I have an application that is communicating with a UDP server. My application listens on one port (say 1234) and sends on another (say 5678). The UDP server I am communicating with also requires a "heartbeat" ever 5 seconds, for which I create another thread. When my application first starts up, I create the listen thread, then create the heartbeat thread, then I start sending the UDP server message packets. The only thing, however, is that it seems like all the packets I send out finish before the heartbeat thread starts.
Here is what I have for my listener:
public class MyListener implements Runnable {
private volatile boolean run = true;
private DatagramSocket myDatagramSocket;
private DatagramPacket myDatagramPacket;
private byte[] receiveBuffer;
private int receiveBufferSize;
#Override
public void run(){
while(run){
try {
myDatagramSocket = new DatagramSocket(null);
InetSocketAddress myInetSocketAddress = new InetSocketAddress(1234);
myDatagramSocket.bind(myInetSocketAddress);
receiveBuffer = new byte[2047];
myDatagramPacket = new DatagramPacket(receiveBuffer, 2047);
myDatagramSocket.receive(myDatagramPacket);
byte[] data = myDatagramPacket.getData();
receiveBufferSize = myDatagramPacket.getLength();
switch(messageID){
...
}
} catch (Exception e){
}
}
}
}
Here is what I have for my heartbeat:
public class MyHeartbeat implements Runnable {
private volatile boolean run = true;
private HeartbeatSenderClass heartbeatSender;
#Override
public void run(){
while(run){
try {
TimeUnit.SECONDS.sleep(5);
heartbeatSender.sendHeartbeat();
} catch(Exception e){
}
}
}
}
Here is what I have for my main class:
public class MyApp {
public static void main(String[] args){
MyListener listener = new MyListener();
Thread listenerThread = new Thread(listener);
listenerThread.setName("Listener Thread");
listenerThread.start();
MyHeartbeat heartbeat = new MyHeartbeat();
Thread heartbeatThread = new Thread(heartbeat);
heartbeatThread.setName("Heartbeat Thread");
heartbeatThread.start();
MySender sender = new MySender();
Thread senderThread = new Thread(sender);
senderThread.setName("Sender Thread");
senderThread.start();
}
}
All of my packets are making it to the UDP server, but not smoothly like I would have thought. I would have thought that while I am sending packets to the server, every 5 seconds my heartbeat would be sent out. However, it seems like my heartbeats are going out only after my packets are done sending. Also, I believe I am not getting all of the messages from the UDP server. I say this because I have sniffed the UDP packets on my machine and I see data coming from the server that my receiver is not receiving/processing. Any suggestions?
You have in heartbeat this:
TimeUnit.SECONDS.sleep(5);
heartbeatSender.sendHeartbeat();
So before sending the very first beat, you wait for 5 seconds. No wonder that the other threads do their job meanwhile.
The DatagramSocket you use to send the packets is a shared resource that is contended between threads, and then if a thread consume too much of that resource, another one may starve. See: Thread starvation
Also if you are loosing packets, it happens because you can't read as fast as you should. If udp packets arrive faster then they can be read, the queue will discard the remaining.
Under linux, for example you can control the receive buffer with:
sudo sysctl -w net.core.rmem_default=26214400
sudo sysctl -w net.ipv4.udp_mem='26214400 26214400 26214400'
sudo sysctl -w net.ipv4.udp_rmem_min=26214400
But anyway if we are talking about a sustained loss, you should consider to have a thread for reading the buffer, a queue and a thread to process the readed data.
Related
I'm writing a program that:
Sends ping packets to other processes running the same program
The process sends a response to the ping from (1)
If a process does not receive a response in X seconds, then call a method to update states
So I currently only have 1 Datagram socket that handles both sending and receiving the packets. The way I handled the ping timeout was calling setSoTimeout on the socket and blocking on the receive right after I send the ping packet to another process.
public MyProgram {
private DatagramSocket socket;
public MyProgram() {
socket = new DatagramSocket("localhost", 12345);
Thread sendingThread = new Thread(new Runnable() {
public void run() {
// call sendLoop method
}
}
Thread receiveThread = new Thread(new Runnable() {
public void run() {
// call receiveLoop method
}
}
}
public void sendLoop() {
while (true) {
/*
send ping to other process
*/
// now wait until it replies with ACK
try {
socket.receive(...);
} catch (SocketTimeoutException e) {
// call method to update state
}
}
}
// ... more code ...
}
Now the problem is when I write receiveLoop (separate thread to sendLoop) which waits for incoming ping packets from other processes
public void receiveLoop() {
while (true) {
socket.receive(...);
// call methods to reply to the ping
}
}
How can I wait for a response to my own ping, whilst waiting for incoming pings on another thread? I figured this approach would be bad because the ping response could arrive on the other thread?
