I am wondering when I doing some example on Wirshark, where TCP sequence number come from
An initial sequence number is generated randomly on handshake, and the number is incremented accordingly for packets sent.
I'm not sure but probably from the system included stack implementation of the OSI model???
http://packetlife.net/blog/2010/jun/7/understanding-tcp-sequence-acknowledgment-numbers/
Related
I'm creating pseudo TCP packet from scratch (based on some existing data) that can be later analysed using Wireshark.
How should I fill out sequence/acknowledgement number so that in Wireshark we can see a clean flow? Currently, I'm seeing TCP retransmission/out of order/previous segment not captured etc.
What I've tried: increment sequence number with the current TCP packet's length, but it didn't work.
(It doesn't necessarily need to makes sense, I should just get rid of the warnings)
If TCP runs out of its sequence number, what will happen?
If it again turns to 0 as the sequence number of the next byte, won't that be considered "duplicate" by the receiver?
If yes, then it has to ignore that byte.
If not, why?
I think, i found the answer.
The answer of this query, lies on one of the TCP option field known as "timestamp". It's in every TCP segment (including data and ACK segments).
Therefore to identify a unique tcp segment, we look for a combination of "timestamp" and "sequence number".
The basic idea is that a segment can be discarded as an old duplicate if it is
received with a timestamp less than some timestamp recently received on this connection.
Example :
Two segments 400:12001 and 700:12001 definitely belongs to two different incarnations.
And this mechanism is known as "PAWS" or protection against wrapped sequence numbers.
Reference: https://www.rfc-editor.org/rfc/rfc1323#page-17
In a situation where both client and server sets their respective sequence number to 0, I read that the following is true:
C-->S: SYN=1, SEQ=0 (No data bytes)
C<--S: SYN=1, SEQ=0, ACK=1 (No data bytes)
C-->S: SEQ=1, ACK=1 (Data bytes optional)
In the third part, I understand the server is expecting the next sequence number to be 1, but aren't sequence numbers supposed to be set to initial_seq_num + sent_data_bytes_num? Since there was no data bytes sent in the first part of the handshake shouldn't the seq # be 0?
Is this just an exception during the handshake or are segments sent to with no data bytes supposed to increment the sequence number by 1 if they can be sent at all?
(There is a similar Q & A but the answer doesn't explain if this is an exception during the handshake phase OR if this happens after a TCP connection has been establish. I'm not even sure if a segment with no data bytes can even be sent. I'm assuming you can't)
ADDED It seems TCP keep-alive packets have no payload either. RFC 1122 says in these packets, SEG.SEQ = SND.NXT-1, and because this sequence number will be an already ACKed number, and a duplicate ACK will be sent, so as to keep the sequence number of the server the same.
Otherwise, I couldn't find any indications of what needs to be done when the sequence number is correct but there is no payload. I might be wrong since I only briefly scanned the document, but there is also no statement of rules of sequence numbering during the handshake except for examples.
In RFC 1122, it says
Unfortunately, some misbehaved TCP implementations fail to respond to a segment with SEG.SEQ = SND.NXT-1 unless the segment contains data.
So I'm assuming it depends on each implementation, but if there is any statement of a) the sequence numbering during handshake, and b) how to behave when the sequence # is correct but there is no payload, I would really appreciate it if someone could point me to that part.
Thanks!
The first ACK (that occurs as part of Handshake) acknowledges the reception of SYN from the other end. The SYN segment does not carry any data. But to allow the provision for acknowledging the reception of SYN, the first ACK is incremented though no payload is present.
Can somebody just simplely describe the retransmission mechanism in TCP?
I want to know how it deal in this situation?
A send a packet to B:
A send a packet.
B receive and send ack,but this ack is lose.
A timeout and resend.
In this situation B will receive 2 same packets, how can B do to avoid dealing the same packet again?
Thanks.
Each packet has a sequence number associated with it. As data is sent, the sequence number is incremented by the amount of original data in the packet. You can think of the sequence number as the offset of the first byte in the packet from the beginning of the data stream although it may not, likely will not, start at zero. When A sends the retry, it will use the same sequence number it used the first time. B tracks the sequence numbers as it receives data and can know that it has seen the retry's sequence number before. If it has already made that data available to the (upper layer) client, then it knows that it should not do so again.
I'm trying to capture packets and reorganize packets for obtaining original HTTP request.
I'm capturing packets by IPQUEUE(by iptables rule), and I figured out that packets are not captured in order.
I already know that in TCP protocol, packets have to be re-sequenced, so I'm trying to re-sequence packets by sequence number.
According to Wikipedia, the sequence number of TCP is 32 bits number. Then, what happens if sequence number reaches to MAX 32bits number?
Because sequence number of SYN packet is random number, I think this limitation can be reached very fast.
If anybody has a commend on it, or has some links helpful, please leave me a answer.
From RFC-1185
Avoiding reuse of sequence numbers within the same connection is
simple in principle: enforce a segment lifetime shorter than the
time it takes to cycle the sequence space, whose size is
effectively 2**31.
If the maximum effective bandwidth at which TCP
is able to transmit over a particular path is B bytes per second,
then the following constraint must be satisfied for error-free
operation:
2**31 / B > MSL (secs)
So In simpler words TCP will take care of it.
In addition of this condition TCP also has concept of Timestamps to handle sequence number wrap around condition. From the same above RFC
Timestamps carried from sender to receiver in TCP Echo options can
also be used to prevent data corruption caused by sequence number
wrap-around, as this section describes.
Specifically TCP uses PAWS mechanism to handle TCP wrap around case.
You can find more information about PAWS in RFC-1323
RFC793 Section 3.3:
It is essential to remember that the actual sequence number space is
finite, though very large. This space ranges from 0 to 2*32 - 1.
Since the space is finite, all arithmetic dealing with sequence
numbers must be performed modulo 2*32. This unsigned arithmetic
preserves the relationship of sequence numbers as they cycle from
2**32 - 1 to 0 again. There are some subtleties to computer modulo
arithmetic, so great care should be taken in programming the
comparison of such values.
Any arithmetics done on the sequence number are modulo 2^32
In simple terms, the 32-bit unsigned number will wrap around:
...
0xFFFFFFFE
0xFFFFFFFF
0x00000000
0x00000001
...
have a look at tcp timestamps section of https://en.wikipedia.org/wiki/Transmission_Control_Protocol