Requisites:
I should open multiple sockets/connections to the same server IP and
Port.
I should detect from which connection a request has come and
reroute the response to the same connection
A connection is represented as server Ip and port + client Ip and
port
Each connection has to be single-use=false, it is expected to have multiple request/replies
I'm using collaborating TcpReceivingChannelAdapter and TcpSendingMessageHandler with TcpNetClientConnectionFactory.
And IntegrationFlow for generating those connections dynamically.
How to create multiple sockets for TcpNetClientConnectionFactory that all point to the same Host and port?
I know how to set destination with new TcpNetClientConnectionFactory(host, port). But I'm unable to see or find how to affect which port is used
Should it be done with multiple TcpNetClientConnectionFactory each being bound to one inbound and outbound TCP adapter?
How can I set the local port for those connections? or at least how to obtain it?
I don't seem to find any documentantion about this option. The most similar would be This question
You need a separate connection factory/adapters for each.
See TCP Connection Events.
Use an ApplicationListener or #EventListener to receive TcpConnectionEvents.
The event has getConnectionId() which contains both the local and remote port; the event also has the connection factory bean name.
Or you can cast getSource() to TcpConnection and call getPort() (but you should not otherwise interact with the TcpConnection object.
This question already has answers here:
Does the port change when a server accepts a TCP connection?
(3 answers)
Closed 4 years ago.
I understand the basics of how ports work. However, what I don't get is how multiple clients can simultaneously connect to say port 80. I know each client has a unique (for their machine) port. Does the server reply back from an available port to the client, and simply state the reply came from 80? How does this work?
First off, a "port" is just a number. All a "connection to a port" really represents is a packet which has that number specified in its "destination port" header field.
Now, there are two answers to your question, one for stateful protocols and one for stateless protocols.
For a stateless protocol (ie UDP), there is no problem because "connections" don't exist - multiple people can send packets to the same port, and their packets will arrive in whatever sequence. Nobody is ever in the "connected" state.
For a stateful protocol (like TCP), a connection is identified by a 4-tuple consisting of source and destination ports and source and destination IP addresses. So, if two different machines connect to the same port on a third machine, there are two distinct connections because the source IPs differ. If the same machine (or two behind NAT or otherwise sharing the same IP address) connects twice to a single remote end, the connections are differentiated by source port (which is generally a random high-numbered port).
Simply, if I connect to the same web server twice from my client, the two connections will have different source ports from my perspective and destination ports from the web server's. So there is no ambiguity, even though both connections have the same source and destination IP addresses.
Ports are a way to multiplex IP addresses so that different applications can listen on the same IP address/protocol pair. Unless an application defines its own higher-level protocol, there is no way to multiplex a port. If two connections using the same protocol simultaneously have identical source and destination IPs and identical source and destination ports, they must be the same connection.
Important:
I'm sorry to say that the response from "Borealid" is imprecise and somewhat incorrect - firstly there is no relation to statefulness or statelessness to answer this question, and most importantly the definition of the tuple for a socket is incorrect.
First remember below two rules:
Primary key of a socket: A socket is identified by {SRC-IP, SRC-PORT, DEST-IP, DEST-PORT, PROTOCOL} not by {SRC-IP, SRC-PORT, DEST-IP, DEST-PORT} - Protocol is an important part of a socket's definition.
OS Process & Socket mapping: A process can be associated with (can open/can listen to) multiple sockets which might be obvious to many readers.
Example 1: Two clients connecting to same server port means: socket1 {SRC-A, 100, DEST-X,80, TCP} and socket2{SRC-B, 100, DEST-X,80, TCP}. This means host A connects to server X's port 80 and another host B also connects to the same server X to the same port 80. Now, how the server handles these two sockets depends on if the server is single-threaded or multiple-threaded (I'll explain this later). What is important is that one server can listen to multiple sockets simultaneously.
To answer the original question of the post:
Irrespective of stateful or stateless protocols, two clients can connect to the same server port because for each client we can assign a different socket (as the client IP will definitely differ). The same client can also have two sockets connecting to the same server port - since such sockets differ by SRC-PORT. With all fairness, "Borealid" essentially mentioned the same correct answer but the reference to state-less/full was kind of unnecessary/confusing.
To answer the second part of the question on how a server knows which socket to answer. First understand that for a single server process that is listening to the same port, there could be more than one socket (maybe from the same client or from different clients). Now as long as a server knows which request is associated with which socket, it can always respond to the appropriate client using the same socket. Thus a server never needs to open another port in its own node than the original one on which the client initially tried to connect. If any server allocates different server ports after a socket is bound, then in my opinion the server is wasting its resource and it must be needing the client to connect again to the new port assigned.
