convert uuid to byte, that works when using UUID.nameUUIDFromBytes(b)

Question

I am converting UUID to byte using this code

public byte[] getIdAsByte(UUID uuid)
{
    ByteBuffer bb = ByteBuffer.wrap(new byte[16]);
    bb.putLong(uuid.ge         


        

Answer 1

public class UuidUtils {
  public static UUID asUuid(byte[] bytes) {
    ByteBuffer bb = ByteBuffer.wrap(bytes);
    long firstLong = bb.getLong();
    long secondLong = bb.getLong();
    return new UUID(firstLong, secondLong);
  }

  public static byte[] asBytes(UUID uuid) {
    ByteBuffer bb = ByteBuffer.wrap(new byte[16]);
    bb.putLong(uuid.getMostSignificantBits());
    bb.putLong(uuid.getLeastSignificantBits());
    return bb.array();
  }
}

---

@Test
public void verifyUUIDBytesCanBeReconstructedBackToOriginalUUID() {
  UUID u = UUID.randomUUID();
  byte[] uBytes = UuidUtils.asBytes(u);
  UUID u2 = UuidUtils.asUuid(uBytes);
  Assert.assertEquals(u, u2);
}

@Test
public void verifyNameUUIDFromBytesMethodDoesNotRecreateOriginalUUID() {
  UUID u = UUID.randomUUID();
  byte[] uBytes = UuidUtils.asBytes(u);
  UUID u2 = UUID.nameUUIDFromBytes(uBytes);
  Assert.assertNotEquals(u, u2);
}

Answer 2

that's because nameUUIDFromBytes constructs a specific kind of UUID (as the javadoc states).

if you want to convert a byte[] back to a UUID, you should use the UUID constructor. Wrap a ByteBuffer around the byte[], read the 2 longs and pass them to the UUID constructor.