This accept() method return a tuple with a new socket and an address but why do i need a new socket if i already have one, so why don't use it?
import socket
sock = socket.socket()
sock.bind(('', 9090))
sock.listen(1)
conn, addr = sock.accept()
print 'connected:', addr
while True:
data = conn.recv(1024)
if not data:
break
conn.send(data.upper())
conn.close()
ps: When i program sockets in Java, i don't really have this kind of accepting stuff and i only need one socket per client and one per server, which makes sense.
You have one listening socket active while the server is running and one new connected socket for each accepted connection which is active until the connection is closed.
Seems like you haven't implemented TCP in Java before.
The example you are providing with, uses a default AF_INET and SOCK_STREAM which by default is TCP:
socket.socket([family[, type[, proto]]])
Create a new socket using the given address family, socket type and protocol number. The address family should be AF_INET (the default), AF_INET6 or AF_UNIX. The socket type should be SOCK_STREAM (the default), SOCK_DGRAM or perhaps one of the other SOCK_ constants. The protocol number is usually zero and may be omitted in that case.
If you were implemented SOCK_DGRAM which is UDP, you wouldn't need to be using sock.accept()
Related
I have a ServerSocketChannel and once this channel accepts a connection I want to "transfer" or "pass" it to a SocketChannel. I have tried, once I have accepted the connection, to simply do a socketchannel.bind(ServerSocketChannel.socket.getLocalSocketAddress) but I get an exception that says "Address already in use". Which makes sense so I have tried to save the socket address in a variable, close the ServerSocketChannel, and then attempt to bind it with the variable but it still complains about already being in use.
//Accept the connection
ServerSocketChannel.accept();
//Save the address and close the connection
address = ServerSocketChannel.socket().getLocalSocketAddress();
ServerSocketChannel.close();
//Bind the address to another channel
SocketChannel.bind(address);//Here is where I get the exception
This is how my solution looks right now(except for the variable names but you get the idea). The reason I can't just accept it with that channel is because it's used for another purpose once this connection has been established, so I need to have dedicated ServerSocketChannel opened just for accepting connections which are then passed to the SocketChannel.
I just solved my problem. I overlooked something simple and tried a longer way instead.
SocketChannel = ServerSocketChannel.accept();
Is what I was looking for all along.
I am using UDP broadcast for interservice communication The server is in Python and I can see the UDP messages using this code:
import sys
import socket
HOST = ''
PORT = 9002
s = socket.socket(socket.AF_INET,socket.SOCK_DGRAM)
s.setsockopt(socket.SOL_SOCKET,socket.SO_REUSEADDR,1)
s.setsockopt(socket.SOL_SOCKET,socket.SO_BROADCAST,1)
s.bind((HOST,PORT))
while True:
try:
message = s.recv(8192)
print("Got data: " + message)
except KeyboardInterrupt:
sys.exit()
I can run as many of these clients simultaneously as I want on the same machine.
I'm trying to implement a similar client in Java using the DatagramSocket class, but I keep getting an "address already in use" error. Evidently I need to construct it differently than I am currently:
DatagramSocket socket = new DatagramSocket(broadcastPort);
Is it possible to get the same behavior as the Python code?
Try this:
// create an unbound socket
DatagramSocket socket = new DatagramSocket(null);
// make it possible to bind several sockets to the same port
socket.setReuseAddress(true);
// might not be necessary, but for clarity
socket.setBroadcast(true);
socket.bind(new InetSocketAddress(9002));
The null argument to the constructor is the key here. That wisdom is hidden in the second paragraph of the javadoc for the DatagramSocket(SocketAddress bindAddress) constructor:
if the address is null, creates an unbound socket.
Address already in use probably means you haven't properly terminated your programs. If you're using Eclipse, make sure you check all your open consoles and terminate all of them. (Top right corner, blue box - click it and it'll show all running programs)
In Eclipse, just because you "run" your program again, it doesn't terminate the previous one(s).
Another possible issue is that you may be using the same port as your python server. 2 Applications can't claim the same port, so just change the port number if that is the case.
Edit: Use a MulticastSocket.
I have a java.nio.channels.ServerSocketChannel which I initialised as follows:
while(true)
{
ServerSocketChannel channel = ServerSocketChannel.open();
InetSocketAddress serverSocket = new InetSocketAddress(host,port);
channel.bind(serverSocket);
SocketChannel ch = channel.accept();
// Later on when I have read off data from a client, I want to shut this
// connection down and restart listening.
channel.socket().close(); //Just trying to close the associated socket
// too because earlier approaches failed
channel.close();
}
When I send the first message from client it is successfully delivered to server and the client program exits. Then trouble begins. When I initialise the client again and try to
establish at the same port and address of the server as I did the first time, I get a
java.net.BindException: Address already in use: connect
exception even though I closed the associated channel/socket.
I have been renewing the ServerSocketChannel and InetSocketAddressobjects because as my client instance has to shut down after a write, I have to disengage that channel and since I cannot reuse a channel after it has been closed, I have to make a new object everytime. My theory is since the channel reference is reassigned each time, the orphaned object becomes GC meat, but since the close() method apparently is not working properly, the channel is still alive and until GC collects it my port will be hogged.
Nevertheless I tried keeping the initialisation of ServerSocketChannel and InetSocketAddress objects before the while loop, but this did not help, and the same exception occurred after the first write, as before.
