Generate 8-byte number in Java

Question

I'm a little bit confused, how to do this. I know I can use Random class to generate random numbers, but I don't know how to specify and generate 8-byte number?

Thanks, Vuk

Answer 1

You should note that the java.util.Random class uses a 48-bit seed, so not all 8-byte values (sequences of 64 bits) can be generated using this class. Due to this restriction I suggest you use SecureRandom and the nextBytes method in this situation.

The usage is quite similar to the java.util.Random solution.

SecureRandom sr = new SecureRandom();
byte[] rndBytes = new byte[8];
sr.nextBytes(rndBytes);

---

Here is the reason why a 48-bit seed is not enough:

• The Random class implements a pseudo random generator which means that it is deterministic.
• The current "state" of the Random determines the future sequence of bits.
• Since it has 2⁴⁸ states, it can't have more than 2⁴⁸ possible future sequences.
• Since an 8-byte value has 2⁶⁴ different possibilities, some of these possibilities will never be read from the Random object.

---

Based on @Peter Lawreys excellent answer (it deserves more upvotes!): Here is a solution for creating a java.util.Random with 2×48-bit seed. That is, a java.util.Random instance capable of generating all possible longs.

class Random96 extends Random {
    int count = 0;
    ExposedRandom extra48bits;

    class ExposedRandom extends Random {
        public int next(int bits) {    // Expose the next-method.
            return super.next(bits);
        }
    }

    @Override
    protected int next(int bits) {
        if (count++ == 0)
            extra48bits = new ExposedRandom();
        return super.next(bits) ^ extra48bits.next(bits) << 1;
    }
}