000");
        hex = hex.replaceAll("9", "1001");
        hex = hex.replaceAll("A", "1010");
        hex = hex.replaceAll("B", "1011");
        hex = hex.replaceAll("C", "1100");
        hex = hex.replaceAll("D", "1101");
        hex = hex.replaceAll("E", "1110");
        hex = hex.replaceAll("F", "1111");
        return hex;
    }

Upvotes: 14

Vaidehi Hemant Hindlekar
Vaidehi Hemant Hindlekar

Reputation: 1

public static byte[] hexToBytes(String string) {
 int length = string.length();
 byte[] data = new byte[length / 2];
 for (int i = 0; i < length; i += 2) {
  data[i / 2] = (byte)((Character.digit(string.charAt(i), 16) << 4) + Character.digit(string.charAt(i + 1), 16));
 }
 return data;
}

Upvotes: -1

rayan
rayan

Reputation: 1

import java.util.*;
public class HexadeciamlToBinary
{
   public static void main()
   {
       Scanner sc=new Scanner(System.in);
       System.out.println("enter the hexadecimal number");
       String s=sc.nextLine();
       String p="";
       long n=0;
       int c=0;
       for(int i=s.length()-1;i>=0;i--)
       {
          if(s.charAt(i)=='A')
          {
             n=n+(long)(Math.pow(16,c)*10);
             c++;
          }
         else if(s.charAt(i)=='B')
         {
            n=n+(long)(Math.pow(16,c)*11);
            c++;
         }
        else if(s.charAt(i)=='C')
        {
            n=n+(long)(Math.pow(16,c)*12);
            c++;
        }
        else if(s.charAt(i)=='D')
        {
           n=n+(long)(Math.pow(16,c)*13);
           c++;
        }
        else if(s.charAt(i)=='E')
        {
            n=n+(long)(Math.pow(16,c)*14);
            c++;
        }
        else if(s.charAt(i)=='F')
        {
            n=n+(long)(Math.pow(16,c)*15);
            c++;
        }
        else
        {
            n=n+(long)Math.pow(16,c)*(long)s.charAt(i);
            c++;
        }
    }
    String s1="",k="";
    if(n>1)
    {
    while(n>0)
    {
        if(n%2==0)
        {
            k=k+"0";
            n=n/2;
        }
        else
        {
            k=k+"1";
            n=n/2;
        }
    }
    for(int i=0;i<k.length();i++)
    {
        s1=k.charAt(i)+s1;
    }
    System.out.println("The respective binary number is : "+s1);
    }
    else
    {
        System.out.println("The respective binary number is : "+n);
    }
  }
}

Upvotes: 0

Imran khan
Imran khan

Reputation: 847

public static byte[] hexToBin(String str)
    {
        int len = str.length();
        byte[] out = new byte[len / 2];
        int endIndx;

        for (int i = 0; i < len; i = i + 2)
        {
            endIndx = i + 2;
            if (endIndx > len)
                endIndx = len - 1;
            out[i / 2] = (byte) Integer.parseInt(str.substring(i, endIndx), 16);
        }
        return out;
    }

Upvotes: 2

villanuevaquelin
villanuevaquelin

Reputation: 944

With all zeroes:

static String hexToBin(String s) {
    String preBin = new BigInteger(s, 16).toString(2);
    Integer length = preBin.length();
    if (length < 8) {
        for (int i = 0; i < 8 - length; i++) {
            preBin = "0" + preBin;
        }
    }
    return preBin;
}

Upvotes: 2

Kumar
Kumar

Reputation: 325

Integer.parseInt(hex,16);    
System.out.print(Integer.toBinaryString(hex));

Parse hex(String) to integer with base 16 then convert it to Binary String using toBinaryString(int) method

example

int num = (Integer.parseInt("A2B", 16));
System.out.print(Integer.toBinaryString(num));

Will Print 

101000101011

Max Hex vakue Handled by int is FFFFFFF

i.e. if FFFFFFF0 is passed ti will give error

Upvotes: 7

Mike Samuel
Mike Samuel

Reputation: 120516

BigInteger.toString(radix) will do what you want. Just pass in a radix of 2.

static String hexToBin(String s) {
  return new BigInteger(s, 16).toString(2);
}

Upvotes: 48

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