I ran the command sudo lsof -i -n -P | grep TCP and I was wondering if I could get some more clarification on its output.
Specifically, in this image:
Why do I have an IP:PORT pointing to another IP:PORT and then back at itself with the label 'ESTABLISHED'? I am confused on what this means exactly.
I'm not sure how familiar you are with networking and TCP in general, so I'll try to provide a brief description with a couple of details. From your question, it appears that you're not very familiar with networking internals, so it may be hard to understand some of these concepts, but I hope this helps:
The TCP protocol has various states. Think of it as a state machine. States on the client side include CLOSED, SYN_SENT, ESTABLISHED, FIN_WAIT_1, FIN_WAIT_2 and TIME_WAIT.
Thus, the ESTABLISHED label means that the TCP connection is in the ESTABLISHED state. Being in the established state means that both hosts successfully completed the TCP 3-way handshake (and in doing so, transitioned from SYN_SENT to ESTABLISHED). The transition from CLOSED to SYN_SENT happens when the client side sends the TCP SYN request to the server.
In an established connection, both sides transmit and receive application specific data. Basically, a session is established and a bidirectional stream of bytes flows between the two end systems.
TCP sockets are uniquely identified by the 4-tuple (source-ip, source-port, destination-ip, destination-port). The IP identifies an end system's network interface, and the port number is used to multiplex and demultiplex packet arrival at that network interface (so that the target system knows which service to deliver the packets to). That's the meaning of the IP:PORT fields.
I'm not sure why you have two entries for the same connection. This might be system-dependent, although it's odd (in my system I get only one entry per socket). But sockets are bidirectional, so it may be the case that your system shows you each packet flow direction as a distinct entry. This might also depend on how the system implements sockets.
ESTABLISHED means that the TCP connection has completed the 3-way handshake. (Not sure though whether accept must have been called). See TCP state diagram.
Why do I have an IP:PORT pointing to another IP:PORT and then back at itself
That mean you have two TCP sockets open in your process. Most likely, one listens on port 9092, and another one that connected from port 57633 to that listening socket. Port 57633 belongs to the ephemeral port range, i.e. the range of ports that the OS automatically assigns to the sockets that call connect but did not call bind to assign a specific port.
Related
My company releases a special TCP stack for special purposes and I'm tasked with implementing RFC793 compliant closing sequence. One of the unit tests has a server working on top of the special TCP stack talking to a normal Linux TCP client, and I'm running into some strange behaviour that I'm not sure whether is caused by programming error on my part or is to be expected.
Before my work started, we used to send a RST packet when the user application calls close(). I've implemented the FIN handshake, but I noticed that in the case of simultaneous TCP termination (FIN_WAIT_1 -> CLOSING -> TIME_WAIT on both ends, see the picture), the standard Linux TCP client cannot connect to the same destination address and port again, with connect() returning with EADDRNOTAVAIL, until after TIME_WAIT passes into CLOSED.
Now, the standard Linux client application sets the option SO_REUSEADDR, binds the socket to port 8888 each time, and connects to destination port 6666. My question is, why does bind() succeed and why does connect() fail? I would have thought SO_REUSEADDR could take over a local TIME_WAIT port, which it did, but what does connect() have against talking to the destination-ip:6666 again?
Is my code doing something it shouldn't or is this expected behaviour?
I can confirm no SYN packet for the failed connect() makes it out of the client machine at all. I've attached a screenshot of the FIN handshake for the above session.
Your previous implementation used RST to end the connection. Receipt of an RST packet immediately removes the connection from the active connection table. That's because there is no further possibility of receiving a valid packet on that connection: the peer has just told your system that that session is not valid.
If you do a proper session termination with FIN, on the other hand, there is the last packet problem: how do you know for sure whether the peer actually received the last acknowledgment you sent to their FIN (this is the definition of TCP's TIME_WAIT state)? And if the peer didn't receive it, they might validly send another copy of the FIN packet which your machine should then re-ACK.
Now, your bind succeeds because you're using SO_REUSEADDR, but you still cannot create a new connection with the exact same ports on both sides because that entry is still in your active connection table (in TIME_WAIT state). The 4-tuple (IP1, port1, IP2, port2) must always be unique.
As #EJP suggested in the comment, it is unusual for the client to specify a port, and there is typically no reason to. I would change your test.
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
Reading the following article: 10M concurrent websockets
So, there are 1000 websocket servers listening on ports 10000-11000. When a connection is made to one of these servers, I assume they continue communication from a random established TCP connection with random ports. So, as one IP is used, and there are 64K ports, how can one maintain 10M connections? Are connections identified by IP-Port pairs? Can two different connections from different IPs to same port be established? How does this work under the hood?
