up vote 11 down vote favorite
3

Again I have some problems with hexadecimal numbers, java and byte arrays. I now have a byte array filled with hex numbers and printing it the easy way is pretty pointless because there are many unprintable elements. What I need is the exact hexcode in the form of: 3a5f771c...

And I have no idea how to convert the byte array into a string like this. It's everything in Java.

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3  
Why not just give it a try first and show us what you've got. You've nothing to lose and all to gain. Integer has a toHexString(...) method that may help if this is what you're looking for. Also String.format(...) can do some neat formatting tricks using the %2x code string. – Hovercraft Full Of Eels Mar 11 '12 at 13:09
right. i don't know, i'im just not capable for that today. bad day for working. vote for delete because it's pretty pointless to ask. – vlad Mar 11 '12 at 13:17

5 Answers

up vote 31 down vote accepted

From the discussion here, and especially this answer, this is the function I currently use:

public static String bytesToHex(byte[] bytes) {
    final char[] hexArray = {'0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F'};
    char[] hexChars = new char[bytes.length * 2];
    int v;
    for ( int j = 0; j < bytes.length; j++ ) {
        v = bytes[j] & 0xFF;
        hexChars[j * 2] = hexArray[v >>> 4];
        hexChars[j * 2 + 1] = hexArray[v & 0x0F];
    }
    return new String(hexChars);
}

My own tiny benchmarks (a million bytes a thousand times, 256 bytes 10 million times) showed it to be much faster than any other alternative, about half the time on long arrays. Compared to the answer I took it from, switching to bitwise ops --- as suggested in the discussion --- cut about 20% off of the time for long arrays.

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14  
I just found javax.xml.bind.DataTypeConverter, part of the standard distribution. Why doesn't this come up when you Google this kind of problem? Lots helpful tools, including String printHexBinary(byte[]) and byte[] parseHexBinary(String). printHexBinary is, however, much (2x) slower than the function in this answer. (I checked the source; it uses a stringBuilder. parseHexBinary uses an array.) Really, though, for most purposes it's fast enough and you probably already have it. – maybeWeCouldStealAVan Mar 31 '12 at 1:31
up vote 18 down vote

If you're not averse to including extra libraries, the commons codec library has a class for doing just this.

String foo = "I am a string";
byte[] bytes = foo.getBytes();
System.out.println( Hex.encodeHexString( bytes ) );

I used it just this morning in a personal project.

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7  
+1 Part of the whole idea of Java is to use libraries when they exist! – Beta033 Sep 26 '12 at 21:15
2  
@Beta033 At first glance I agreed and though this was a great answer and was surprised no one else suggested it on account of its elegance and short length. Then I realized it relied on an external library - if you're not already using the library for some other reason, this is a terrible solution. -1 – ArtOfWarfare Dec 6 '12 at 14:11
up vote 13 down vote

Simplest solution, no external libs, no digits constants:

String bytArrayToHex(byte[] a) {
   StringBuilder sb = new StringBuilder();
   for(byte b: a)
      sb.append(String.format("%02x", b&0xff));
   return sb.toString();
}
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up vote 12 down vote

This simple oneliner works for me
String result = new BigInteger(1, inputBytes).toString(16);

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3  
This oneliner drops leading zero bytes. – Voicu Jan 17 at 0:21
up vote 1 down vote

I found three different ways here: http://www.rgagnon.com/javadetails/java-0596.html

The most elegant one, as he also notes, I think is this one:

static final String HEXES = "0123456789ABCDEF";
public static String getHex( byte [] raw ) {
    if ( raw == null ) {
        return null;
    }
    final StringBuilder hex = new StringBuilder( 2 * raw.length );
    for ( final byte b : raw ) {
        hex.append(HEXES.charAt((b & 0xF0) >> 4))
            .append(HEXES.charAt((b & 0x0F)));
    }
    return hex.toString();
}
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