I have some troubles understanding send (2) syscall on my linux x86 box.
Consider I established an SSH connection in my app with the other host in LAN. Then I put down the network (e.g. unplug the cable) and call the function (from my app) that sends some SSH packets trough the connection. This function inside calls send like
w = send(s->fd_out,buffer, len, 0);
In debugger I found that send returns len (i.e. w == len after the call).
How this can be if network is unreachable? When I call netstat it says my SSH connection is in state ESTABLISHED even though the network is down.
Can't understand why send executes normally and don't return any error (like EPIPE or ECONNRESET). May be an SSH connection lives some time after the network put down?
Thanks to all.
It's due to the implementation of TCP (and ssh uses TCP). Your send() just writes to a socket, which is just a file descriptor, and return means this operation is successful. It doesn't mean the data has been sent. A file descriptor is just some pointer with state for kernel after all. It's implemented in the kernel to keep TCP state a bit longer before failing a session. In fact, kernel is allowed to indefinitely keep this session until you explicitly call close() or kill your process. So your data is actually buffered in kernel space for network card to deliver it later.
Here is a quick experiment you can do:
Write a server that keeps receiving messages after establishing a connection
socket();
bind();
listen();
while (1) {
accept();
recv();
}
Write a client establishes a connection, takes cin inputs, and send a message to server whenever you hit return.
socket();
connect();
while (1) {
getline();
send();
}
Be careful that you NEVER call close() in while loop on either side. Now, if you unplug your cable AFTER you've established a connection, send a message, reconnect again, and send another message, you will find both messages on the server side.
What you will NEVER observe is that you receive the second message before the first one. You either lose them all, or receive them in order.
Now let me explain why it behaves like this. This is the state diagram of a TCP session.
https://dl.dropbox.com/u/17011409/TCP_State.png
You can see clearly that until you explicitly call close(), the connection will always be in established state. That's expected behavior of TCP. Establishing TCP connection is expensive, and keeping a session alive is good for performance. (That's partially how those TCP DOS works. Attackers keep establishing connections until server runs out of resources to keep TCP state information.)
In this state, your send() will be delegated to kernel for actual sending. TCP guarantees in-order, reliable delivery, but network can lose packets at any time. So TCP HAVE TO buffer your packets, and keep trying. There are algorithms to throttle this retry, but it's buffered for quite a very long time before it declares failure. The default time out to assume a packet loss is 3 seconds in Linux. But after a loss, TCP will retry. Then try again after certain seconds. The fact you unplugged your cable is just the same situation as a packet loss along the way to the destination. Once you plug in your cable again, a retry succeeds, and TCP will start sending remaining messages in order.
I know I must have failed to explain it thoroughly. You really need to know the details of TCP to reason about this behavior. It's required for the properties TCP is giving you. And it's not acceptable to expose internal implementations to programmer. (How about a send call that sometimes returns within milliseconds, and sometimes returns after 10 seconds? I bet no one will want this performance bomb in their code. The point of having a TCP library is exactly to hide this ugly nature of networks.) In fact, you even need to understand multiple RFCs and algorithms of how TCP realize in-order reliable delivery over a lossy network. Congestion control comes into the play of how long the buffer will be there as well. Wikipedia is a good starting point, but it's a full semester's undergraduate course if you really want to understand the details.
With a zero flags argument, send() is equivalent to write(2). And it will write your data on file descriptor (stores in kernel space to deliver).
You have to use other types of flag: MSG_CONFIRM may help you.
Related
My Linux application needs to receive a single UDP flow with modestly-sized packets (~1 KB) at a rate on the order of ~600,000 packets per second. My current implementation is naive: it has a single thread that simply calls recv() repeatedly, placing the received data in a queue to be processed by another thread. Therefore, the receiver thread is only in charge of pulling in the packets.
In some initial testing that I've done, I'm only able to receive between 200,000-300,000 packets per second before the thread reaches full utilization of its CPU core. This obviously isn't good enough to meet the goal of ~600,000 packets per second.
Ideally, I would find some way of distributing the packet reception load across multiple threads. In looking for a solution to the problem, I came across the SO_REUSEPORT socket option, which allows multiple TCP/UDP threads to be bound to the same IP/port combination. At first, this seemed to be exactly what I wanted.
