I have an application that uses QUdpSocket to send broadcast packets. The machine sending the packets has several network interfaces.
Unfortunately, I haven't been able to figure out how to get QUdpSocket to use the right network interface. Is there an option somewhere that allows me to specify exactly which interface to use for sending the packets?
Every subnet has its own broadcast address. So I think you can send broadcast packets 'directly' to the addresses like 10.255.255.255 or 192.168.255.255.
You are actually choosing which network to listen to in the first argument of bind function, address. If you are choosing for example QHostAddress::AnyIPv4 it means that no matter from which interface (better to say as a result of which IP address) the packet is received, the program should catch it. Otherwise you can enter an IP address (or a broadcast address) to which the packet should be sent.
For debugging and verification of it you can use netstat to list open ports as explained here:
https://superuser.com/questions/529830/get-a-list-of-open-ports-in-linux
In the result of netstat command, Local Address is what you are looking for, specifying which address is being listened to.
NOTE: The address of localhost in Local Address (e.g.: 0.0.0.0 or 127.0.0.1) means any Local Address is acceptable.
If two processes both use a shared local port to connect to the same remote port of the server server, what happens when the server tries to respond to one? Or is there a mechanism to prevent this?
I assume you're asking about TCP here. When the two processes connect to a single remote port, they will be using different local ports. That is how the server distinguishes the connections. A connection has four parts that uniquely identify it: source port, source IP address, destination port, and destination IP address.
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.
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
From what I understand, each HTTP request uses its own TCP connection (please correct me if i'm wrong). So, let's say that there are two current connections to the same server. For example, client side javascript code triggering a couple of AJAX POST requests using the XMLHttpRequest object, one right after the other, before getting the response to the first one. So we're talking about two connections to the same server, each waiting for a response in order to route it to each separate callback function.
Now here's the thing that I don't understand: The TCP packet includes source and destination ip and port, but won't both of these connections have the same src and dest ip addresses, and port 80? How can the packets be differentiated and routed to appropriately? Does it have anything to do with the packet sequence number which is different for each connection?
When your browser creates a new connection to the HTTP server, it uses a different source port.
For example, say your browser creates two connections to a server and that your IP address is 60.12.34.56. The first connection might originate from source port 60123 and the second from 60127. This is embedded in the TCP header of each packet sent to the server. When the server replies to each connection, it uses the appropriate port (e.g. 60123 or 60127) so that the packet makes it back to the right spot.
One of the best ways to learn about this is to download Wireshark and just observe traffic on your own network. It will show you this and much more.
Additionally, this gives insight into how Network Address Translation (NAT) works on a router. You can have many computers share the same IP address and the router will rewrite the request to use a different port so that two computers can simultaneously connect to places like AOL Instant Messenger.
They're differentiated by the source port.
The main reason for each HTTP request to not generate a separate TCP connection is called keepalives, incidentally.
A socket, in packet network communications, is considered to be the combination of 4 elements: server IP, server port, client IP, client port. The second one is usually fixed in a protocol, e.g. http usually listen in port 80, but the client port is a random number usually in the range 1024-65535. This is because the operating system could use those ports for known server protocols (e.g. 21 for FTP, 22 for SSH, etc.). The same network device can not use the same client port to open two different connections even to different servers and if two different clients use the same port, the server can tell them apart by their IP addresses. If a port is being used in a system either to listen for connection or to establish a connection, it can not be used for anything else. That's how the operating system can dispatch packets to the correct process once received by the network card.