What's real about SO_REUSEADDR - tcp

In man page about SO_REUSEADDR(man 7 socket):
When the listening socket is bound to INADDR_ANY with a specific port then it is not possible to bind to this port for any local address. Argument is an integer boolean flag.
But in Unix Network Programming, I find:
SO_REUSEADDR allows a new server to be started on the same port as an existing server that is bound to the wildcard address, as long as each instance binds a different local IP address. This is common for a site hosting multiple HTTP servers using the IP alias technique (Section A.4). Assume the local host's primary IP address is 198.69.10.2 but it has two aliases: 198.69.10.128 and 198.69.10.129. Three HTTP servers are started. The first HTTP server would call bind with the wildcard as the local IP address and a local port of 80 (the well-known port for HTTP). The second server would call bind with a local IP address of 198.69.10.128 and a local port of 80. But, this second call to bind fails unless SO_REUSEADDR is set before the call. The third server would bind 198.69.10.129 and port 80. Again, SO_REUSEADDR is required for this final call to succeed.
Aren't they contradictory?

They are not in conflict. The second quote provides a way around the default behaviour specified by the first paragraph.

Related

Port Forward based on condition in Twisted

I am trying to write a port-forwarder in Twisted, that will forward to port 8000 if an IP address is already in the cache & to another port - say 4000 if not. I already have the cache written, but am having issues with where to add logic to the portforwarding with Twisted.
Take this simple example:
class LoggingProxyServer(portforward.ProxyServer):
def dataReceived(self, data):
portforward.ProxyServer.dataReceived(self, data)
class LoggingProxyFactory(portforward.ProxyFactory):
protocol = LoggingProxyServer
What twisted method do I override to add the cache checking?
ProxyServer.connectionMade is responsible for setting up the TCP connection that is outgoing from the proxy process. It uses the host and port attributes of its factory to decide what it's going to use as the destination of that connection attempt.
If you want to vary the behavior of the proxy, that's the code you'll need to override.
You can easily find the IP address of the client which has connected to ProxyServer. The ProxyServer instance has a transport attribute that refers to an ITransport provider (probably an ITCPTransport provider if your proxy is listening for incoming TCP connections).
Transports have methods to tell you the addresses of their two endpoints. getHost tells you the local address and getPeer tells you the remote address.
So, for example, you could write a conditional that had one behavior for all TCP clients with IP addresses starting with a 1 and something else for all TCP clients with other IP addresses:
if self.transport.getPeer().host.startswith("1"):
...
else:
...

Why is the 'foreign address' not IP address of external computer?

I am checking the internet connection of my computer and do not understand a few points.
The following is a result from $ netstat:
I do not understand why for some items in the list have localhost:xxxxx as their foreign address.
since netstat shows the internet connections, shouldn't this show IP addresses of outside my computer?
netstat per default tries to convert the data it finds into host names. It does so by reverse resolving the IP addresses via DNS.
For example, 127.0.0.1 gets replaced with localhost in this case.
The same happens with the ports, but there it doesn't happen via DNS, but via a file like /etc/services which provides a mapping between port numbers and service names.
You can prevent this with the netstat option -n.

IP Address when Multiple NICs are Involved

I am writing a TCP server application using Winsock. The machine on which my server will run and the machine on which the client will run both have 2 NICs. The IP addresses involved are:
Server NIC 1: 192.168.132.14 <-- This is the one I want to bind
Server NIC 2: 192.168.132.15
Client NIC 1: 192.168.132.16
Client NIC 2: 192.168.132.17
QUESTION 1:
In my server application, if I use INADDR_ANY when binding my listening socket, which of my two IP addresses will be used? Would I be correct to assume that there's no telling and that I should just use inet_addr("192.168.132.14") in place of INADDR_ANY?
QUESTION 2:
How can the client control which IP address he uses when connecting to me? Would he simply call bind() before calling connect()? Am I liable to see him as connecting from either address (no telling which one) if he does not?
When calling listen() on a server socket, binding to INADDR_ANY will bind the socket to all available local IPs on the machine. That allows a client to connect to any of the server's IPs. If the server uses inet_addr() instead, that will be the only IP that the server can accept client connections on.
When calling connect() on a client socket, it has to indicate a specific IP that the server is listening on. If the client wants to pick which local IP it binds to for its endpoint of the connection, it can call bind() on itself before calling connect(). If the client does not bind to a specific IP, or it binds to INADDR_ANY, the socket will use the first IP it finds that has an available route to the server IP being connected to.
Once the connection has been established, both parties can call getsockname() and getpeername() on their respective socket endpoints at any time to discover which IPs (and ports) are actually in use for that connection.

Why two HTTP and TCP addresses can use the same port and two IPC addresses cannot use the same named pipe?

What I think of a port is: Whenever a message arrives to a machine, it is copied to a memory area which is mapped to the port specified and the concerned application or service is notified that a message has arrived for it.
If this is true, then what happens if two messages arrive for two different services listening on the same port ? ( either http or tcp )
And why can not two named pipe addresses use the same named pipe ?
TCP identifies "connections" via a tuple of { local ip, local port, remote ip, remote port }. Therefore, since each incoming connection has a different remote ip/port pair, your local machine can distinguish between them.
HTTP uses TCP for its transport. Thus, an HTTP port is a TCP port.
If you've ever had your machine get a new IP address while you had connections open, you'll note that they break the first time they send any data out since the remote host does not recognize the (new) address and sends a RST response.
A pipe has only its name to distinguish it so there is only one "connection" no matter how many writers it has.
Your description is one way to handle incoming messages.
In the case of two web sites listening on the same port, there is one web server listening on that port, which then looks at the http host header to find the correct web site to forward the request to.
The same is true for named pipes, the RPC listener listens on the TCP port, and then finds out that it is a named pipe message and then forwards the message to the right named pipe.

How do multiple clients connect simultaneously to one port, say 80, on a server? [duplicate]

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

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