Convert a string representation of a hex dump to a byte array using Java?

Question

I am looking for a way to convert a long string (from a dump), that represents hex values into a byte array.

I couldn't have phrased it better than the person that posted the same question here:

http://www.experts-exchange.com/Programming/Programming_Languages/Java/Q_21062554.html

But to keep it original, I'll phrase it my own way: suppose I have a string "00A0BF" that I would like interpreted as the byte[] {0x00,0xA0,0xBf} what should I do?

I am a Java novice and ended up using BigInteger and watching out for leading hex zeros. But I think it is ugly and I am sure I am missing something simple...

Answer 1

The Hex class in commons-codec should do that for you.

http://commons.apache.org/codec/

Answer 2

I've always used a method like

public static final byte[] fromHexString(final String s) {
    String[] v = s.split(" ");
    byte[] arr = new byte[v.length];
    int i = 0;
    for(String val: v) {
        arr[i++] =  Integer.decode("0x" + val).byteValue();

    }
    return arr;
}

this method splits on space delimited hex values but it wouldn't be hard to make it split the string on any other criteria such as into groupings of two characters.

Answer 3

I think will do it for you. I cobbled it together from a similar function that returned the data as a string:

private static byte[] decode(String encoded) {
    byte result[] = new byte[encoded/2];
    char enc[] = encoded.toUpperCase().toCharArray();
    StringBuffer curr;
    for (int i = 0; i < enc.length; i += 2) {
        curr = new StringBuffer("");
        curr.append(String.valueOf(enc[i]));
        curr.append(String.valueOf(enc[i + 1]));
        result[i] = (byte) Integer.parseInt(curr.toString(), 16);
    }
    return result;
}

Answer 4

EDIT: as pointed out by @mmyers, this method doesn't work on input that contains substrings corresponding to bytes with the high bit set ("80" - "FF"). The explanation is at Bug ID: 6259307 Byte.parseByte not working as advertised in the SDK Documentation.

public static final byte[] fromHexString(final String s) {
    byte[] arr = new byte[s.length()/2];
    for ( int start = 0; start < s.length(); start += 2 )
    {
        String thisByte = s.substring(start, start+2);
        arr[start/2] = Byte.parseByte(thisByte, 16);
    }
    return arr;
}

Answer 5

Here is a method that actually works (based on several previous semi-correct answers):

private static byte[] fromHexString(final String encoded) {
    if ((encoded.length() % 2) != 0)
        throw new IllegalArgumentException("Input string must contain an even number of characters");

    final byte result[] = new byte[encoded.length()/2];
    final char enc[] = encoded.toCharArray();
    for (int i = 0; i < enc.length; i += 2) {
        StringBuilder curr = new StringBuilder(2);
        curr.append(enc[i]).append(enc[i + 1]);
        result[i/2] = (byte) Integer.parseInt(curr.toString(), 16);
    }
    return result;
}

The only possible issue that I can see is if the input string is extremely long; calling toCharArray() makes a copy of the string's internal array.

EDIT: Oh, and by the way, bytes are signed in Java, so your input string converts to [0, -96, -65] instead of [0, 160, 191]. But you probably knew that already.

Answer 6

Actually, I think the BigInteger is solution is very nice:

new BigInteger("00A0BF", 16).toByteArray();

Edit: Not safe for leading zeros, as noted by the poster.

Answer 7

Here's a solution that I think is better than any posted so far:

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

Reasons why it is an improvement:

• Safe with leading zeros (unlike BigInteger) and with negative byte values (unlike Byte.parseByte)

• Doesn't convert the String into a char[], or create StringBuilder and String objects for every single byte.

Feel free to add argument checking via assert or exceptions if the argument is not known to be safe.

Answer 8

Hi.

The BigInteger() Method from java.math is very Slow and not recommandable.

Integer.parseInt(HEXString, 16)

can cause problems with some characters without converting to Digit / Integer

a Well Working method:

Integer.decode("0xXX") .byteValue()

Function:

public static byte[] HexStringToByteArray(String s) {
    byte data[] = new byte[s.length()/2];
    for(int i=0;i < s.length();i+=2) {
        data[i/2] = (Integer.decode("0x"+s.charAt(i)+s.charAt(i+1))).byteValue();
    }
    return data;
}

Have Fun, Good Luck

Answer 9

public static byte[] hex2ba(String sHex) throws Hex2baException {
    if (1==sHex.length()%2) {
        throw(new Hex2baException("Hex string need even number of chars"));
    }

    byte[] ba = new byte[sHex.length()/2];
    for (int i=0;i<sHex.length()/2;i++) {
        ba[i] = (Integer.decode(
                "0x"+sHex.substring(i*2, (i+1)*2))).byteValue();
    }
    return ba;
}

Answer 10

I like the Character.digit solution, but here is how I solved it

public byte[] hex2ByteArray( String hexString ) {
    String hexVal = "0123456789ABCDEF";
    byte[] out = new byte[hexString.length() / 2];

    int n = hexString.length();

    for( int i = 0; i < n; i += 2 ) {
        //make a bit representation in an int of the hex value 
        int hn = hexVal.indexOf( hexString.charAt( i ) );
        int ln = hexVal.indexOf( hexString.charAt( i + 1 ) );

        //now just shift the high order nibble and add them together
        out[i/2] = (byte)( ( hn << 4 ) | ln );
    }

    return out;
}