How to convert two longs to a byte array = how to

问题:

I am using Javas UUID and need to convert a UUID to a byte Array. Strangely the UUID Class does not provide a "toBytes()" method.

I already found out about the two methods:

UUID.getMostSignificantBits()
and
UUID.getLeasSignificantBits()

But how to get this into a byte array? the result should be a byte[] with those tow values. I somehow need to do Bitshifting but, how?

update:

I found:

 ByteBuffer byteBuffer = MappedByteBuffer.allocate(2);
 byteBuffer.putLong(uuid.getMostSignificantBits());
 byteBuffer.putLong(uuid.getLeastSignificantBits());

Is this approach corret?

Are there any other methods (for learning purposes)?

Thanks very much!! Jens

回答1:

You can use ByteBuffer

 byte[] bytes = new byte[16];
 ByteBuffer bb = ByteBuffer.wrap(bytes);
 bb.order(ByteOrder.LITTLE_ENDIAN or ByteOrder.BIG_ENDIAN);
 bb.putLong(UUID.getMostSignificantBits());
 bb.putLong(UUID.getLeastSignificantBits());

 // to reverse
 bb.flip();
 UUID uuid = new UUID(bb.getLong(), bb.getLong());

回答2:

One option if you prefer "regular" IO to NIO:

ByteArrayOutputStream baos = new ByteArrayOutputStream();
DataOutputStream dos = new DataOutputStream(baos);
dos.write(uuid.getMostSignificantBits());
dos.write(uuid.getLeastSignificantBits());
dos.flush(); // May not be necessary
byte[] data = dos.toByteArray();

回答3:

For anyone trying to use this in Java 1.7, I found the following to be necessary:

<!-- language: lang-java -->
ByteArrayOutputStream baos = new ByteArrayOutputStream();
DataOutputStream dos = new DataOutputStream(baos);
dos.writeLong(password.getMostSignificantBits());
dos.writeLong(password.getLeastSignificantBits());
dos.flush(); // May not be necessary
return baos.toByteArray();