A bit more for completeness:
Example 2: It's a very interesting question: "can two different processes on a server listen to the same port". If you do not consider protocol as one of the parameters defining sockets then the answer is no. This is so because we can say that in such a case, a single client trying to connect to a server port will not have any mechanism to mention which of the two listening processes the client intends to connect to. This is the same theme asserted by rule (2). However, this is the WRONG answer because 'protocol' is also a part of the socket definition. Thus two processes in the same node can listen to the same port only if they are using different protocols. For example, two unrelated clients (say one is using TCP and another is using UDP) can connect and communicate to the same server node and to the same port but they must be served by two different server processes.
Server Types - single & multiple:
When a server processes listening to a port that means multiple sockets can simultaneously connect and communicate with the same server process. If a server uses only a single child process to serve all the sockets then the server is called single-process/threaded and if the server uses many sub-processes to serve each socket by one sub-process then the server is called a multi-process/threaded server. Note that irrespective of the server's type a server can/should always use the same initial socket to respond back (no need to allocate another server port).
Suggested Books and the rest of the two volumes if you can.
A Note on Parent/Child Process (in response to query/comment of 'Ioan Alexandru Cucu')
Wherever I mentioned any concept in relation to two processes say A and B, consider that they are not related by the parent-child relationship. OS's (especially UNIX) by design allows a child process to inherit all File-descriptors (FD) from parents. Thus all the sockets (in UNIX like OS are also part of FD) that process A listening to can be listened to by many more processes A1, A2, .. as long as they are related by parent-child relation to A. But an independent process B (i.e. having no parent-child relation to A) cannot listen to the same socket. In addition, also note that this rule of disallowing two independent processes to listen to the same socket lies on an OS (or its network libraries), and by far it's obeyed by most OS's. However, one can create own OS which can very well violate this restriction.
TCP / HTTP Listening On Ports: How Can Many Users Share the Same Port
So, what happens when a server listen for incoming connections on a TCP port? For example, let's say you have a web-server on port 80. Let's assume that your computer has the public IP address of 24.14.181.229 and the person that tries to connect to you has IP address 10.1.2.3. This person can connect to you by opening a TCP socket to 24.14.181.229:80. Simple enough.
Intuitively (and wrongly), most people assume that it looks something like this:
Local Computer | Remote Computer
--------------------------------
<local_ip>:80 | <foreign_ip>:80
^^ not actually what happens, but this is the conceptual model a lot of people have in mind.
This is intuitive, because from the standpoint of the client, he has an IP address, and connects to a server at IP:PORT. Since the client connects to port 80, then his port must be 80 too? This is a sensible thing to think, but actually not what happens. If that were to be correct, we could only serve one user per foreign IP address. Once a remote computer connects, then he would hog the port 80 to port 80 connection, and no one else could connect.
Three things must be understood:
1.) On a server, a process is listening on a port. Once it gets a connection, it hands it off to another thread. The communication never hogs the listening port.
2.) Connections are uniquely identified by the OS by the following 5-tuple: (local-IP, local-port, remote-IP, remote-port, protocol). If any element in the tuple is different, then this is a completely independent connection.
3.) When a client connects to a server, it picks a random, unused high-order source port. This way, a single client can have up to ~64k connections to the server for the same destination port.
So, this is really what gets created when a client connects to a server:
Local Computer | Remote Computer | Role
-----------------------------------------------------------
0.0.0.0:80 | <none> | LISTENING
127.0.0.1:80 | 10.1.2.3:<random_port> | ESTABLISHED
Looking at What Actually Happens
First, let's use netstat to see what is happening on this computer. We will use port 500 instead of 80 (because a whole bunch of stuff is happening on port 80 as it is a common port, but functionally it does not make a difference).
netstat -atnp | grep -i ":500 "
As expected, the output is blank. Now let's start a web server:
sudo python3 -m http.server 500
Now, here is the output of running netstat again:
Proto Recv-Q Send-Q Local Address Foreign Address State
tcp 0 0 0.0.0.0:500 0.0.0.0:* LISTEN -
So now there is one process that is actively listening (State: LISTEN) on port 500. The local address is 0.0.0.0, which is code for "listening for all". An easy mistake to make is to listen on address 127.0.0.1, which will only accept connections from the current computer. So this is not a connection, this just means that a process requested to bind() to port IP, and that process is responsible for handling all connections to that port. This hints to the limitation that there can only be one process per computer listening on a port (there are ways to get around that using multiplexing, but this is a much more complicated topic). If a web-server is listening on port 80, it cannot share that port with other web-servers.
So now, let's connect a user to our machine:
quicknet -m tcp -t localhost:500 -p Test payload.
