Socket TCP/IP Octave - python

I have created a socket TCP/IP server in C:
On the client-side, I used a script Octave to implement socket TCP/IP client
Even though the script has been terminated, but the server is still open, I would like to know what is the element which can keep the connection alive?
I saw that the Octave window is still open which contains the variables also, and I think that is the reason makes the connection is still alive. But I need some explanations about this problem.
I did the same things with a client in Python but the server is closed even I did not close the socket.

Related

Python Socket reconnect after connection failure [duplicate]

Okay, I've read this post in search for the right answer, but it does not seem to serve my purpose.
This Question
Now, getting to the trouble:
I have a conventional client-server architecture in C (all sockets are non-blocking), where the server is listening for incoming connections and the client tries to connect. The first connect succeeds and everything goes on just fine until I press Ctrl + C on my server.
The client side of the code detects that the connection is lost and arms a retry timer.
The client code is supposed to try a reconnect on the server again and again by using the POSIX interval timers on each timer popping. It however, does not close the socket or start out afresh. Now, every time it retries the connection, the connect() returns
Transport endpoint is already connected
Even after restarting the server, which uses the SO_REUSEADDR and successfully starts, the connect does not complete.
One thing that I will need to implement is the signal handler on the server for the shutdown on Ctrl+C.
But still, do I need to close the socket descriptor on the client side and start afresh every time a disconnect happens, or is there a way out of this?
sockets cannot be reused.
Once the connection a socket served has gone down in both directions, the socket is unusable.
close() the client socket on loss of connection and create a new socket for a new connection.
Update (based on the comments below):
In the OP's case one side (the server side) went down (by means of the server process ending). This implies all sockets held by this process are implicitly close()ed and therefore shutdown() in both directions.

What does "Only one usage of each socket address is normally permitted" tell me when using sockets under Python?

I have a socket server under Python:
sock= socket.socket(socket.AF_INET, socket.SOCK_STREAM)
sock.setblocking(0)
sock.bind((self._ipadress, port))
later I'm accepting incoming requests in a loop, using select.select:
connection, client_address = sock.accept()
...
select.select(...)
Note that connections can be closed by clients when they not need it anymore.
I tested my code with a python client and was able to observe that multiple connections can be easily handled simultaneously as expected.
However, very sporadically I get error:
“Only one usage of each socket address is normally permitted”
What does it tell me and when does it happen?
Multiple connections on the same port are definitely possible (I tested it), so why should there be only one usage permitted? This is against the principle, that multiple clients can be accepted by the same server.
I learned from
Python server "Only one usage of each socket address is normally permitted"
that it can be avoided by using SO_REUSEADDR.
But why is it required, since even after closing a connection by a client, the socket should still be able to accept other connections. Otherwise my program wouldn't work at all.
I'm at home now and not in the office, so I cannot test it, but I have even problems to understand the principles behind...
.

Keeping python sockets alive in event of connection loss

I'm trying to make a socket connection that will stay alive so that in event of connection loss. So basically I want to keep the server always open (also the client preferably) and restart the client after the connection is lost. But if one end shuts down both ends shut down. I simulated this by having both ends on the same computer "localhost" and just clicking the X button. Could this be the source of my problems?
Anyway my connection code
m.connect(("localhost", 5000))
is in a if and try and while e.g.
while True:
if tryconnection:
#Error handeling
try:
m.connect(("localhost", 5000))
init = True
tryconnection = False
except socket.error:
init = False
tryconnection = True
And at the end of my code I just a m.send("example") when I press a button and if that returns an error the code of trying to connect to "localhost" starts again. And the server is a pretty generic server setup with a while loop around the x.accept(). So how do keep them both alive when the connection closes so they can reconnect when it opens again. Or is my code alright and its just by simulating on the same computer is messing with it?
I'm assuming we're dealing with TCP here since you use the word "connection".
It all depend by what you mean by "connection loss".
If by connection loss you mean that the data exchanges between the server and the client may be suspended/irresponsive (important: I did not say "closed" here) for a long among of time, seconds or minutes, then there's not much you can do about it and it's fine like that because the TCP protocol have been carefully designed to handle such situations gracefully. The timeout before deciding one or the other side is definitely down, give up, and close the connection is veeeery long (minutes). Example of such situation: the client is your smartphone, connected to some server on the web, and you enter a long tunnel.
But when you say: "But if one end shuts down both ends shut down. I simulated this by having both ends on the same computer localhost and just clicking the X button", what you are doing is actually closing the connections.
If you abruptly terminate the server: the TCP/IP implementation of your operating system will know that there's not any more a process listening on port 5000, and will cleanly close all connections to that port. In doing so a few TCP segments exchange will occur with the client(s) side (it's a TCP 4-way tear down or a reset), and all clients will be disconected. It is important to understand that this is done at the TCP/IP implementation level, that's to say your operating system.
If you abruptly terminate a client, accordingly, the TCP/IP implementation of your operating system will cleanly close the connection from it's port Y to your server port 5000.
In both cases/side, at the network level, that would be the same as if you explicitly (not abruptly) closed the connection in your code.
...and once closed, there's no way you can possibly re-establish those connections as they were before. You have to establish new connections.
If you want to establish these new connections and get the application logic to the state it was before, now that's another topic. TCP alone can't help you here. You need a higher level protocol, maybe your own, to implement stateful client/server application.
The issue is not related to the programming language, in this case python. The oeprating system (Windows or linux), has the final word regarding the resilience degree of the socket.