Can you even simultaneously call receive on the same UDP socket in different threads?
Is there a way to solve this without having to create another UDP socket? Many thanks!
I face this problem in Java.
I have a server class named MyServer and I want to implement a thread pool where each thread runs a method of MyServer when a request comes. I have created another class that implements a server pool named MultiThreadedSocketServer. The class is this:
public class MultiThreadedSocketServer {
public void startServer(MyServer s, int localport, int threadPoolSize) {
final ExecutorService clientProcessingPool = Executors.newFixedThreadPool(threadPoolSize);
Runnable serverTask = new Runnable() {
#Override
public void run() {
try {
ServerSocket serverSocket = new ServerSocket(localport);
System.out.println("Waiting for clients to connect...");
while (true) {
Socket clientSocket = serverSocket.accept();
clientProcessingPool.submit(new ClientTask(clientSocket, s));
}
} catch (IOException e) {
System.err.println("Unable to process client request");
e.printStackTrace();
}
}
};
Thread serverThread = new Thread(serverTask);
serverThread.start();
}
}
the class named MultiThreadedSocketServer has an argument named Server s which passes it in client Task class which a thread is created. The client task class is this:
class ClientTask implements Runnable {
private final Socket clientSocket;
private MyServer s;
public ClientTask(Socket clientSocket, MyServer s) {
this.s = s;
this.clientSocket = clientSocket;
}
#Override
public void run() {
System.out.println("Got a client !");
String inputLine = null;
try {
BufferedReader in = new BufferedReader(new InputStreamReader(clientSocket.getInputStream()));
// Do whatever required to process the client's request
inputLine = in.readLine();
if (inputLine.equals("Bye")) {
System.out.println("Bye");
System.exit(0);
}
s.handleRequest(inputLine);
clientSocket.close();
} catch (IOException e) {
e.printStackTrace();
}
}
}
As you can see when a request comes the handleRequest method of class MyServer is invoked. I want to make this method to run synchronized, meaning only one thread at a time to be able to run this method. Adding synchronized before the method implementation does not achieve anything.
Can anybody give me the proper way to do this?
Thanks in advance for your time.
PS: I added the whole code
MyServer Class
http://pastebin.com/6i2bn5jj
Multithreaded server Class
http://pastebin.com/hzfLJbCS
As it is evident in main I create three requests with handleRequest with arguments Task, task2 and Bye.
The correct output would be
Waiting for clients to connect...
Got a client !
This is an input Task
Request for Task
Got a client !
This is an input task2
Request for task2
Got a client !
This is an input
Bye
But Instead the order is mixed. Sometimes Bye which shuts the server can be executed first. I want to ensure that the order is the one where the requests are created in the main.
But Instead the order is mixed. Sometimes Bye which shuts the server can be executed first. I want to ensure that the order is the one where the requests are created in the main.
You say that you want the server to handle requests in order. This is hard to ensure because you are opening up 3 sockets and writing them to the server but not waiting for any response. This is implementation dependent but I'm not sure there is any guarantee that when the client returns from doing a socket InputStream write, that the server has received the bytes. This means that from the client side, there is no guarantee that the IO completes in the order that you want.
To see if this is the problem, I would remove the System.exit(0) to see if the other lines make it, just after the "Bye" string does. Or you could put a Thread.sleep(5000); before the exit(0).
A simple sort-of fix would be to make sure your PrintStream has auto-flush turned on. That at least will call flush on the socket but even then there are race conditions between the client and the server. If the auto-flush doesn't work then I'd have your client wait for a response from the server. So then the first client would write the first command and wait for the acknowledgement before going to the 2nd command.
In terms of your original question, locking on the server wouldn't help because of the race conditions. The "Bye" might make it first and lock the server fine.
These sorts of questions around how to synchronize the threads in a multi-threaded program really make no sense to me. The whole point of threads is that they run asynchronously in parallel and don't have to operate in any particular order. The more that you force your program to spit out the output in a particular order, the more you are arguing for writing this without any threads.
Hope this helps.
If the problem is that the bye message kills the server before other requests can be handled, one solution could be to not call System.exit(0); on bye.
The bye message could set a flag block further requests from being handled and also notify some other mechanism to call System.exit(0); when the thread pool is idle with no requests left to handle.
First question here on StackOverflow, so please excuse me if I ask this incorrectly.
