I want to use a real time TCP connection, I have a streaming of data from server , and I receive it by a client, but this client is too slow to receive as fast as the sender is, so the server buffer the data until it's reach the destination, for example if I "produce" data at time t, and suppose that the client are 10 time slower, then the data produced at time t, will arrive at time 10t.
I want to make the server "drop" the data that can't reach the client at the present time, and send the new data which is expected to arrive at the time?
B.S : I know that UDP protocol do this, but I want to do this by TCP.
I've done this sort of thing in the past, and got reasonably good results. Here's how I did it:
1) On the sending side, use setsockopt(SOL_SOCKET, SO_SNDBUF) to make the server's TCP socket's send buffer as small as you can get away with (since you can't drop data once it's already in the socket's send buffer, you want to keep as little data there as possible)
2) On the sending side, never proactively send() any outgoing data into the socket at all. Instead, write a function (we'll call it DumpCurrentStateToBuffer()) that writes the "current state" bytes (that you want to send to the client) into an in-memory buffer.
3) When the client's socket select()'s (or poll()'s, or whatever mechanism you use) as ready-for-write, call DumpCurrentStateToBuffer() to create a memory-buffer of bytes that are to be sent to the client. Now send that data to the client (if you're using blocking I/O you can do it synchronously, at the cost of potentially stalling your server until the data can be sent; OTOH if you're using non-blocking I/O, you may need to keep the memory-buffer and your current sent-bytes index into the buffer around as state variables, so you can keep sending more sub-chunks of the memory buffer over time, whenever the socket indicates that it can receive more bytes)
4) Once the memory-buffer's contents have been fully sent, you can free the memory buffer, and then wait for the socket to select as ready-for-write again; when it does, goto (3).
This technique doesn't solve all of TCP's non-real-time issues; for example, a dropped TCP packet will still have to be resent to the client. What it does do is guarantee that the client-to-server data backlog will never be more than one or two "states" long, because you never generate any new data unless/until there is at least some room in the socket's output buffer.
Related
I saw the socket option TCP_NODELAY, which is used to turn on or off the Nagle alorithm.
I checked what the Nagle algorithm is, and it seems similar to 'stop and wait'.
Can someone give me a clear difference between these two concepts?
In a stop and wait protocol, one
sends a message to the peer
waits for an ack for that message
sends the next message
(i.e. one cannot send a new message until the previous one has been acknowledged)
Nagle's algorithem as used in TCP is orthoginal to this concept. When the TCP application sends some data, the protocol buffers the data and waits a little while to see if there's more data to be sent instead of sending data to the peer immediately.
If the application has more data to send in this small timeframe, the protocol stack merges that data into the current buffer and can send it as one large message.
This concept could very well be applied to a stop and go protocol as well. (Note that TCP is not a stop and wait protocol)
The Nagle Algorithm is used to control whether the socket provider sends outgoing data immediately as-is at the cost of less efficient network transmissions (off), or if it buffers outgoing data so it can make more efficient network transmissions at the cost of speed (on).
Stop and Wait is a mechanism used to ensure the integrity of transmitted data, by making the sender send a frame of data and then wait for an acknowledgement from the receiver before sending another frame, thus ensuring frames are received in the same order in which they are sent.
These two features operate independently of each other.
I have an application whose main purpose is to transform a RTP stream into an HTTP stream. One thread is receiving RTP packets and write them into a circular buffer and another thread acts as a mini webserver and answers HTTP request by reading from that buffer (only one GET request can happen at a time).
This HTTP thread, once the GET has been received is a simple loop that call send() whenever there is something in the circular buffer. But sometimes, the send() blocks for an insane amount of time (like >1s), creating audio dropout.
To be clear, RTP packets arrive in due real time, no over or underflow here. The HTTP socket is, on purpose, blocking as it is expected that the receiver regulates its flow using TCP when it does not need audio (enough on its own buffers). But the HTTP client is not overwhelmed by audio as the RTP source is, again, just doing realtime.
But obviously, something else happens and I've observed that on Linux, MacOS and Windows (the code works on all these) and on two different network topologies.
