How will you identify the next layer protocol after Ethernet ? IS there any provision for same in Ethernet frame ?
The ethernet frame contains an Ethertype, a 2 byte field designating the upper layer protocol. For example IP has 0x800. When a network engine receives a frame from the network interface it checks this field and forwards it to the appropriate handler.
The ethertype (2 bytes, the 13th and 14th bytes) is generally right behind destination mac and source mac.
The case is true for Ethernet II.
For 802.3, there is a case where the ethertype is encapsulated as part of the LLC SNAP.
So in general, you can check if the 13th and 14th bytes are indeed ethertype, then use it.
If not ethertype (ethertype must be greater than 0x05DC), then you know it's payload length.
Then parse the LLC SNAP to obtain the ethertype within.
This is Ethernet header
DstMacAddr // 6 bytes
SrcMacAddr // 6 bytes
EthType // 2 bytes -> ARP(0x0806), IPv4(0x0800), IPv6(0x86dd), VLAN(0x8100), etc
Payload // ARP, IPv4, IPv6, etc
FSC // CRC32
Related
For example, one 2000 bytes UDP package(contains UDP header) and network MTU is 1500. So this UDP package should be split to two IP fragments. Only the first IP package contains the UDP header.
What value should be filled into UDP length in the UDP header of first IP package? 1480 or 2000?
Is there any document to confirm this?
UDP Length is the length of the UDP header AND the UDP data, in bytes. The fragmentation will/should NOT change this.
[UDP] "Length is the length in octets of this user datagram including this header and the data" --RFC768, an Internet Standard. It is also in the Stevens link referenced: "Referring to Figure 10-2, the UDP Length field is the length of the UDP header and the UDP data in bytes."
I think you may be overthinking this. Just set the UDP length to however big your data is.
IP fragmenting can occur at any router hop, without your knowledge (assuming you are the sender). It's the IP layer's responsibility to fragment and re-assemble the packet before the UDP datagram is delivered to the application. Unless you're plugged in to the NIC driver, you won't be able to reliably tell if the packet was fragmented on the way.
If there's an IP fragment lost on the way, you won't get any UDP datagram at all, it will simply be lost from the application's point of view, even though the IP layer may have gotten 9 out of 10 fragments.
Everything about this is brilliantly described in Richard Stevens et al's seminal work: TCP/IP Illustrated, Volume 1: The protocols
, section 10.7 IP Fragementation.
I think at L2 it will be ether type and destination MAC address.
at L3 it can be protocol field and Destination IP Address.
I want to know what are the other fields by which we can differentiate between control and data packet.
#Anupam Thakur: In pure L2 infrastructure, on receiving frame, node does the bridge-table/l2-table/cam lookup based on destination mac address. For different BPDUs (almost always reserved multicast mac dst addresses, few exceptions beyond the scope of this question) wherever required cam table is programmed with punt path to control-plane module (/also). Note, punt path doesn't distinguish between different protocols. It's job is to just push the packet in Q towards control-plane module. When cp module receives such packet, based on ethertype or LLC content it determines the further protocol processing of the packet.
"Router Alert Option" in IP header can also force packet to be punted to Control Plane, even though it is not destined for recipient device. (i.e. transit node)
I've encountered the following statement somewhere in the net: "Although
theoretically only 8 bytes of L4 info may be guaranteed in a fragment,
assume complete L4 info is available...". I don't understand how possible
that only 8 bytes of a transport are guaranteed in a fragment, as the IP
fragment can't be less than 46 bytes (minimum payload size for Ethernet
frame), and this includes 20 bytes of IP header and 20bytes of TCP header
(not considering variable-lngth options), UDP will be less.
Thus for the first IP fragment we always can expect IP header at tcp header,
while other fragments will carry only IP header + payload.
I believe I'm missing something, but I still can't understand why only 8
bytes can be guaranteed in a fragment? I would appreaciate if someone helps
to clarify this issue. Thanks !
Mark
Imagine a router receives a TCP packet containing one more byte of data than will fit in the MTU of the destination network. It must split on an 8-byte boundary because that's an IP fragmentation rule.
It won't split into more than two fragments because that would be silly. So it must include at least one byte of data in the first fragment..
Thus the minimum number of data bytes you can place in the first fragment of an IP datagram is 8.
My application connects to the server using TCP and system is working fine. Using Wireshark I noticed that some TCP packets of 60 bytes are sent with no data. Is this normal?
as part of TCP transmissions and handshake are there some packets sent with no data?
Thanks
Kumar
There are the ACK packets that carry no data and only acknowledge received data. When using Wireshark it should display these "flags" on the empty packets.
To be more accurate you should show a screenshot of the wireshark capture, so we know what the size you mean is.
Meanwhile I dissected an ACK packet of IPv4 TCP traffic here and these are my results:
Protocol Size Description
Ethernet protocol 14 byte 2 MAC adresses (6 byte), Protocoll type (2 byte)
Internet protocol 20 byte The default header size
TC protocol 32 byte The default header size (20 byte) + options (12 byte)
_____________________________________________________________________________________
Total 66 byte
without EP 52 byte (Probably the size the OP is talking about)
without EP, Opts 40 byte (The size talked about in the comments)
The options in my case were 2 NOPs (each 1 byte) and two timestamps (each 5 byte??). The order of the protocols is the same as in Wireshark.
Wireshark splits the information down to each field, so you can see what takes up how much space.
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