How do I store an Int, a short, and multiple bytes from file in byte array

Question

I'm attempting to do this for a project but can't figure out how to store the three values in one byte array. This is what i've tried
public void send(byte[] input)
          // TODO
          ByteArrayInputStream bais = new ByteArrayInputStream(input);
          ByteArrayOutputStream baos = new ByteArrayOutputStream();
          DataOutputStream dos = new DataOutputStream(baos);
          byte[] pData = new byte[100];
          try {
               dos.writeShort(checkSum);
               dos.writeInt(nextSeqNum);
               pData=baos.toByteArray();
               bais.read(pData);
          } catch (IOException e) {
          DatagramPacket dgp = new DatagramPacket(pData, pData.length);When i set pData=baos.toByteArray() it changes the size and then stops me from being able to read the other 94 bytes into the array to send. Any ideas?

Answer 1

You don't need the ByteArrayInputStream at all.
dos.writeShort(checkSum);
dos.writeInt(nextSeqNum);
dos.write(input);
pData = baos.toByteArray();
When i set pData=baos.toByteArray() it changes the sizeNo, it makes it refer to a new byte[] array containing what you've written to the BAOS. That's what it's supposed to do.