How is the fastest way to check each digit of a byte array in Java?

Question

Given an arbitrary byteArray I need to check if a digit is greater than 9, if so the method returns false. I came to this solution, but I think it's a better way of doind that (maybe with some unary operations or other tricks):

for (int i : byteArray)
    if (i/16 > 0x09 || i%16 > 0x09) return false;

I tried to convert the number to a string, but it was slower.

why converting byte back to int? why not compare it directly? — office.aizaz, Nov 06 '22 at 10:58
You can replace `if (i/16 > 0x09 || i%16 > 0x09) return false;` with `return i%16 <= 0x09;` — Arvind Kumar Avinash, Nov 06 '22 at 10:58
Have you tested your current code? What does it say for the byte 0xFF? — tgdavies, Nov 06 '22 at 11:07
Related and necessary: [How do I write a correct micro-benchmark in Java?](https://stackoverflow.com/questions/504103/how-do-i-write-a-correct-micro-benchmark-in-java) — Ole V.V., Nov 06 '22 at 12:05
@office.aizaz you was right about that, IDK why I converted back to int — NotInCheck, Nov 06 '22 at 13:04
@ArvindKumarAvinash That won’t give the desired result since it’s in a loop. — Ole V.V., Nov 06 '22 at 15:49

office.aizaz · Answer 1 · 2022-11-06T11:32:28.710

1

How about this?

byte[] arr = new byte[] {0x00, 0x01, 0x02};
for (byte b : arr) {
      if (b > 0x09) {
          return false;
      }
}

As mentioned in comment, OP is interested in comparing each 'nibble':

byte[] arr = new byte[] {0x00, 0x01, 0x02};
for (byte b : arr) {
      if ((b & 0xF0) > 0x90 || (b & 0x0F) > 0x09) {
          return false;
      }
  }

edited Nov 06 '22 at 11:32

answered Nov 06 '22 at 11:04

office.aizaz

179
11

1

The OP wants to check each 4 bit "nybble" of the byte. – tgdavies Nov 06 '22 at 11:04
1

`(b & 0xF0) > 0x09` -> `(b & 0xF0) > 0x90` – Nov 06 '22 at 11:30
@OleV.V. The maximum byte value is 0x7F, so it can never exceed 0x9F. – Nov 06 '22 at 12:13

How is the fastest way to check each digit of a byte array in Java?

1 Answers1