How do you get the maximum number of bytes that can be passed to a sendto(..) call for a socket opened as a UDP port?
Use getsockopt(). This site has a good breakdown of the usage and options you can retrieve.
In Windows, you can do:
int optlen = sizeof(int);
int optval;
getsockopt(socket, SOL_SOCKET, SO_MAX_MSG_SIZE, (int *)&optval, &optlen);
For Linux, according to the UDP man page, the kernel will use MTU discovery (it will check what the maximum UDP packet size is between here and the destination, and pick that), or if MTU discovery is off, it'll set the maximum size to the interface MTU and fragment anything larger. If you're sending over Ethernet, the typical MTU is 1500 bytes.
On Mac OS X there are different values for sending (SO_SNDBUF) and receiving (SO_RCVBUF).
This is the size of the send buffer (man getsockopt):
getsockopt(sock, SOL_SOCKET, SO_SNDBUF, (int *)&optval, &optlen);
Trying to send a bigger message (on Leopard 9216 octets on UDP sent via the local loopback) will result in "Message too long / EMSGSIZE".
As UDP is not connection oriented there's no way to indicate that two packets belong together. As a result you're limited by the maximum size of a single IP packet (65535). The data you can send is somewhat less that that, because the IP packet size also includes the IP header (usually 20 bytes) and the UDP header (8 bytes).
Note that this IP packet can be fragmented to fit in smaller packets (eg. ~1500 bytes for ethernet).
I'm not aware of any OS restricting this further.
Bonus
SO_MAX_MSG_SIZE of UDP packet
IPv4: 65,507 bytes
IPv6: 65,527 bytes
Related
I've noticed in wireshark that I'm able to send 4096 bytes of data to a HTTP webserver (from uploading a file) however the server only seems to be acknowledging data 1460 bytes at a time. Why is this the case?
The size of TCP segments is restricted to the MSS (Maximum Segment Size), which is basically the MTU (Maximum Transmission Unit) less the bytes comprising the IP and TCP overhead. On a typical Ethernet link, the MTU is 1500 bytes and basic IP and TCP headers comprise 20 bytes each, so the MSS is 1460 (1500 - 20 - 20).
If you're seeing packets indicated with a length field of 4096 bytes, then it almost certainly means that you're capturing on the transmitting host and Wireshark is being handed the large packet before it's segmented into 1460 byte chunks. If you were to capture at the receiving side, you would see the individual 1460 byte segments arriving and not a single, large 4096 byte packet.
For further reading, I would encourage you to read Jasper Bongertz's blog titled, "The drawbacks of local packet captures".
TCP by default uses path MTU discovery:
When system send packet to the network it set don't fragment flag (DF) in IP header
When IP router or you local machine see DF packet that should be fragmented to match MTU of the next hop link it sends feedback (RTCP fragmentation need) that contains new MTU
When system receives fragmentation needed ICMP it adjusts MSS and send data again.
This procedure is performed to reduce overall load on the network and increase probability of each packet delivery.
This is why you see 1460 packets.
Regarding to you question: the server only seems to be acknowledging data 1460 bytes at a time. Why is this the case?
TCP keep track window that defines "how many bytes of data you can send without acknowledge". Its purpose is to provide flow control mechanisms (sender can't send too much data that can't be processed) and congestion control mechanisms (sender can't send too much data to overload network). Window is defined by receiver side and may be increased during connection when TCP will estimate real channel bandwidth. So you may see one ACK that acknowledges several packets.
As we all perfectly know, UDP does not support retransmission along with some other things.
We also aware of such thing like MTU that works basically in the following way -- when one of the network devices on the path between source and destination points does not support packet of some size, it just drops it.
In case of TCP, it's not a problem -- it already knows MSS after handshake that is always less than MTU (am I right?), so there's no possibility to send a packet with the size greater than MTU.
However, I wonder how does it work in case of UDP? As I already said, there's no retransmission in this protocol and there's no such thing like MSS. So what happens when the packet is dropped due to exceeding MTU?
Or it just works because of the MTU nature (it actually belongs to the IP layer, not the transport layer protocols like UDP or TCP)? So the IP layer reconstruct the dropped packet in smaller units and send it again?
First of all, you must distinguish between the local MTU, which is just the MTU of the local link, and the path MTU (PMTU), which is the smallest MTU of the local link. Consider the following topology:
1500 1480 1500
A -------- B -------- C -------- D
then A's local MTU is 1500, but the PMTU is just 1480.
When router B receives a packet of size 1500 which it needs to forward, and the DF bit is set, it sends an ICMP packet back to the sender with the next hop's MTU, 1480 in this case. The sender can then reduce the packet size.
In TCP, this is done transparently by the network stack. In UDP, the application needs to deal with it. There are three ways to do that:
always send packets that are small enough; 1024 is always safe over IPv6, and 512 is usually (but not always) safe over IPv4;
use a connected UDP socket, and react to an EMSGSIZE error by reducing the packet size; or
use any kind of UDP socket, request the PMTU ancillary data, and use the data provided.
