I am writing the network stack for my os.
While receiving the data from ethernet, I have to detect and remove the padding if data is less than 64 bytes and set the data length, as the standard says.
What is the algorithm for doing this?
If the data contains 'a0' , it is less than 64 bytes and it will be padded. If I am checking bytewise, I will get the second bye 0, so it will also be counted as padded data, however it is not.
The Type/Length field will tell you how large the payload is. Wireshark has a pretty good explanation:
Therefore, if the type/length field has a value 1500 or lower, it's a
length field, and is followed by an 802.2 header, otherwise it's a
type field and is followed by the data for the upper layer protocol
(XXX - slight duplicate of sentence above?). Note that when the
length/type field is used as a length field the length value specified
does not include the length of any padding bytes (e.g. if a raw
ethernet frame was sent with a payload containing a single byte of
data the length field would be set to 0x0001 and 45 padding bytes
would be appended to the data field to bring the ethernet frame up to
the required minimum 64-byte length).
Related
I am trying to debug a RTMP client that fails to connect to some servers. I'm using Wireshark to capture the packets and compare them with a client that connects successfully (in this case, ffmpeg).
Looking at the captured packets for a successfull connection, I noticed that, when viewing at TCP level, there is an extra byte in the payload (see pics below). The extra byte has value 0xc3 and is placed at byte 0xc3 in the payload.
I Googled the best I could to find information about extra bytes in the TCP payload, but I didn't find anything like this. I tried to look in the TCP spec but no luck either. Where can I find information about this ?
TCP-level view
RTMP-level view
This happens because the message length is larger than the maximum chunk size (as per the RTMP spec, the default maximum chunk size is 128). So if no Set Chunk Size control message was sent before connect (in your case), and the connect message is greater than 128 bytes, the client will split the message into multiple chunks.
0xC3 is the header of the next chunk, looking at the bits of 0xC3 we would have 11 000011. The highest 2 bits specify the format (fmt = 3 in this case, meaning that this next chunk is a type 3 chunk as per the spec). The remaining 6 bits specify the chunk stream ID (in this case 3). So that extra byte you're seeing is the header of a new chunk. The client/server would then have to assemble these chunks to form the complete message.
By IEEE 802.3, an Ethernet frame has to carry a payload of at least 46 bytes. This is for collision detection-- collisions of smaller frames (may) go undetected.
The Q is: what if the payload to be carried is shorter? what kind of padding is used to scale the frame up to the slot size-- 64 bytes?
TIA.
To quote from Data and Computer Network Communication (emphasis mine);
If the network layer wishes to send less than 46 bytes of data the MAC protocol adds sufficient number of zero bytes (0x00, is also known as null padding characters) to satisfy the requirement.
Some buggy drivers fail to do this though as noted by Adaptec.
If I encrypt emails so that I can store them in a database, the resulting string is longer than the email itself. Is there a maximum length to this resulting coded string? if so, does it depend on both key length and the email length? I need to know this so I can set my database fields to the correct length.
Thanks.
As Alex K. notes, for block ciphers (like DES), common modes will pad them out to a multiple of the block size. The block size for 3DES is 64-bits (8 bytes). The most common padding scheme is PKCS7, which pads the block with "n x n bytes." This is to say, if you need one bytes of padding, it pads with 0x01. If you need four bytes of padding, it pads with 0x04040404 (4x 4s). If your data is already the right length, it pads with a full block (8 bytes of 0x08 for 3DES).
The short version is that the padded cipher text for 3DES can be up to 8 bytes longer than the plaintext. If your encryption scheme is a typical, insecure implementation, this is the length. The fact that you're using 3DES (an obsolete cipher) makes it a bit more likely that it's also insecurely implemented, and so this is the answer.
But if your scheme is implemented well, then there could be quite a few other things attached to the message. There could be 8 bytes of initialization vector. There could be a salt of arbitrary length if you're using a password. There could be an HMAC. There could be lots of things that could add an arbitrary amount of space. (The RNCryptor format, for example, adds up to 82 bytes to the message.) So you need to know how your format is implemented.
I am currently going through my networking slides and was wondering if someone could help me with the concept of fragmentation and reassembly.
I understand how it works, namely how datagrams are split into smaller chunks because network links have a MTU. However the example in the picture is confusing me.
So the first two sections show a length of 1500, because this is the MSU, but shouldn't this mean that the last one should have 1000 (for a total of 4000 bytes) and not 1040? Where did these extra 40 bytes come from? My guess is that because the previous two fragments both had a header of 20 bytes, this extra 40 bytes of data needed to go somewhere, so it will arrive in the last fragment?
Fragflag essentially means that there is another fragment, so all of them will have a Fragflag of 1 except the last fragment which will be at zero. However I don't understand what offset is or how it is calculated. Why is the first offset at zero? Why did we divide the bytes in the datafield (1480) by 8 to get the second offset? Where did this 8 come from? Aside from that, I am assuming that each fragments offset will just increase by this value?
