Order byte array by the DataInputStream int

Question

I have audio packets with a sequence number at the start, which is 4 bytes.

ByteArrayOutputStream baos = new ByteArrayOutputStream();
DataOutputStream dos = new DataOutputStream(baos);
dos.writeInt(sequenceNumber);

I receive the packets in a random order, which I then place into a buffer, so an array of arrays.

I was wondering what would be the best way to order the packets by the sequence number.

I retrieve the sequence number like:

 ByteArrayInputStream bos = new ByteArrayInputStream(udpPacketBytes);
 DataInputStream ds = new DataInputStream(bos);
 int receivedValue = ds.readInt();

Is there a way without removing the sequence number that I can order the entire byte array by said sequence number?

Answer 1

You can do

byte[] bytes = new byte[ds.available()];
ds.readFully(bytes);

to get the remaining bytes.

To ensure the packets are in the original order, you need to check the sequence number is 1 more than the previous ones. If not, you need to save the packets by sequence number and re-order them. The more challenging problem is when packet are dropped you need to ask for the again.

You could use something like

public class OrderManager {
    int nextSequence = 0;
    final SortedMap<Integer, byte[]> buffered = new TreeMap<>();
    final Consumer<byte[]> consumer;

    public OrderManager(Consumer<byte[]> consumer) {
        this.consumer = consumer;
    }

    public void accept(int num, byte[] bytes) {
        if (num == nextSequence) {
            consumer.accept(bytes);
            nextSequence++;
            while (buffered.firstKey() == nextSequence) {
                consumer.accept(buffered.remove(buffered.firstKey()));
                nextSequence++;
            }
        } else {
            buffered.put(num, bytes);
        }
    }
}

Since out of order packets are rare but lost packets are fairly common you could treat an out of order packet as if it had been lost and send a packet to the producer to sent it again.