udp socket stops receiving data

I am learning network programing in python and I'm trying to write a Toy vpn forked from android sdk https://github.com/android/platform_development/tree/master/samples/ToyVpn.
My Toy vpn is https://github.com/325862401/ToyVPN.
It's only for Linux.
My home network is behind NAT.
I can use this vpn to surf the internet after connect to remote sever.
But about half an hour or some time later the client udp socket stops receiving any data but the server can receive and send normally.
At this point I must terminate my client and run ToyVpnClient again.
It works normal for some time until it stop receiving again.
Please help me check the client logs.
>2013-08-24 11:42:38 INFO receive data from the tunnel timeout`
you can see that when problem happens, the socket always sends, not receive.
> means send, < means receive
I want to know why the udp socket stops receiving data.
Is there any debug method to find the cause?
For now I've just used logging to debug my program.
Since you're trying your client on the Internet, there is the whole universe of possible causes represented by all the Internet newtwork.
There's not a simple way of debugging here. Possible causes could be of course a software error but also some intermediate network configurations between you and the remote server.
You should capture the udp traffic using the good wireshark or the commandline tcpdump between you and the server and check if you're stopping sending packets or if the server is stopping receiving them.
If you send packets but your server doesn't receive them ( tcpdump on the server ) then there is something on the network which decides to filter your packets. And if it's not on the server (firewall rules to rate limit packets for example or something like that) then there's nothing you can do to that without modifying the logic of your program. Like changing UDP port every X seconds or using a persistent tcp connection.
A udp socket is not stable and may become null once a scanning or other event occupy your network interface for a while (especially true on Android). Using tcp avoids this problem. If you wants to maintain a stable udp, keep monitoring the status of your udp socket; if it becomes null or any unusual things happens, delete this socket and create a new one. Put this reactivating staff in a loop so that your udp socket is always alive.

How to close a socket left open by a killed program?

I have a Python application which opens a simple TCP socket to communicate with another Python application on a separate host. Sometimes the program will either error or I will directly kill it, and in either case the socket may be left open for some unknown time.
The next time I go to run the program I get this error:
socket.error: [Errno 98] Address already in use
Now the program always tries to use the same port, so it appears as though it is still open. I checked and am quite sure the program isn't running in the background and yet my address is still in use.
SO, how can I manually (or otherwise) close a socket/address so that my program can immediately re-use it?
Update
Based on Mike's answer I checked out the socket(7) page and looked at SO_REUSEADDR:
SO_REUSEADDR
Indicates that the rules used in validating addresses supplied in a bind(2) call should
allow reuse of local addresses. For AF_INET sockets this means that a socket may bind,
except when there is an active listening socket bound to the address. When the listen‐
ing 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.
Assume your socket is named s... you need to set socket.SO_REUSEADDR on the server's socket before binding to an interface... this will allow you to immediately restart a TCP server...
s.setsockopt(socket.SOL_SOCKET, socket.SO_REUSEADDR, 1)
s.bind((ADDR, PORT))
You might want to try using Twisted for your networking. Mike gave the correct low-level answer, SO_REUSEADDR, but he didn't mention that this isn't a very good option to set on Windows. This is the sort of thing that Twisted takes care of for you automatically. There are many, many other examples of this kind of boring low-level detail that you have to pay attention to when using the socket module directly but which you can forget about if you use a higher level library like Twisted.
You are confusing sockets, connections, and ports. Sockets are endpoints of connections, which in turn are 5-tuples {protocol, local-ip, local-port, remote-ip, remote-port}. The killed program's socket has been closed by the OS, and ditto the connection. The only relic of the connection is the peer's socket and the corresponding port at the peer host. So what you should really be asking about is how to reuse the local port. To which the answer is SO_REUSEADDR as per the other answers.

Categories

Resources