However, the article also points out this detail:
Incoming connections and datagrams are distributed to the server sockets using a hash based on the 4-tuple of the connection—that is, the peer IP address and port plus the local IP address and port. This means, for example, that if a client uses the same socket to send a series of datagrams to the server port, then those datagrams will all be directed to the same receiving server (as long as it continues to exist). This eases the task of conducting stateful conversations between the client and server.
Therefore, if I only have a single UDP flow, the above hashing implementation would yield all of the packets being directed to the same receiver thread, thwarting my attempt at parallelizing the work. Therefore, the question is: is there a way to receive a single flow of UDP packets from multiple threads, using SO_REUSEPORT or some other mechanism?
Note that my application can handle reordering of packets; the protocol that the datagrams are formatted with contains sequencing information that I can use to reorder them properly afterward.
If you didn't find the solution for last 3 years take a look at SO_ATTACH_REUSEPORT_CBPF. We had exactly the same issue and we solved it by attaching simple BPF program which distributes datagrams randomly mod n.
I'm working on an application where I need to ensure that even if the network goes down, messages will still arrive at their destination reliably, in-order, and unmodified. I've been using TCP, and up until now, I was just using a strategy of:
If a send/receive fails, do it again until no error.
If the remote disconnects, wait until the next connection and replace the socket I was send/receiving from with this new one (achieved through some threading and blocking to ensure it's swapped cleanly).
I recently realised that this doesn't work, as send can't report errors indicating that the remote hasn't received the message (cite eg. here).
I did also learn that TCP connections can survive brief network outages, as the kernel buffers the packets until the connection is declared dead after the timeout period (cite.
here).
The question: Is it a feasible strategy to just crank the timeout period waaaay higher on both client/server side (using setsockopt and the SO_KEEPALIVE options), so that a connection "never times out"? I'd have to handle errors related to the kernel's buffer filling up, but that should be relatively simple.
Are there any other failure cases?
If both ends doesn't explicitly disconnect, the tcp connection will stay open forever even if you unplug the cable. There is no timeout in TCP.
However, I would use (or design) an application protocol on top of tcp, making it possible to resume data transmission after re-connects. You may use HTTP for example.
That would be much more stable because depending on buffers would, as you say, at some time exhaust the buffers but the buffers would also being lost on let's say a power outage.
I'll post as much as possible here and I'll add more info upon your requests in the comments
I have a mule flow that takes SNMP packets over UDP (Inbound UDP Endpoint) and then passes the message to a transformer that transforms the byte array (The packet) into an Trap object then if this trap of some specific kind I'll just log it and ignore it, otherwise, it will be logged and inserted into DB in addition to some updates (The figure below illustrates the my flow).
The application will be listening on port 17985 for coming SNMP traps from an agent, now if the agent have lot of traps the UDP socket's Recv-Q will be like in the figure below and the UDP endpoint will stop to log any events (Traps to DB nor Log messages to log file)
What did I do?
I tried to increase endpoint buffer size to be 10MB.
I tried to increase system buffer size to 25MB but no matter how much is it, it will be filled up as soon as the agent starts to get crazy
Additional Info
The agent may sometimes send up to 400~600 traps per second.
Trap's packet size is ~1500 bytes max.
The database is very fast, no more optimization is needed there I think.
I know that if the remote host gracefully shuts down a connection, epoll will report EPOLLIN, and calling read or recv will not block, and will return 0 bytes (i.e. end of stream).
However, if the connection is not closed gracefully, and a write or send operation fails, does this cause epoll to subsequently return EPOLLIN for that socket, producing the same/similar end of stream scenario?
I've tried to find documentation on this behaviour, but have not succeeded, and while I could test it, I'm not interested in what happens on a specific distribution with a specific kernel version.
It is indeed not entirely obvious from the specification, but it works as follows for poll():
If there is data available to be read, even if the connection is closed, POLLIN is returned.
If neither reading or writing is possible because of a closed connection, POLLHUP or POLLERR is returned.
If reading is no longer possible but writing is (such as if the other side did shutdown(SHUT_WR)), POLLIN is returned and POLLHUP and POLLERR are not returned. (This allows waiting for POLLOUT normally.)
The simple thing to do is to try a read when any of POLLIN, POLLHUP and POLLERR are set.
In kqueue(), there is just an EVFILT_READ filter that may be triggered. This is described in the man page and should be clear enough.
