I have created this game bot where it connects to the game, and starts playing.
My problem is that i can't start more than one of these as the other then won't work.
Is is possible that if i run 2 instances of the same program the sockets are interfering with each other ? After all, they do connect to the same IP with the same port ?
And sometimes after i close(just closing cmd) the program is unable to connect again. Is that cause i didn't close the connections right ?.
I hope this is enough else i'll just have to post my source code
Best regards.
It's possible to connect to the same socket/port several times. Actually a socket is a double peer: {client ip/ client port}{server ip/server port}. When you connect to a server, your client port is assigned dynamically. You will have a new and different client port per client. So it should work unless the server side forbid it.
You should have a server that listens for multiple connections. A server is bound to a port and once that port is in use another application cannot use it. So for the server just have one instance. Multiple clients can connect to this IP/Port as long as the Server accepts multiple connections.
If a client connects to the server and the other clients stop working this may be because the server does not support multiple clients. To do this you need to use multi threading in the server. The server should accept a client socket and create a new instance of a client with it's own StreamReader/Writer objects.
http://tutorials.jenkov.com/java-multithreaded-servers/multithreaded-server.html
if you are working with a specific TCP port, then there is a close-wait period that this port cannot be claimed temporariliy for some time. also multiple programs cannot listen the same TCP port. Use threads.
Related
I am creating a multi threaded app, where server should handle multiple client requests simultaneously. Further, a client should communicate with other clients. The server will hold the port address of various clients and so each client must contact the server in order to know the port address of its peers.
My understanding about the programs:
client.java ( has a socket and a server socket)
socket - used for talking to the server in order to register its serversocket no, which will help other peers to identify. (Note: unlike server socket, where we might have to mentiond the port at the time of creation, socket # is uniquely assigned by the OS and so we don't need to worry about this.)
serversocket - used to talk to other peers and this number should be known by other peers.
server.java
serversocket - used for communicating with various clients
Now, I have two programs server.java and client.java. Of course, I will run client.java more than once in order to have multiple clients. Here are my questions:
will my client.java have two ports? one socket to talk to the server and one server socket to listen to other clients?
if so, should I pass the port no as a parameter to client.java so that it can be used as a unique server socket # for each client instance?
Please help me out!
yes you can pass them as runtime args or even better let the server decide it for you that way you wont have to worry about assigning unique port numbers.
Also i believe you will have to pass the client port and ip for clients to talk to each other, unless all your clients will be on the same machine.
I've got a server written on Java with ServerSocket.
And I have a client that is over a corporative firewall that is blocking everything except common ports.
I've started server on SMTP port (#25).
The user with firewall connects to it and so far everything is ok.
Then the server processes ServerSocket.accept(). And as far as I understand it creates a socket on a random port (every time the port number is different). And fails because of a firewall.
My question is - how can I make ServerSocket.accept() to choose a port for a socket from my white-list? I understand that it will not suite for massive online, but I want to make one my friend to be able to connect to my server.
Is it possible? And how?
server socket doesn't choose random port. it is the client socket which chose the random port. my guess is that your fire wall is smart and it knows to detect if the connection is approved by some sort of dpi (deep packet inspection).
if you want to mislead it, you can try to run data which looks like smtp in the first 2-3 packets and then switch to your protocol.
Another option it to use a kind of a connector outside the system, in this case both machines are clients which are connected through a 3rd client (there are many such proxies)
I am writing this game in Java and have problems with networking architecture.
I decided I will UDP packets. I am just at the beginning, but the problem I am facing is that it seems to be that server have to respond from exactly same IP/Port to client (which is behind router which uses NAT) as client connected that server.
For example I have client A behind router. Client A has IP (local) 192.168.8.100 and it connects server B from port 1234. Server is on 11.11.11.11:2345.
When client A connects to server B it uses 192.168.8.100:1234 but router converts that to (for example) 22.22.22.22:6789.
Now, when server wants to send packets to that client it has to be from 11.11.11.11:2345.
I would like to send data from another port like 11.11.11.11:2222, but this does not seem to work, at least not with my router.
I want to use different port because I want to have two threads one for listening and one for sending data, and each thread would have it's own DatagramSocket. But, as i said once client A connects to server on port 2345, I can not send data from port 2222.
Does anyone know how is this handled? I am doing it in Java, but it's not really a language specific problem.
UPDATE
After #Perception commented I have some more questions regarding his comments:
OK, so if I understand this correctly, if I have server which is hosting 1000 games, each with 2 players, all sending/receiving will have to be done through the same DatagramSocket.