I'm wondering if the send() long blocks are not due to something else than the TCP flow control, like something I'm missing with what happens when a thread blocks in a send()
Get a wireshark trace so you can see where the TCP stall is happening. I suspect what is happening is any of the following:
You're actually sending faster than client is consuming. I think you've already ruled that out...
The more likely case is that an IP packet is getting lost and TCP is stuck waiting for the ACK, times out, and then retransmits. Meanwhile your sending thread is trying to stuff more data into the socket and it's getting backed up and eventually blocks.
One simple things you can do is to try increasing the send buffer (SO_SNDBUF) on the socket you send with. This value specifies how many untransmitted bytes that the app can write to the socket before blocking. And if possible, increase the receive buffer (SO_RCVBUF) on the client side. That way, if the network takes a burp for a couple of seconds, your socket will take longer to fill up before blocking.
int size = 512*1024;
setsockopt(sock, SOL_SOCKET, SO_SNDBUF, &size, sizeof(size));
It is my understanding that a write to a TCP/IP socket will be atomic if the amount of data written is small. By atomic, I mean that the receiver will receive all of the data or none of the data. However, it is not atomic, if the amount of the data written is large. Am I correct? and if so, what counts as large?
Thanks,
Bob
No. TCP is a byte-stream protocol. No messages, no datagram-like behaviour.
For UDP, that is true, because all data written by the app is sent out in one UDP datagram.
For TCP, that is not true, unless the application sends only 1 byte of data at a time. A write to a TCP socket will write all of the data to a buffer that is associated with that socket. TCP will then read data from that buffer in the background and send it to the receiver. How much data TCP actually sends in one TCP segment depends on variables of its flow control mechanisms, and other factors, including:
Receive Window published by the other node (receiver)
Amount of data sent in previous segments in flight that are not acknowledged yet
Slow start and congestion avoidance algorithm state
Negotiated maximum segment size (MSS)
In TCP, you can never assume what the application writes to a socket is actually received in one read by the receiver. Data in the socket's buffer can be sent to the receiver in one or many TCP segments. At any moment when data is made available, the receiver can perform a socket read and return with whatever data is actually available at that moment.
Of course, all sent data will eventually reach the receiver, if there is no failure in the middle preventing that, and if the receiver does not close the connection or stop reading before the data arrives.
how can I transfer large data without splitting. Am using tcp socket. Its for a game. I cant use udp and there might be 1200 values in an array. Am sending array in json format. But the server receiving it like splitted.
Also is there any option to send http request like tcp? I need the response in order. Also it should be faster.
Thanks,
You can't.
HTTP may chunk it
TCP will segment it
IP will packetize it
routers will fragment it ...
and TCP will reassemble it all at the other end.
There isn't a problem here to solve.
You do not have much control over splitting packets/datagrams. The network decides about this.
In the case of IP, you have the DF (don't fragment) flag, but I doubt it will be of much help here. If you are communicating over Ethernet, then 1200 element array may not fit into an Ethernet frame (payload size is up to the MTU of 1500 octets).
Why does your application depend on the fact that the whole data must arrive in a single unit, and not in a single connection (comprised potentially of multiple units)?
how can I transfer large data without splitting.
I'm interpreting the above to be roughly equivalent to "how can I transfer my data across a TCP connection using as few TCP packets as possible". As others have noted, there is no way to guarantee that your data will be placed into a single TCP packet -- but you can do some things to make it more likely. Here are some things I would do:
Keep a single TCP connection open. (HTTP traditionally opens a separate TCP connection for each request, but for low-latency you can't afford to do that. Instead you need to open a single TCP connection, keep it open, and continue sending/receiving data on it for as long as necessary).
Reduce the amount of data you need to send. (i.e. are there things that you are sending that the receiving program already knows? If so, don't send them)
Reduce the number of bytes you need to send. (The easiest way to do this is to zlib-compress your message-data before you send it, and have the receiving program decompress the message after receiving it. This can give you a size-reduction of 50-90%, depending on the content of your data)
Turn off Nagle's algorithm on your TCP socket. That will reduce latency by 200mS and discourage the TCP stack from playing unnecessary games with your data.