ServerSocketChannel channel = ServerSocketChannel.open();
InetSocketAddress serverSocket = new InetSocketAddress(host,port);
channel.bind(serverSocket);
while (true)
{
SocketChannel ch = channel.accept();
//read from a client
}
For clarity , here is how I connect from the client:
SocketChannel ch=SocketChannel.open();
ch.bind(new InetSocketAddress("localhost", 8077));
InetSocketAddress address=new InetSocketAddress("localhost",8079);
//the address and port of the server
System.out.print(ch.connect(address));
ByteBuffer buf=ByteBuffer.allocate(48);
buf.clear();
buf.put("Hellooooooooooooooooooooooooo".getBytes());
buf.flip();
while(buf.hasRemaining()) {
ch.write(buf);
}
ch.close();
It looks like you're confusing client and server. Normally, server starts only once and binds to s port. Usually, there's no need to close there anything as the port gets freed when the program exits. Obviously, you must close the Sockets obtained by ServerSocket.accept(), but that's another story.
I guess you've got confused by your variable names (just like it happened to me as I started with this). Try to call all things according to their type, here was Hungarian really helpful for me.
The code I wrote for testing this is long, stupid, and boring. But it seems to work.
It may also be helpful to do:
channel.setOption(StandardSocketOptions.SO_REUSEADDR, true);
Search for information about this option to learn more.
do ch.close() as well to GC the client socket.
I'm trying to receive UDP data broadcast by PlayCap to network address 192.168.103.255 port 3000 in Java, but I'm having trouble setting things up. Here's what I have:
DatagramSocket socket = new DatagramSocket();
InetSocketAddress address = new InetSocketAddress("192.168.103.255", 3000);
socket.bind(address);
I'm getting "java.net.SocketException: already bound" from the bind call. I'm pretty inexperienced with networking, so I may be doing something way wrong here. Any help is appreciated.
Here is the stacktrace:
java.net.SocketException: already bound
at java.net.DatagramSocket.bind(Unknown Source)
at runner.main(runner.java:16)
I dont want to revive and old thread but i don't think the answer to this question is correct. I faced the same issue when i used the similar code to create a DatagramSocket.
DatagramSocket socket = new DatagramSocket();
socket.setReuseAddress(true);
socket.bind(new InetSocketAddress(InetAddress.getByName("localhost"), 5566));
This results in a SocketException
Exception in thread "main" java.net.SocketException: already bound
at java.net.DatagramSocket.bind(DatagramSocket.java:376)
at testapplication.TestApplication.main(TestApplication.java:25)
Java Result: 1
Not because there is another process occupying the same port but i have created an already BOUND datagram socket when i use the default constructor.
new DatagramSocket()
According to javadoc:
DatagramSocket()
Constructs a datagram socket and binds it to any available port on the
local host machine.
So the reason for the exception is you are trying to bind an already bound socket. To make it work you need to create an unbond socket with below constructor
DatagramSocket socket = new DatagramSocket(null);
InetSocketAddress address = new InetSocketAddress("192.168.103.255", 3000);
socket.bind(address);
Hope this helps...
Do netstat -a -o -n and from this you can find that either this port is already bind or not(even from this you can get all the bound ports).If yes , then try any other port :)
Most probably your application is running twice. Or you might be executing the same code twice. Even the same application may fail when binding twice.
Happens a lot for beginners that they didn't shut down their previous attempt (happened to me, too), and then their port is already in use. Make sure to add proper exception handling, e.g. by popping up a message "Port already in use."
Note that for listening you usually will bind a port only, without an explicit address (you might need to use "0.0.0.0" for this). Then you can receive both broadcast and unicast.
The code I use for listening to broadcasts is simply:
DatagramSocket s = new DatagramSocket();
s.bind(new InetSocketAddress(port))
Note that I'm not binding to a particular address, but only to a port.
Check the port 3000 it may be already used by another application. Try using a different port.
what will happen to the serversocket in my app when I suddenly change the wifi network? I guess it will shut down since my device will get a new IP, at least in TCP, is the UDP MulticastSocket prone to this as well? And how to end the previous Server socket thread and start a new one when the network changes? One solution is using time outs, another is using a flag that will indicate whether the infinite loop should end or not but since listening to a socket is a blocking function it will produce an exception/error anyways.
Any thoughts will be appreciated! :)
EDIT: sample of my server thread.
ServerSocket ss = new ServerSocket(4445);
while(true){
Socket socket = ss.accept();
ObjectInputStream in = new ObjectInputStream(socket.getInputStream());
Object obj = in.readObject();
Log.i("TAG", "Received: " + obj.toString());
in.close();
socket.close();
}
TCPIP connection will break. So client would have to connect again.
UDP will be ok provided your IP does not change after reconnection. Of course if you transmit UDP its not going to make a difference for that machine.
You should get an exception in case of TCPIP which you can handle.
UDP sockets that are not bound to the address will remain open, as they are stateless. TCP listening sockets not bound to the address will remain open as well.
Conntected TCP sockets may be severed (RST) or just linger until a timeout hits.
It is a little known fact that IP mandates it that a device by default will accept packets directed to any address it has configured on any interface, no matter on which interface the packet arrives. If this were not so, routing would be broken. One can use packet filters to filter out packets with non-matching addresses depending on the interface.