As I understand DatagramSocket is thread safe so I guess I can have one thread doing:
datagramSocket.receive();
while at the same time second thread is doing
datagramSocket.send(.....);
Correct?
Also, two threads can send data at the same time through the same DatagramSocket? Is sending in any way serialized, meaning that second send() starts only after previous send() is finished or is data being sent at the same time?
gorann, I'm not sure if I'm understanding you correctly, but it sounds like you're trying to control the port on which the server communicates with the client. There's no way to control this, and for good reasons.
This is one of the trickier differences between TCP and UDP.
When a new TCP session is initiated, the server side call to accept() gives you a new socket and the OS handles multiplexing the various sessions for you. With UDP, you need to handle the multiplexing yourself. But you need to do so in a way that works with NATs and other firewalls.
The way NAT works is that when it sees an outgoing packet, it creates a temporary rule allow packets to return along the same port pair. Data returning from a port that the client has not yet sent to will likely be blocked.
This gives you two choices:
You could do all of your communication through a single port. This is not a bad option, it just means that you need a way to identify client sessions and route them to the appropriate thread.
You could create a separate port and instruct the client to send to that one instead. Have the server listen on a fixed port. The client sends a message to there, the server then sets up a new session port and sends that number back to the client using the server's listen port. The client then sends a message to the session port, which causes the NAT to open up that port and allow return traffic. Now the client and server thread have their own private port pair.
Option 1 is a bit more work because it requires data to be exchanged between threads, but it scales up better. Option 1 is easier and more CPU efficient because each session thread can be independent, but there are a finite number of ports available.
Either way, I recommend that you have the client include a semi-unique session id in each packet so that the server has more than just the client address and port number to verify who belongs to each session.
I'm making a java program & I want this to be both as server and a client (using sockets). How is this best achieved?
If you mean that you want to both send and receive data, a single regular socket (on each computer) will do just fine. See Socket.getInputStream and Socket.getOutputStream.
The usual "server" / "client" distinction just boils down which host is listening for incoming connections, and which hosts connect to those hosts. Once the connection is setup, you can both send and receive from both ends.
If you want both hosts to listen for incoming connections, then just set up a ServerSocket and call accept on both hosts.
Related links:
Official trail: The Java™ Tutorials, Lesson: All About Sockets
If you want each station to function as a server and a client, like a p2p chat,
you should implement a thread with a ServerSocket, listening for incoming connections, and once it got a connection, open a new thread to handle it so the current one will keep on listening for new connections.
For it to be able to connect to others, simple use SocketAddress and Socket, in a different thread to try to connect to a specified server address (e.g. by a list of the user's friends)
you can find plenty of chat examples by googling.
cheers.
If you want the program to perform the same operations regardless of whether it is a server or a client for a certain connection, I could imagine handing off both the client Socket and the ServerSocket.accept()-produced socket to the same method for processing.
Have a look at jgroups it's a library that allows the creation of groups of processes whose members can send messages to each other. Another option would be to use hazelcast...
You may also look at this question.
The best way to do this is to run the server on a thread:
You run server.accept() therefore while your program is listening for a connection on that thread you can do whatever you want on the main thread, even connect to another server therefore making the program both a server & a client.
I've a server (Java) and a number of clients (c++), connected by sockets.
I would like to set the ports automatically.
Assuming the IP is already known.
In the Java side I can make :
ServerSocket s = new ServerSocket(0);
So now I've a random free port on the server.
How can I know in the C++ side, what port is the server listening to?
I think is not possible, if you want establish a connection with a server, you must know in which port is the server listening, there are programs like nmap that shows you a list of opened ports in a server, but a server can have many opened ports at the same time and then, How do you know what is the port opened by your server? and in any case, is too slow and inefficient to call external tool, read and parse its output. For what reason do you need a random port service?
Other option can be get the opened socket in the server side calling to s.getLocalPort() and send it via UDP for any listening node in the network with broadcasting, and re-program the client side to listen in broadcast and when it receives a message, check if it is a port number and connect to the server using that port.
You can't, not reliably. In IP, a machine is identified by an address. A server (ie, a service) is identified by an address and a port. You clients need some form of "known service" that they can connect to.
If you, for whatever reason, absolutely want to have dynamic listening port, you could combine it with a "locator" service on a known port. For instance, have a web service/servlet on the standard http port (80). Your clients connect to the "locator" service (always on port 80) and asks which port your application is currently listening on. This is a not entirely uncommon pattern. RMI works is a similar way where you have a registry on a known port. Clients connect to the registry and asks for the location of RMI endpoints.