When a connection is made to one of these servers, I assume they continue communication from a random established TCP connection with random ports.
Wrong assumption. They communicate with the clients using the same local port number they are listening on.
So, as one IP is used, and there are 64K ports, how can one maintain 10M connections?
Not a problem.
Are connections identified by IP-Port pairs?
Yes.
Can two different connections from different IPs to same port be established?
Yes.
How does this work under the hood?
See above. IP:port pairs. You answered your own question.
Sorry for totally changing my answer.
Linux can easily support millions of open sockets if the machine has enough memory and processing power. The TCP/IP stack allows this because the socket the OS targets for a given TCP packet is determined by the source and destination IP and port tuple.
The server implementing the websocket protocol need only listen to a single TCP socket, often defined by the HTTP or HTTPS port number, but not in this example. As part of standard TCP handshaking, the server OS and application open a unique socket for the TCP connection to the new client when the HTTP request which is a websocket request is received. The websocket package takes care of upgrading the protocol used on this new socket from standard HTTP to websocket.
In the example, a goroutine is started for each websocket socket.
The client side, the side initiating the TCP connections, is limited by the number of ephemeral ports its OS can open for a given destination host and port. Honestly, I don't know if this is a limitation of the client OS or the TCP/IP specification itself.
I think the part you are missing is a TCP connection is actually two pairs of IP:PORT.
One for the server, one for the client.
The listening side of a tcp socket is generally always the same IP/Port pair.
Example: net.Listen("tcp", ":8080") is listening on port 8080 (on all interfaces in this case)
The connecting (client) side is usually uses a single outgoing IP along with a random port.
Example: net.Dial("tcp","server:8080) Selects a random available ephemeral port and then attempts to connect to server:8080.
So, in the above example, that connection is: client.ip:32768 -> server.ip:8080 (where 32768 is the ephemeral port selected)
the two pairs combined make a unique connection.
The server side can take as many connections from a single client as there are available (client side) ports. It can also take as many clients are there are IP addresses.
Think of it as, for one listening socket, you can theoretically have 2^16(ports) * 2^32(ipv4 addrs) connections.
In reality, there are reserved IPs, ports, memory limitations, etc so the number is far smaller.
For exmaple, the ephemeral port range on Linux is 32768 - 61000. Which means I'll start getting errors if I net.Dial("tcp", "server:8080") more than 28232 times as I will have exhausted my ephemeral port range for the given server address. But if the server is listening on 2 separate ports, I can do 28232 to the first port, and another 28232 to the second port.
When you see people do the 10MM connection tests, they have to use multiple client IPs or multiple server IPs/Ports to achieve this (or a combo of both to get 10MM unique client:ip/server:ip pairs)
I'm trying to write a TCP transparent proxy to run on Linux.
I want to, upon receipt of an incoming connection, initiate a corresponding outgoing connection, but only accept (SYN|ACK) the incoming connection if the outgoing connection is successful.
TCP_DEFERRED_ACCEPT doesn't do what I want -- it always sends a SYN|ACK.
The question is: how do I accept TCP connections, but defer the SYN|ACK, with the Linux sockets API?
You can do that with Linux, but not via the socket API. You would use the NFQUEUE target which allows you to redirect some packets to userspace and decide their fate from within your program.
Obiously, you'd still have to parse the packet in userspace, but searching for a few TCP flags should not be that hard and not require a complete TCP stack. And this way Linux still does the whole network job.
In your case, it would seem possible that you both use NFQUEUE and classical sockets API. The first will give you early decisions, the latter TCP stream data access. Although I never tried it.
See https://home.regit.org/netfilter-en/using-nfqueue-and-libnetfilter_queue/ for instance.
How to open a raw socket for sending from specific TCP port? I want to have all my connections always go from a range of ports below ephemerals.
If you are using raw sockets, then just fill in the correct TCP source port in the packet header.
If, instead, you are using the TCP socket interface (socket(), connect() and friends), then you can set the source port by calling the bind() system call for the client socket - exactly as you would to set the listening port for the server socket.
Making a tcp connection using raw sockets is somewhere between difficult and impossible; you'd need to implement the entire tcp protocol in your program AND also stop the kernel from sending its own replies to the packets (if the kernel has IP bound on that address on that interface).
This is probably not what you want. However, if you did want it, it is trivial to send tcp frames with any source port you want, as you get to specify it in the tcp header, which of course, if you're implementing your own TCP layer, you'll need to understand.