For example, the first fragment will have a offset of 0, the second 185, the third 370 and the fourth 555? (370+185)
Thanks for any help!
There is a 20 byte header in each packet. So the original packet contains 3,980 bytes of data. The fragments contain 1480, 1480, and 1020 bytes of data. 1480 + 1480 + 1020 = 3980
Every bit in the header is precious. Dividing the offset by 8 allows it to fit in 13 bits instead of 16. This means every packet but the last must contain a number of data bytes that is a multiple of 8, which isn't a problem.
The fragmentation and Reassembly has been exclusively explained in the RFC 791. Do go through the Internet Protocol Specification RFC. The RFC has various sections explaining the sample fragmentation and reassembly. All your doubts and questions are well catered in it.
Ans 1: Regarding the lengths of the packet: The original Packet contains 4000 Bytes. This packet is a fully IP packet and hence contains the IP header as well . Thus the payload length is actually 4000 - ( IP Header Length i. e. 20 ).
Actual Payload Length = 4000 - 20 = 3980
Now the packet is fragmented owing to the fact that the length is greater than the MTU ( 1500 Bytes).
Thus the 1st packet contains 1500 Bytes which includes IP header + Payload Fraction.
1500 = 20 ( IP header ) + 1480 ( Data Payload )
Similarly for the other packet.
The third packet shall contain remaining left over data ( 3980 - 1480 -1480 ) = 1020
Thus length of the packet is 20 ( IP Header ) + 1020 ( payload ) = 1040
Ans 2: The offset is the address or the locator from where the data starts with reference to the original data payload. For IP the data payload comprises all the data thats after the IP header and Options header. Thus the system/router takes the payload and divides it into smaller parts and keeps the track of the offset with reference to the original packet so that reassembly can be done.
As given in the RFC Page 12.
"The fragment offset field tells the receiver the position of a fragment in the original datagram. The fragment offset and length determine the portion of the original datagram
covered by this fragment. The more-fragments flag indicates (by being reset) the last fragment. These fields provide sufficient information to reassemble datagrams. "
The fragment offset is measured in Units of 8 bytes each. It has 13 bit field in the IP header. As said in the RFC page 17
"This field indicates where in the datagram this fragment belongs.The fragment offset is measured in units of 8 octets (64 bits). The first fragment has offset zero."
Thus as you asked in the question where did this 8 come from, its the standard thats been defined for IP protocol specification, where 8 octets are taken as one value. This also helps us to transmit large packets via this.
Page 28 of the RFC writes:
*Fragments are counted in units of 8 octets. The fragmentation strategy is designed so than an unfragmented datagram has all zero fragmentation information (MF = 0, fragment offset =
0). If an internet datagram is fragmented, its data portion must be
broken on 8 octet boundaries. This format allows 2**13 = 8192 fragments of 8 octets each for a
total of 65,536 octets. Note that this is consistent with the the
datagram total length field (of course, the header is counted in the
total length and not in the fragments).*
the offset size is 13 bits in the IP header but we need 16 bits as in worst case. So we use a scaling factor of 8 i.e. (2^16/2^13).
those are not extra bits but the total length of last fragment.
as 1500 is MTU this means there can be 1500 byte of data in one fragment including header. Header is appended with every fragment. this means in fragment we are capable of sending 1500-20 =1480 byte of data.
it is given there is 4000B datagram .datagram is nothing but a packet encapsulation of data at network layer.so the total data we have to send is 4000-20=3980 . then it is fragmented into 3parts (ceil(3980/1480)) each of length 1480,1480,1020 respectively . hence when 20B header is appended to last fragment its length becomes 1020+20=1040 .
What are the size limits on DNS responses? For instance how many 'A' resource records can be present in a single DNS response? The DNS response should still be cache-able.
According to this RFC, the limit is based on the UDP message size limit, which is 512 octets. The EDNS standard supports a negotiated response with a virtually unlimited response size, but at the time of that writing (March 2011), only 65% of clients supported it (which means you can't really rely on it)
The largest guaranteed supported DNS message size is 512 bytes.
Of those, 12 are used up by the header (see §4.1.1 of RFC 1035).
The Question Section appears next, but is of variable length - specifically it'll be:
the domain name (in wire format)
two bytes each for QTYPE and QCLASS
Hence the longer your domain name is, the less room you have left over for answers.
Assuming that label compression is used (§4.1.4), each A record will require:
two bytes for the compression pointer
two bytes each for TYPE and CLASS
four bytes for the TTL
two bytes for the RDLENGTH
four bytes for the A record data itself
i.e. 16 bytes for each A record (§4.1.3).
You should if possible also include your NS records in the Authority Section.
Given all that, you might squeeze around 25 records into one response.