Note that if you don't enable TCP keepalives (FreeBSD enables them by default but most other operating systems do not), waiting for data to read may get stuck forever if the network breaks in certain ways. Even if TCP keepalives are on, it tends to take a few hours to detect a broken connection.
It may not return EPOLLIN when the peer machine is closed unexpectly. In the past, I encounted this kind of phenomenon by VirtualBox as following steps:
Launch server on one VM.
Launch client on the other VM, connect the server and keep the connection without doing anything.
Save client VM state (something like hibernate).
And I saw the connection was still established in Server VM by
netstat -anp --tcp
In other words, EPOLLIN was not triggered in server.
http://tldp.org/HOWTO/html_single/TCP-Keepalive-HOWTO/ says that it will keep about 7200 seconds by default.
Of course, you can change keep alive timeout value by setsockopt or kernel parameters.
But some books says the better solution is to detect it in application layer, e.g. design the protocol that make sure sending some dummy messages periodically to detect the connection state.
epoll() is basically poll() but it scales better when you increase number of fds. I am not sure what it does when you are using it as edge-triggered interface. But for level triggered - yes, it will always return EPOLLIN, provided you are listening to this event, if end of stream is detected.
Though you must know TCP is not perfect. If connection is terminated abnormally (physycal link is down) by the other side, your side may never detect this until you write to the socket. TCP_KEEPALIVE may help, but not much.
However, if the connection is not closed gracefully, and a write or send operation fails, does this cause epoll to subsequently return EPOLLIN for that socket, producing the same/similar end of stream scenario?
No. That would imply receipt of a FIN, which means normal termination of the connection, which didn't happen. I would expect you would get an EPOLLERR or maybe EPOLLHUP.
But I'm curious why you wouldn't have already closed the socket on getting the write error, and why you would still be polling it. That's not correct behaviour.
I'm trying to get an HTTP server I'm writing on to behave well when under heavy load, but I'm getting some weird behavior that I cannot quite understand.
My testing consists of using ab (the Apache benchmark program) over the loopback interface at a concurrency level of 1000 (ab -n 50000 -c 1000 http://localhost:8080/apa), while straceing the server process. Strace both slows processing down well enough for the problem to be readily reproducible and allows me to debug the server internals post completion to some extent. I also capture the network traffic with tcpdump while the test is running.
What happens is that ab stops running a while into the test, complaining that a connection returned ECONNRESET, which I find a bit weird. I could easily buy into a connection timing out since the server might simply not have the bandwidth to process them all, but shouldn't that reasonably return ETIMEDOUT or even ECONNREFUSED if not all connections can be accepted?
I used Wireshark to extract the packets constituting the first connection to return ECONNRESET, and its brief packet list looks like this:
(The entire tcpdump file of this connection is available here.)
As you can see from this dump, the connection is accepted (after a few SYN retransmissions), and then the request is retransmitted a few times, and then the server resets the connection. I'm wondering, what could cause this to happen? Normally, Linux' TCP implementation ACKs data before the reading process even chooses to receive it so long as their is space in the TCP window, so why doesn't it do that here? Are there some kind of shared buffers that are running out? Most importantly, why is the kernel responding with a RST packet all of a sudden instead of simply waiting and letting the client re-transmit further?
For the record, the strace of the process indicates that it never even accepts a connection from the port in this connection (port 56946), so this seems to be something Linux does on its own. It is also worth noting that the server works perfectly well as long as ab's concurrency level is low enough (it works perfectly well up to about 100, and then starts failing intermittently somewhere between 100-500), and that its request throughput is rather constant regardless of the concurrency level (it processes somewhere between 6000-7000 requests per second as long as it isn't being straced). I have not found any particular correlation between the frequency of the problem occurring and my backlog setting to listen() (I'm currently using 128, but I've tried up to 1024 without it seeming to make a difference).
In case it matters, I'm running Linux 3.2.0 on this AMD64 box.
The backlog queue filled up: hence the SYN retransmissions.
Then a slot became available: hence the SYN/ACK.
Then the GET was sent, followed by four retransmissions, which I can't account for.
Then the server gave up and reset the connection.
I suspect you have a concurrency or throughput problem in your server which is preventing you from accepting connections rapidly enough. You should have a thread that is dedicated to doing nothing else but calling accept() and either starting another thread to handle the accepted socket or else queueing a job to handle it to a thread pool. I would then speculate that Linux resets connections on connections which are in the backlog queue and which are receiving I/O retries, but that's only a guess.