Convert hexadecimal string (hex) to a binary string

Question

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.

Would a long not work for what you are wanting? It also has the ToBinaryString(). That would support... Ranges from -9,223,372,036,854,775,808 to 9,223,372,036,854,775,807. — Jared
– Jared, Commented Feb 12, 2012 at 4:07
if that does what you'd like let me know so I can post as the answer please. — Jared
– Jared, Commented Feb 12, 2012 at 4:07
Your code looks good to me. You may also look at following link: java2everyone.blogspot.in/2009/04/… — Kuldeep Jain
– Kuldeep Jain, Commented Feb 12, 2012 at 4:18

Mike Samuel · Accepted Answer · 2012-02-12 04:05:10Z

48

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);
}

answered Feb 12, 2012 at 4:05

Mike Samuel

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7 Comments

Louis Wasserman Over a year ago

Indeed. The only issue here is that the numbers grow too big to fit in an int.

Louis Wasserman Over a year ago

That's the point, sorry. Overflow is the problem with the OP's code, and this solution fixes it.

Mark Over a year ago

Thanks, this is just what I was missing.

Andrei Botalov Over a year ago

Unfortunately, toString(2) doesn't pad with zeroes

wmassingham Over a year ago

For a width of 8 bits: String.format("%8s", Integer.toBinaryString(s)).replace(' ', '0');

|

Aleksander Wostal · Accepted Answer · 2019-07-24 07:29:08Z

14

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;
    }

answered Jul 24, 2019 at 7:29

Aleksander Wostal

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3 Comments

Adam Over a year ago

Works good, keeps leading 0s as well.

NLxDoDge Over a year ago

This is the only solution that keeps all leading and trailing zero's the rest doesn't.

tej shah Over a year ago

thank you so much. only one solution work prefectly

Kumar · Accepted Answer · 2015-04-15 11:11:26Z

7

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

edited Apr 15, 2015 at 11:11

answered Apr 15, 2015 at 11:05

Kumar

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Comments

villanuevaquelin · Accepted Answer · 2015-11-26 12:58:33Z

2

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;
}

answered Nov 26, 2015 at 12:58

villanuevaquelin

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Comments

Imran khan · Accepted Answer · 2016-02-03 09:30:38Z

2

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;
    }

answered Feb 3, 2016 at 9:30

Imran khan

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Comments

rayan · Accepted Answer · 2017-11-01 06:32:12Z

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);
    }
  }
}

Raduan Santos · Accepted Answer · 2019-01-17 10:55:58Z

-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;
}

edited Jan 17, 2019 at 10:55

Raduan Santos

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answered Jan 17, 2019 at 9:23

Vaidehi Hemant Hindlekar

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