Question

I have a hex string which looks like:

String hexImage ="0xFFD8FFE000104A46494600010200006400640000FFEC00114475636B79000100040000003C..."

Which I need to convert to an image using Java. I’ve tried converting to a byte array first with org.apache.commons.codec.binary.Hex class. FileOutputStream later dumps this to a file.

char[] charArr= hexImage.toCharArray();
byte[] byteArray = Hex.decodeHex(charArr);

however the parser in the Hex class chokes, with an Decoder exception : Odd number of characters. No matter what method I use to convert to a byte array, it fails. The fileformat is definitely a JPEG.

I’ve tried removing this 0xFF from the string, but the image is corrupt at this point. 0xFF I assume is a the culprit – anyone any ideas on what do I need to do to correct this?

Answer 1

Normally, each byte is represented by 2 hex digits, therefore, if you have an odd number of digits in your HEX string, then something is wrong with it. You can try padding it with 0 in the beginning, such as this:

String hexImage ="0xFFD8FFE000104A46494600010200006400640000FFEC00114475636B79000100040000003C...";
if(hexImage.length()%2 == 1)
    hexImage = "0x0" + hexImage.substring(2);

or at the end, such as this:

String hexImage ="0xFFD8FFE000104A46494600010200006400640000FFEC00114475636B79000100040000003C...";
if(hexImage.length()%2 == 1)
    hexImage += "0";

However this is not guaranteed to produce a proper image.

On the whole, you should check how you get your hex string. A proper byte sequence should always contain an even number of hex digits.

EDIT: In addition, as Peter Lawrey indicated in his comment, you should check whether the decode method expects 0x in front of the string.