This might be a very basic question but it confuses me.
Can two different connected sockets share a port? I'm writing an application server that should be able to handle more than 100k concurrent connections, and we know that the number of ports available on a system is around 60k (16bit). A connected socket is assigned to a new (dedicated) port, so it means that the number of concurrent connections is limited by the number of ports, unless multiple sockets can share the same port. So the 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 in the book "UNIX Network Programming: The sockets networking API" by
W. Richard Stevens, Bill Fenner, Andrew M. Rudoff 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.
I guess none of the answers tells every detail of the process, so here it goes:
Consider an HTTP server:
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)).
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).
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).
The port 80 is then released for the next connection (first, it will accept the first one in the backlog).
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).
After that waiting, that random port is free again to receive other connections.
So, TCP cannot even share a port amongst two IP's!
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.
Absolutely not, because even multiple connections may shave same ports but they'll have different IP addresses
RELATED POST
The post here In UNIX forum describes
The server will keep on listeninig on a port number.
The server will accept a clients connect() request using accept(). As soon as the server accepts the client request, the kernel allocates a random port number for the server for further send() and receive(), since the same port number on the server can't be used for sending as well as listening, and the previous port is still listening for new connections
QUESTION
I have a server application S which is constantly listening on port 18333 (this is actually bitcoind testnet). When another client node C connects with it on say 53446 (random port). According to the above post, S will be able to send/receive data of 'C' only from port 53446.
But when I run a bitcoind testnet. This perfectly communicates with other node with only one socket connection in port 18333 without need for another for sending/receiving. Below is snippet and I even verified this
bitcoin-cli -testnet -rpcport=16591 -datadir=/home/user/mytest/1/
{
"id": 1,
"addr": "178.32.61.149:18333"
}
Can anyone help me understand what is the right working in TCP socket connection?
A TCP connection is identified by a socket pair and this is uniquely identified by 4 parameters :
source ip
source port
dest ip
dest port
For every connection that is established to a server the socket is basically cloned and the same port is being used. So for every connection you have a socket using the same server port. So you have n+1 socket using the same port when there are n connections.
The TCP kernel is able to make distinction between all these sockets and connections because the socket is either in the listening state, or it belongs to the socket pair where all 4 parameters are considered.
Your second bullet is therefore wrong because the same port is being used as i explained above.
The server will accept a clients connect() request using accept(). As
soon as the server accepts the client request, the kernel allocates a
random port number for the server for further send() and receive().
On normal TCP traffic this is not the case. If a webserver is listening on port 80, all packets sent back to the client wil be over server port 80 (this can be verified with WireShark for example) - but there will be a different socket for each connection (srcIP:port - dstIP:port). That information is sent in the headers of the network packets - IP and protocol code (TCP, UDP or other) in the IP header, port numbers as part of the TCP or UDP header).
But changing ports can happen when communicating over ftp, where there can be a control port (ususally 21) and a negotiated data port.
When I open TCP with the server (on 7 layer of OSI), the layer 5 create socket with port number and IP.
I want to know if this socket include my IP/the server IP, and my (random) port or the server port (e.g. 80 for HTTP)
And when I open TCP with server we open TCP together
So it's mean we have common socket?
When I open TCP with the server (on 7 layer of OSI)
Forget about OSI. It is obsolete, and TCP/IP doesn't follow it. It has its own layer model.
The layer 5 create socket with port number and IP
TCP creates it at the TCP layer.
I want to know if this socket include my IP/the server IP, and my (random) port or the server port (80 for HTTP for ex.)
All of the above.
And when I open TCP with server we open TCP together So it's mean we have common socket?
No. A socket is an endpoint of a connection. There are two ends, and two sockets.
TCP is a Layer 4 as it is called - or a Transport Layer, so ignore the OSI model for the time being.
Generally - 'a socket' is just an end point without any identity. The socket gets it's identity when you bind to an address or connect to an address.
When you bind to an address - you only get your local port and local IP address in it's end point, but not the remote IP and port address. As such such socket is not very useful unless you listen on it. This is typically done on the server. Also note that you can bind to 'All Addresses on the machine' and then you really don't have any one end-point per se.
When you connect to a server (a TCP server # port 80 say), your OS TCP/IP stack makes use of a local IP address and chooses a random port to connect to a sever socket (like say one listening above). This is when all the 4 addresses come into picture. This socket is a connected socket and all 4 values will be present.
My question is very simple and straight to the point. In my readings, is it okay to say that the port 80 on a web server is a passive socket that simply creates an active socket for every connection?
Server is a passive listener (listening on a specific port), and will create a new socket when accept a new connection with unreserved port number.
For example web server (TCP server)
Listen on port 80 and accept many connections on different ports (e.g has 5 accepted sockets on port 5001, 5002, 5003, 5004, 5005)
Is it okay to say that the port 80 on a web server is a passive socket
It is a passive port. It is represented inside the web server process by a socket.
that simply creates an active socket for every connection?
Ports don't create sockets. TCP creates sockets. The web server process accepts a connection and this is delivered to the process in the form of an active socket.
I am not sure how to use a single port at server side simultaneously for multiple sockets. How can we do it in node.js. I am currently using socket.io and have one socket per port. In case solutions do not exist but is possible then also please give your suggestion to achieve the same. Also what issues can be there if we share a port? What could be other related options considering the situation that clients can be idle but will consume a port on server as we need to maintain a socket connection for each client?
Assuming your server is running on port 80, here is what happens underneath:
Server listens port 80.
Client1 connects to server port 80 from its port 12345
Server accepts client1's connection request and assigns port 9876 to commune with client1.
Server continues listening port 80.
So despite what you think, the port 80 is not consumed, it is a listener. Your computer probably has 50000 ports at free, so there is no problem.
FYI: Ports cannot be shared among other processes. Only Node's child processes can be shared, have a look at how it can be: http://nodejs.org/docs/latest/api/cluster.html