TCP: can two different sockets share a port?

Question

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 ip addresses". An easy mistake to make is to only listen on port 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/quickweb) 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.

A server socket listens on a single port. All established client connections on that server are associated with that same listening port on the server side of the connection. An established connection is uniquely identified by the combination of client-side and server-side IP/Port pairs. Multiple connections on the same server can share the same server-side IP/Port pair as long as they are associated with different client-side IP/Port pairs, and the server would be able to handle as many clients as available system resources allow it to.

On the client-side, it is common practice for new outbound connections to use a random client-side port, in which case it is possible to run out of available ports if you make a lot of connections in a short amount of time.

A connected socket is assigned to a new (dedicated) port

That's a common intuition, but it's incorrect. A connected socket is not assigned to a new/dedicated port. The only actual constraint that the TCP stack must satisfy is that the tuple of (local_address, local_port, remote_address, remote_port) must be unique for each socket connection. Thus the server can have many TCP sockets using the same local port, as long as each of the sockets on the port is connected to a different remote location.

See the "Socket Pair" paragraph at: http://books.google.com/books?id=ptSC4LpwGA0C&lpg=PA52&dq=socket%20pair%20tuple&pg=PA52#v=onepage&q=socket%20pair%20tuple&f=false

Theoretically, yes. Practice, not. Most kernels (incl. linux) doesn't allow you a second bind() to an already allocated port. It weren't a really big patch to make this allowed.

Conceptionally, we should differentiate between socket and port. Sockets are bidirectional communication endpoints, i.e. "things" where we can send and receive bytes. It is a conceptional thing, there is no such field in a packet header named "socket".

Port is an identifier which is capable to identify a socket. In case of the TCP, a port is a 16 bit integer, but there are other protocols as well (for example, on unix sockets, a "port" is essentially a string).

The main problem is the following: if an incoming packet arrives, the kernel can identify its socket by its destination port number. It is a most common way, but it is not the only possibility:

• Sockets can be identified by the destination IP of the incoming packets. This is the case, for example, if we have a server using two IPs simultanously. Then we can run, for example, different webservers on the same ports, but on the different IPs.
• Sockets can be identified by their source port and ip as well. This is the case in many load balancing configurations.

Because you are working on an application server, it will be able to do that.

No. It is not possible to share the same port at a particular instant. But you can make your application such a way that it will make the port access at different instant.

I guess none of the answers tells every detail of the process, so here it goes:

Consider an HTTP server:

1. It asks the OS to bind the port 80 to one or many IP addresses (if you choose 127.0.0.1, only local connections are accepted. You can choose 0.0.0.0 to bind to all IP addresses (localhost, local network, wide area network, both IP versions)).

2. When a client connects to that port, it WILL lock it up for a while (that's why the socket has a backlog: it queues a number of connection attempts, because they ARE NOT instantaneous).

3. The OS then chooses a random port and transfer that connection to that port (think of it as a temporary port that will handle all the traffic from now on).

4. The port 80 is then released for the next connection (first, it will accept the first one in the backlog).

5. When client or server disconnects, the random port is held open for a while (CLOSE_WAIT in the remote side, TIME_WAIT in the local side). That allows flushing some lost packets along the path. The default time for that state is 2 * MSL seconds (and it WILL consume memory while is waiting).

6. After that waiting, that random port is free again to receive other connections.

So, TCP cannot even share a port amongst two IP's!