Basically, I'm writing a Multicast Client that indefinitely listens to a multicast address until the user types "quit" into the console. I've found that setting SO_TIMEOUT for the MulticastSocket, checking if "quit" has been typed, and then returning to the receive method call doesn't really work since a packet could be sent right after the timeout and the check of the console blocks. So I believe the best option is to simply have 2 threads going where one listens on the socket and blocks until it receives something, and the other thread listens to the console until told to quit. The only issue I have is that I'm unsure of how to go about having the console listening thread tell the socket thread to close the socket and terminate. System.end() would work but I fear that I'd leave a socket open, etc.
TLDR; Is there a way for the main method of a class to start a thread, and then respond a specific way once that thread ends? I need to listen to the console on one thread and a MulticastSocket on another, or just in the main of the client class.
Thanks everyone.
I would call Socket.close() to close the socket. This will produce an IOException in that thread. so before doing this I would set a flag like closed = true; and have the other thread check this before printing the error i.e. don't print an IOException if you have been closed. Something like this.
public class SocketListener implements Runnable, Closeable {
final MulticastSocket socket;
final Consumer<DatagramPacket> packetConsumer;
volatile boolean closed;
public SocketListener(MulticastSocket socket, Consumer<DatagramPacket> packetConsumer) {
this.socket = socket;
this.packetConsumer = packetConsumer;
}
#Override
public void run() {
DatagramPacket packet = new DatagramPacket(new byte[1024], 1024);
try {
while(!closed) {
socket.receive(packet);
packetConsumer.accept(packet);
}
} catch (IOException e) {
if (!closed)
e.printStackTrace();
}
}
#Override
public void close() throws IOException {
closed = true;
socket.close();
}
}
for example, in your main thread you can do
MulticastSocket socket = ...
Consumer<DatagramPacket> packetConsumer = ...
try (SocketListener listener = new SocketListener(socket, packetConsumer)) {
boolean finished = false;
do {
// read from the console
if (some condition)
finished = true;
} while(!finished);
} // calls close() for you.
I have the following code in my main application:
package acast;
import java.net.SocketException;
import java.util.concurrent.ConcurrentLinkedQueue;
public class ACast {
private ConcurrentLinkedQueue<String> queue;
public ACast() throws SocketException{
queue = new ConcurrentLinkedQueue<String>();
UDPServer srv = new UDPServer(4321);
srv.addUDPacketListener(new UDPPacketListener() {
#Override
public void onPacketReceived(String packet) {
ACast.this.queue.offer(packet);
}
});
srv.start();
}
public static void main(String[] args) throws SocketException {
try{
new ACast();
}
catch(SocketException e){
//e.printStackTrace();
System.out.println("Socket allready opened. Can't start application");
System.exit(1);
}
}
}
My UDPServer extends Thread and calls onPacketReceived every time it receives an UDP datagram. I want my main app to do something every time a configured timeout passes since the last received datagram. I would like to avoid running a Thread that just checks the timeout from second to second. I would like to start a countdown thread exactly on the moment of the last received datagram and cancel any other ongoing timeout threads if available. Any help ?
A simple solution would be to start a Timer with the timeout task, and every time a new datagram is received, cancel the currently running timer and start a new one.
I would lose the asynchronicity altogether, and use blocking I/O with a read timeout.
am currently working on a project where I have to build a multi thread server. I only started to work with threads so please understand me.
So far I have a class that implements the Runnable object, bellow you can see the code I have for the run method provided by the Runnable object.
public void run() {
while(true) {
try {
clientSocket = serversocket.accept();
for (int i = 0; i < 100; i++) {
DataOutputStream respond = new DataOutputStream(clientSocket.getOutputStream());
respond.writeUTF("Hello World! " + i);
try {
Thread.sleep(1000);
} catch(InterruptedException e) {
//
}
}
} catch(IOException e) {
System.out.println(e.getMessage());
}
}
}
Bellow is the main method that creates a new object of the server class and creates a threat. initializing the Thread.
public static void main(String args[]) {
new Thread(new Server(1234, "", false)).start();
}
I know this creates a new thread but it does not serve multiple clients at once. The first client need to close the connection for the second to be served. How can I make a multi threated server that will serve different client sockets at once? Do I create the thread on the clientSocket = serverSocket.accept();
yes.
from the docs:
Supporting Multiple Clients
To keep the KnockKnockServer example simple, we designed it to listen for and handle a single connection request. However, multiple client requests can come into the same port and, consequently, into the same ServerSocket. Client connection requests are queued at the port, so the server must accept the connections sequentially. However, the server can service them simultaneously through the use of threads—one thread per each client connection.
The basic flow of logic in such a server is this:
while (true) {
accept a connection;
create a thread to deal with the client;
}
The thread reads from and writes to the client connection as necessary.
https://docs.oracle.com/javase/tutorial/networking/sockets/clientServer.html