How to format string in hex in Java?

Question

I want to format string in hex in Java.

e.g. input: 123123, output: 1E 0F 03

Here is my code:

String.format("%02X", integer)

Does this answer your question? [Converting A String To Hexadecimal In Java](https://stackoverflow.com/questions/923863/converting-a-string-to-hexadecimal-in-java) — alcatraz, Jul 07 '21 at 17:12

score 1 · Answer 1 · answered Jul 07 '21 at 17:29

1

Try something like:

        int n = 123123;
        StringBuilder sb = new StringBuilder();
        String sep = "";
        while (n > 0) {
            int nybble = n & 0xFF;
            sb.insert(0, sep);
            sb.insert(0, String.format("%02X", nybble));
            sep = " ";
            n >>>= 8;
        }
        System.out.println(sb.toString());

answered Jul 07 '21 at 17:29

g00se

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Did you know that you can append StringBuilder calls? So you can do `sb.insert(0, sep).insert(0, String.format("%02X", nybble));` – WJS Jul 07 '21 at 21:30
1

> Did you know that you can append StringBuilder calls?< You mean *chain* calls. Yes, I did but I thought the code might be a bit clearer for not doing that in this particular case. – g00se Jul 07 '21 at 23:17

WJS · Answer 2 · 2021-07-07T18:21:11.750

You could do it like this

convert the value to uppercase hex
prepend a "0" based on length
then split and rejoin (the regex splits on the zero width area after every two characters).

for (int val : new int[] { 123123, 122, 94392, 9200812 }) {
    String valStr = String.format("%X",val);
    valStr = (valStr.length() % 2 == 0 ? "" : "0") + valStr;
    String result = String.join(" ", valStr.split("(?<=\\G(..))"));
    System.out.printf("%7d = %10s = %s%n", val, valStr, result);
}

Prints

 123123 =     01E0F3 = 01 E0 F3
    122 =         7A = 7A
  94392 =     0170B8 = 01 70 B8
9200812 =     8C64AC = 8C 64 AC

How to format string in hex in Java?

2 Answers2