What does no timestamps option in SYN-ACK packet mean? - tcp

I'm working on a tool to test a tcp proxy and I see the following:
Client sends a SYN to the server with timestamps option (TSval=12345, TSecr=0).
Server sends a SYN-ACK and omits the timestamps option.
The problem is that I'm not sure if a TCP is supposed to interpret that to mean that the timestamps option should not be used.
I've been poring over rfc 1323 for a while now, and this is all I could find on the subject:
A TCP may send the Timestamps option
(TSopt) in an initial segment
(i.e., segment containing a SYN bit
and no ACK bit), and may send a TSopt
in other segments only if it received a TSopt in the initial
segment for the connection.
I infer from this that the fact that the SYN-ACK is missing the timestamps option doesn't mean anything about whether or not it's valid later in the session. However, doing so requires me to assume a bit more than I'd like. Does anyone have an authoritative source on the subject or personal experience with how different TCP stacks behave in this situation?

The peer TCP is indicating that it does not support (or wish to use) the option. You should not send any further Timestamp options on that session.
This is fairly standard practice for TCP options.

Related

Why can't tcp use two way handshake?

Client Send SYN with client isn.
Server reply SYN ACK with server's isn.
Client resend SYN if timeout?
When client send data, it can accumulated confirm the server's isn.
I try to search, but can't find the answer.
I know how the tcp is designed now, I just don't know why it's designed like this. Why can't use a two way handshake.
It cannot use a two way handshake by definition. TCP/IP is formalized as a standard for communication across networks (Internetworking). Specifically, RFC 793 requires that:
The "three-way handshake" is the procedure used to establish a
connection. This procedure normally is initiated by one TCP and
responded to by another TCP. The procedure also works if two TCP
simultaneously initiate the procedure. When simultaneous attempt
occurs, each TCP receives a "SYN" segment which carries no
acknowledgment after it has sent a "SYN". Of course, the arrival of
an old duplicate "SYN" segment can potentially make it appear, to the
recipient, that a simultaneous connection initiation is in progress.
Proper use of "reset" segments can disambiguate these cases.
If you think about how the protocol works, you do not actually have a single full duplex connection. Instead, you have two simplex connections, each going in one direction. This is why the proper response to a SYN is a SYN-ACK. The server is acknowledging the synchronization request by sending the originators sequence number as the acknowledgement number plus one; it is simultaneously attempting to open its own connection to the client by sending a SYN request to synchronize.
The proper answer to a SYN will always be an ACK, even if the connection fails.
Regarding your question about sending a retry, yes; the client will send a retry (typically up to three, but it could be as many as eight in practice... There's no limit defined) attempting to elicit a response of some kind (preferably a SYN-ACK, but possibly a RST-ACK).

About HTTP 1.1 persistent connection and TCP

When I was introduced to HTTP 1.0 and 1.1, it was emphasized that the main difference was that 1.1 allows a single TCP connection for all of the objects as opposed to 1.0 where a new connection was made for each object download.
My question is, since a connection isn't really continuous but discrete (i.e in packets), then how come each different packet of each object that is being downloaded, doesn't need to go through the ACK, SYN TCP protocol?
How do they all know about the first ACK, SYN that was made? (perhaps made even to a different server than the objects?)
Not much of your question makes sense. HTTP keepalive only operates on connections to the same target, not 'for all of the objects'. The part about packets doesn't have anything to do with HTTP: you're really asking how TCP works; and the answer to that is that every TCP segment contains the source and destination IP addresses and ports, which are unique to a connection, and a sequence number, for data ordering purposes.