This is a simple script (https://github.com/grokit/dcore/tree/master/apps/quicknet) that opens a TCP socket, sends the payload ("Test payload." in this case), waits a few seconds and disconnects. Doing netstat again while this is happening displays the following:
Proto Recv-Q Send-Q Local Address Foreign Address State
tcp 0 0 0.0.0.0:500 0.0.0.0:* LISTEN -
tcp 0 0 192.168.1.10:500 192.168.1.13:54240 ESTABLISHED -
If you connect with another client and do netstat again, you will see the following:
Proto Recv-Q Send-Q Local Address Foreign Address State
tcp 0 0 0.0.0.0:500 0.0.0.0:* LISTEN -
tcp 0 0 192.168.1.10:500 192.168.1.13:26813 ESTABLISHED -
... that is, the client used another random port for the connection. So there is never confusion between the IP addresses.
Normally, for every connecting client the server forks a child process that communicates with the client (TCP). The parent server hands off to the child process an established socket that communicates back to the client.
When you send the data to a socket from your child server, the TCP stack in the OS creates a packet going back to the client and sets the "from port" to 80.
Multiple clients can connect to the same port (say 80) on the server because on the server side, after creating a socket and binding (setting local IP and port) listen is called on the socket which tells the OS to accept incoming connections.
When a client tries to connect to server on port 80, the accept call is invoked on the server socket. This creates a new socket for the client trying to connect and similarly new sockets will be created for subsequent clients using same port 80.
Words in italics are system calls.
Ref
http://www.scs.stanford.edu/07wi-cs244b/refs/net2.pdf
Ive searched internet but didnt got the answer can any1 explain me the difference between them
A TCP "connection" is a 4-tuple. Local IP, Local Port, Remote IP, and Remote Port. Each end maintains this identification within its TCP stack, with the senses reversed (Local vs. Remote).
The combination of these 4 values must be unique. This explains the problems people often have writing a TCP client that reuses a socket to reconnect to the same server.
A "closed" connection leaves this ID in the tables at each end for some time, in TIME_WAIT state. This is an artifact of a TCP mechansim that deals with maintaining connection integrity even if the physical layer connection breaks, keeps pending packets from being recevied by a second connection, etc. TIME_WAIT can last up to 4 minutes.
Unless the client resets its socket's LocalPort to 0 (which is a request for an automatic ephemeral port assignment) it can fail if it tries to reconnect before TIME_WAIT expires. Since this is 0 for a newly created socket, programmers often overlook this requirement prior to calling Connect.
LocalPort isn't just an issue for listening sockets.
A server listens on a localport, while a client sends data from the localport.
The client remoteport should be the same as the server localport.
i.e.:
Server listens on port n (local port relative to server)
Client connects to server on port n (remote port relative to client)
To answer your question, the difference is in name, based on perspective.
This seems to be a good place to start with VB6 socket communication
I have a server application that binds to a port and listens on it. I've set up the router to forward the data on this port to the server.
Now, on the client side, I don't actually bind() the socket to any port, and I usually end up with a different port everytime. In that case, how can I prepare the router to forward that port to the client? Or am I supposed to use bind() with the client socket as well? (I remember reading that you're not supposed to do that.)
Firewalls are usually stateful - meaning if TCP connection request into the protected network is allowed, then the packets back to the client are matched (and passed through) automatically. That is to say you don't worry about the client, just setup port forwarding to the server app.
In order to receive datagrams through an UDP connection I have created an object of type UDPClient.
receivedNotificationSock = new UdpClient();
However once done and on using the receive method:
receivedHostNameBuffer=receivedNotificationSock.Receive(ref receivedNotificationIP);
I am getting an exception saying that I must call the bind method.
But there is no bind method in the UDPClient class.
Could You guys please provide me with the code if possible as to what should be done to overcome this exception.
You need I think to know some more about sockets.
All sockets possess a port number. First, you create a socket - which is almost useless on its own. It just floats there. But then you bind it - you assign it a port number. Now it's useful - now you can send and receive data on it.
Remember, all UDP communications are defined by the quad data set of the IP and port of the source and the IP and port of the destination. A freshly created socket doesn't have an IP address or port; binding gives it an IP address and port.
Unfortunately, I'm not a C# programmer, so I can't properly answer your question. But at least you know why it's important.
Pass the port number into the constructor of your UDP client.
receivedNotificationSock = new UdpClient(21000);
You may need to change firewall settings to allow the bind, though a popup window normally opens when you first run this on your dev machine.
For Socket proramming you need to know the sequence of syscalls you need to do on client side and on the server side.
If you are writting a client :
you open a socket with a socket call.
you then connect to the server port with a connect call
once connect is successful
then you send the request to the server using either a send or sendto or a write
which results in reception of data that you can read using a receive or read
On Server Side
you create a socket
bind it to a port
start listening on the socket for incoming connections from various clients using a listen.
There is a non blocking way of listening for connections as well with a select syscall.
Once the you establish a connection you can essentially read the request and start processing it.
Here's an example in C# that may be useful to you.
http://www.developerfusion.com/article/3918/socket-programming-in-c-part-1/