Send each data packet with a single send() call (if that means manually copying all of the data items into a separate memory buffer before calling send(), then so be it).
Note that even after you do all of the above, the TCP layer will still sometimes spread your messages across multiple packets, etc -- that's just the way TCP works. And even if your local TCP stack never did that, the receiving computer's TCP stack would still sometimes merge the data from consecutive TCP packets together inside its receive buffer. So the receiving program is always going to "receive it like splitted" sometimes, because TCP is a stream-based protocol and does not maintain message boundaries. (If you want message boundaries, you'll have to do your own framing -- the easiest way is usually to send a fixed-size (e.g. 1, 2, or 4-byte) integer byte-count field before each message, so the receiver knows how many bytes it needs to read in before it has a full message to parse)
Consider the idea that the issue may be else where or that you may be sending too much unnecessary data. In example with PHP there is the isset() function. If you're creating an internet based turn based game you don't (need to send all 1,200 variables back and forth every single time. Just send what changed and when the other player receives that data only change the variables are are set.
Does a successful call to send() with the number returned equal to the amount specified in the size parameter guarantee that no "partial sends" will occur?
Or is there some way that the OS might be interrupted while servicing the system call, send part of the data, wait for a possibly long time, then send the rest and return without notifying me with a smaller return value?
I'm not talking about a case where there is not enough room in the kernel buffer; I realize that I would then get a smaller return value and have to try again.
Update:
Based on the answers so far, my question could be rephrased as follows:
Is there any way for packets/data to be sent over the wire before the call to send() returns?
Does a successful call to send() with the number returned equal to the amount specified in >the size parameter guarantee that no "partial sends" will occur?
No, it's possible that parts of your data gets passed over the wire, and another part only goes as far as being copied into the internal buffers of the local TCP stack. send() will return the no. of bytes passed to the local TCP stack, not the no. of bytes that gets passed onto the wire (and even if the data reaches the wire, it might not reach the peer).
Or is there some way that the OS might be interrupted while servicing the system call, send part of the data, wait for a possibly long time, then send the rest and return without notifying me with a smaller return value?
As send() only returns the no. of bytes passed into the local TCP stack, not whether send() actually sends anything, you can't really distinguish these two cases anyway. But yes, it's possibly only some data makes it over the wire. Even if there's enough space in the local buffer, the peer might not have enough space. If you send 2 bytes, but the peer only has room for 1 more byte, 1 byte might be sent, the other will reside in the local tcp stack until the peer has enough room again.
(That's an extreme example, most TCP stacks protects against sending such small segments of data at a time, but the same applies if you try to send 4k of data but the peer only have room for 3k).
I'm not talking about a case where there is not enough room in the kernel buffer; I realize that I would then get a smaller return value and have to try again
That will only happen if your socket is non-blocking. If it's blocking and the local buffers are full, send() will wait until there's room in the local buffers again (or, it might return
a short count if parts of the data was delivered, but an error occured in the mean time.)
Edit to answer:
Is there any way for packets/data to be sent over the wire before the call to send() returns?
Yes. That might happen for many reasons.
e.g.
The local buffers gets filled up by that recent send() call, and you use blocking I/O.
The TCP stack sends your data over the wire but decides to schedule other processes to
run before that sending process returns from send().
Though this depends on the protocol you are using, the general question is no.
For TCP the data gets buffered inside the kernel and then sent out at the discretion of the TCP packetization algorithm, which is pretty hairy - it keeps multiple timers, minds path MTU trying to avoid IP fragmentation.
For UDP you can only assume this kind of "atomicity" if your datagram does not exceed link frame size (usual value is 1472 = 1500 of ethernet frame - 20 bytes of IP header - 8 bytes of UDP header). Otherwise your sending host will have to IP-fragment the datagram.
Then intermediate routers can still IP-fragment the passing packet if their outgoing link MTU is less then the packet size.