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.

My confusion is that at server side, there is a socket which is bound to a UDP port and "continuously" listening for any UDP client request, but this is not true at client side, UDP client will open a socket to send the request to UDP server and then that's it, I think it cannot keep that port hanging for UDP server to respond, and if that port closes then how client will receive the response back.

I agree. This is your confusion. Why do you think can't keep the socket open and do a receive on it? It can.

I mean ofcourse, UDP server's response will reach back the UDP client because IP address is there, but once that response has reached UDP module of the client, even though there will be a port but how UDP module can send it to the client who originally sent the request because it would have closed the socket bound to that port?

Why?

Or it will not?

Not.

The client:

• creates a socket
• sends a datagram
• calls recvfrom() or friends to receive the response.

Of course if the client isn't interested in the response it can close the socket, but this is not the normal case.

I am looking for answer which clearly describes the UDP communication (I am not interested in contrasting it with TCP or explaining TCP since I have already fair understanding of TCP), especially how the response will reach back the UDP client.

So don't tag your question with the tcp tag.