How to convert string to hex bytes in Java

Question

I want to split each of two chars in string and convert it to hex byte array representation, i am just lost how to do this.

in string a= hex a which is 10 in decimal in string b= hex b which is 11 in decimal

String toConvert = "abbbbbbbbbbbbbbbbbbbbbbc";
byte[] output = new byte[12];





                          Input
 ab    bb   bb    bb  bb   bb   bb   bb   bb   bb   bb   bc
                          output
[-85, -69, -69, -69, -69, -69, -69, -69, -69, -69, -69, -68]

no i want a byte array ass output not string and its string to hex not hex to string. — H4SN, Nov 22 '14 at 15:09
http://stackoverflow.com/questions/5886619/hexadecimal-to-integer-in-java — code monkey, Nov 22 '14 at 15:16
example (ab) a is msb and b is lsb i want to combine them and make a 16bit no than convert it to hex — H4SN, Nov 22 '14 at 15:18
`in string a= hex a which is 10 in decimal in string b= hex b which is 11 in decimal` You mean in `String toConvert`? What a terrible example! Better `String toConvert = "0344A7DF";`. And you will not convert to hex but decode from hex representation. — greenapps, Nov 22 '14 at 15:24

August · Accepted Answer · 2014-11-22T15:28:28.680

Takes the first character in a group of two, multiplies its hex value by 16 (it's in the 16¹ place). That result is added to the second character's hex value.

String toConvert = "abbbbbbbbbbbbbbbbbbbbbbc";
byte[] output = new byte[toConvert.length() / 2];

for (int i = 0; i < output.length; i++) {
    output[i] |= Character.digit(toConvert.charAt(i * 2), 16) * 16;
    output[i] |= Character.digit(toConvert.charAt(i * 2 + 1), 16);
}

score 0 · Answer 2 · answered Nov 22 '14 at 15:19

Apache Common Codec's Hex class does exactly what you need:

byte[] bytes = Hex.decodeHex("abbbbbbbbbbbbbbbbbbbbbbc".toCharArray());

If you can't/won't use third parties, you can always "borrow" their implementation (slightly simplified - I omitted correctness checks for simplicity):

public static byte[] decodeHex(final char[] data) {

  final int len = data.length;

  // Handle empty string - omitted for clarity's sake

  final byte[] out = new byte[len >> 1];

  // two characters form the hex value.
  for (int i = 0, j = 0; j < len; i++) {
      int f =  Character.digit(data[j], 16) << 4;
      j++;
      f = f | Character.digit(data[j], 16);
      j++;
      out[i] = (byte) (f & 0xFF);
  }

  return out;
}

Dima · Answer 3 · 2014-11-22T15:39:59.827

0

for(int i = 0; i < 12; i++) {
    output[i] = (byte) Integer.parseInt(toConvert.substring(i*2, i*2+2), 16);
}

edited Nov 22 '14 at 15:39

answered Nov 22 '14 at 15:22

Dima

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Primitives don't have methods. Maybe you meant `Integer.valueOf`? – August Nov 22 '14 at 15:30
No, I meant what I wrote :) Integer is not a primitive. Just try running it, you'll see. – Dima Nov 22 '14 at 15:31
@Dima I'd run it, if I could (does not compile, because of what has been stated above). – August Nov 22 '14 at 15:33
sure it does, c'mon. Try it. – Dima Nov 22 '14 at 15:34
Oops, I see what you meant. Fixed the answer. Thanks – Dima Nov 22 '14 at 15:40
1

Code only answers are discouraged. It is always better to explain the what, why, and how. – RubberDuck Nov 22 '14 at 15:52
Hmmm. I think, sometimes seeing the code is clearer than any human-language explanation. Can you suggest what in particular about this snipped deserves to be explained in more detail? – Dima Nov 22 '14 at 15:59

How to convert string to hex bytes in Java

3 Answers3