Technique (3) is the most efficient. For IPv6, it is described in Section 11.3 of RFC 3542.
As far as we know the absolute limitation on TCP packet size is 64K (65535 bytes), and in practicality this is far larger than the size of any packet you will see, because the lower layers (e.g. ethernet) have lower packet sizes. The MTU (Maximum Transmission Unit) for Ethernet, for instance, is 1500 bytes.
I want to know, Is there any any way or any tools, to send packets larger than 64k?
I want to test a device in facing with packet larger than 64k! I mean I want to see, if I send a packet larger than 64K, how it behave? Does it drop some part of it? Or something else.
So :
1- How to send this large packets? What is the proper layer for this?
2- How the receiver behave usually?
The IP packet format has only 16 bit for the size of the packet, so you will not be able to create a packet with a size larger than 64k. See http://en.wikipedia.org/wiki/IPv4#Total_Length. Since TCP uses IP as the lower layer this limit applies here too.
There is no such thing as a TCP packet. TCP data is sent and received in segments, which can be as large as you like up to the limits of the API you're using, as they can be comprised of multiple IP packets. At the receiver TCP is indistinguishable from a byte stream.
NB osi has nothing to do with this, or anything else.
TCP segments are not size-limited. The thing which imposes the limit is that IPv4 and IPv6 packets have 16 bit length fields, so a size larger than this limit is not possible to express.
However, RFC 2675 is a proposed standards for IPv6 which would expand the length field to 32 bits, allowing much larger TCP segments.
See here for a talk about why this change could help improve performance and here for a set of (experimental) patches to Linux to enable this RFC.
I need to send and receive very large data using udp. Unfortunately udp provides 8192 bytes per diagram, so there is need to divide data into smaller pieces.
I'm using Qt and QUdpSocket. There is a QByteArray with length of 921600 I want to send to client. I want to send 8192 bytes each time.
What is the fast way to split a QByteArray?
You should never need to explicitly split the data, just step through it 8 KB at a time. Typically the functions that write data to a socket (including QUdpSocket::writeDatagram, it seems) accept a pointer to the first byte and a byte count, so you can just provide a pointer into the array.
Do note that sending 8 KB datagrams is quite aggressive; they will very likely be fragmented at the IP layer which can affect delivery speed and reliability negatively.
Research the concept of "path MTU", and try to use that for the sends, it might be faster although it will result in more datagrams.
Actually the length field on a UDP header is 16 bits so a UDP datagram can be up to ~65k (minus the size of the header stuff).
However, as unwind pointed out, it will most likely be fragmented along the way to fit the smallest MTU on the path to the destination.
8192 bytes is the default receive buffer size for Windows operating systems. The MTU is likely 1500 bytes if you are using Ethernet. Any UDP datagram larger than that will be fragmented. If your datagram encounters a device along the route with an even smaller MTU, it will be fragmented yet again.
You can use the QByteArray.mid(int start, int len) method (see documentation here) to get a QByteArray of length len starting from start.
Just make len your datagram size and start with 0*len, 1*len, 2*len, ... until everything is sent.
I am analyzing wireshark log files, when I make a request to a web page using firefox through a proxy server.
Following are details of connection establishment:
I have noted "maximum segment size" when I open options branch in the TCP segment details of the [SYN] message from my PC to the proxy server - it says 1460 bytes
Similarly, maximum segment size eof the [SYN,ACK] message from the proxy server to my PC - it says 1460 bytes
After establishing the TCP connection, should not each of the TCP frames sent from proxy server to my PC be of 1460 bytes? I am puzzled that why are they 590 bytes. Please advice how the 590 size is being set
A plausible explanation is that 590 turns out to be the Path MTU for the particular connection.
In other words whereby the client (one of the end nodes of the connection)accepts packets of a maximum of of 1460 bytes payload, some node(s) on the way accepts smaller packets. For efficiency purposes, the Path MTU Discovery allows the originator of a packet to size it so that it would fit the smaller MTU encountered on the path, and hence avoid fragmentation.
BTW:
1460 is a very common MTU (well MSS), because it it corresponds to 1500, Ethernet v2's maximum, minus 20+20= 40 bytes for the IP header overhead)
See the following Wikipedia entry for an overview of MTU (Maximum Transmission Unit) and a basic description of the Path MTU Discovery method (Basically setting the the DF i.e. do-not-fragment flag and relying on the ICMP ""Destination Unreachable (Datagram Too Big)" messages to detect that some node on the way couldn't handle the packet, and hence try with smaller size until it goes through).
Also, I suggest inspecting the packets when the connection is to a different host, maybe a peer on the very same network segment, without going through the proxy mentioned. Chances are you will then start seeing 1460 bytes frames.