Filtering UDP packets accidentally sent to my port

I am designing a simple protocol on top of UDP and now I've realized that someone else can send a packet to the port I'm listening on. Such packet will obviusly be incorrect for my application (I'm not concerned about security now)
Is there the way of filtering these invalid messages? I was thinking about adding some fixed magic number at the beginning of each packet, but how large should it be? Is 16 bits enough?
I believe the typical solution is to require a handshake (that could include a "long enough" magic number) in the beginning of the session. Of course the "session" is something your protocol needs to keep track of, UDP does not have the concept. Keeping a list ip, port and last packet receive time for all current sessions should do it: then you can drop all packets from a peers that have not done the handshake beforehand. This not only prevents random unknown app traffic from breaking your application but will also prevent multiple legitimate peers from screwing up each others traffic.
Additionally you could add either a session id or an increasing packet sequence number (with allowances for missing packets) to packets if you need to ensure that the peer agrees with you on which session this is.
Probably. Java uses 32 for .class files (0xcafebabe). But you only have 534 bytes in practical UDP so you might need to save a couple.

tcpip 3-way handshake

Why is data not transferred during the 3rd part of TCP 3-way handshake?
e.g.
(A to B)SYN
(B to A)ACK+SYN
(A to B) ACK.... why cant data be transferred along with this ACK?
I've always believed it was to keep the session establishment phase separate from the data transfer phase so that no real data is transferred until both ends of the session have agreed on the sequence numbers and session options, especially since packets arriving may be from a totally different, previous, session that just happens to have the same endpoints.
However, on further investigation, I'm not entirely certain that transmitting data with the handshake packets is disallowed. The section on TCP connection establishment in my Internetworking with TCP/IP1 book contains the following snippet:
Because of the protocol design, it is possible to send data along with the initial sequence numbers in the handshake segments. In such cases, the TCP software must hold the data until the handshake completes. Once a connection has been established, the TCP software can release data being held and deliver it to a waiting application program quickly.
Since it's certainly possible to construct a TCP packet with SYN (or ACK) and data, this may well be allowed. I've never seen it happen in the wild but, then again, I've never seen a hairy-eared dwarf lemur in the wild either, though I'm assured they exist.
It may be that it's the sockets software that prevents data going out before the session is fully established but TCP appears to consider it valid. It appears you can send data with a SYN-ACK packet (phase 2 of the connection establishment) since you have the other end's sequence number and options. Likewise, sending data with the phase 3 ACK packet appears to be possible as well.
The reason the TCP software holds on to the data until the handshake is fully complete is probably due to the reason mentioned above - only once both ends have agreed on the sequence numbers can you be sure that the data is not from a previous session.
1 Internetworking with TCP/IP Volume 1 Principles, Protocols and Architecture, 3rd edition, Douglas E. Comer, ISBN 0-13-216987-8.

TCP Connection Persistent State

Is there any field/option/anything that I can put in a TCP packet (be it a syn or an ack or just plain data) that I can be sure will be returned by the other end intact?
For eg. I want to "tag" a particular connection (src, srcport, dst, dstport) with a number that I can always read from a packet belonging to that connection. That means I can identify the connection without using the 4-tuple (as given above).
Yes: it is called a Client protocol encapsulated in the TCP server protocol.
In other words: define the Client protocol to meet your needs. Don't try to "shove" extra bits in the TCP overhead.
There are of course the 'options' overhead in TCP but I doubt you'll find an easy way to access these... and in any case, you shouldn't.
You could possibly abuse the TCP Timestamp option for this. It does not seem like a great idea, though.
You can have a lookup table in your application where you associate your tag with the socket.
No, there isn't any facility for what you describe.
Typically what you would do if you're writing a socket application with multiple connections to other systems, is keep track of the socket handle that belongs to each remote system. When receiving data, you are using the socket handle (in some form, don't know which OS or language you're using) so you can take appropriate action based on whichever socket handle that is.
I've never seen a server application that keeps track of connections based on the 4-tuple of address/ports. That seems like way too much work.
On rereading your question, it seems like you may be asking this from the point of view of the TCP driver level. What sort of software are you writing here?
In UDP, destination IP and destination port number are used to demultiplex the packets, but in TCP destination IP, source IP, destination port number and source port numbers (4-tuple) all needed to distinguish between the connections why reasoning for this usage.

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