I use raw socket to build a tcp client program and run it on machine A
and I run a regular tcp server program on machine B
the raw socket-based client program first send a SYN packet
and then it receives a SYN/ACK packet from the remote tcp server
then the kernel of machine A sends a RST to the remote tcp server
the sequence number and ack-sequence number is fine
what are potential reasons?
and how to deal with it? thanks!
BTW: I used tcpdump to capture packets on the remote machine B
and it shows "TCP port numbers reused" for the SYN packet from client,
actually before the client send the SYN, I used
netstat -tnp
to check on-going tcp sessions, and it shows nothing
This is perfectly normal. If a machine receives a SYN/ACK packet it doesn't expect, it should respond with a RST to let the other side know that it has no knowledge of or interest in that connection. The kernel sent a RST because that's what it's supposed to do -- it has no idea what your program is doing.
If you're trying to run your own TCP stack on a machine that already has a TCP stack, you'll have to prevent the regular TCP stack from responding to machines your stack is trying to talk to -- otherwise, they'll be talking to two TCP stacks which can't possibly work.
Related
Which software libraries does exist for such task for Linux, Windows OS?
Does it exist some info in RFC how people should do it?
I'm interesting how can I create functionality for my C++ project like presented here in that software: https://secure.logmein.com/ru/products/hamachi/download.aspx
There is not much difference if you want to make a connection through TURN relay server. The only difference is how TCP and UDP creates connection and nothing else.
There are some big differences if you want to make P2P connection.
If you are in same network(behind same NAT): In UDP you send a stun binding request to your peer candidate and then if you get a response back then you know you are connected. Same in TCP you have to create one active socket on one side and one passive socket on another. And then send syn from active socket and receive it from passive socket and then send syn ack to the active socket. And then active socket send an ack and the connection is established.
If you are in different Network(behind different NAT): You have to employ TCP hole punching technique for making a connection. Because your NAT won't allow a TCP syn packet through if previously no packet was sent to the address the syn is coming from.
TCP hole punching in details:
You have to use a TCP simultaneous open socket. This socket acts in both active and passive mode. Both end needs to know each others private and public IP:Port.
TCP simultaneous open will happen as follows:
Peer A keeps sending SYN to Peer B
Peer B keeps sending SYN to Peer A
When NAT-a receives the outgoing SYN from Peer A, it creates a mapping in its state machine.
When NAT-b receives the outgoing SYN from Peer B, it creates a mapping in its state machine.
Both SYN cross somewhere along the network path, then:
SYN from Peer A reaches NAT-b, SYN from Peer B reaches NAT-a
Depending on the timing of these events (where in the network the SYN cross),
at least one of the NAT will let the incoming SYN through, and map it to the internal destination peer
Upon receipt of the SYN, the peer sends a SYN+ACK back and the connection is established.
From WIKI.
Also to learn about TCP simultaneous open connection read from here. To learn about NAT filtering behavior see this answer.
I'm trying to send a SYN packet that has been received in one network interface into another one on the same host (while changing the IP addresses), using a raw socket.
I'm on Windows Server 2008 R2, and while sending and receiving UDP packets is going very well, when I'm trying to send a TCP packet, the other side just not receiving it (although it shown correctly in Wireshark). The other side normally listens to SOCK_STREAM data.
I've just read in this site that:
Even though you can send TCP packets with a raw socket, the operating system's TCP/IP stack will process incoming TCP and UDP packets. When you send a TCP SYN packet with a raw socket, the destination's response will be rejected by the operating system because it is unaware of the SYN you sent. Therefore, you cannot complete a three-way handshake.
However, Microsoft noted that:
These above restrictions do not apply to Windows Server 2008 R2, Windows Server 2008 , Windows Server 2003, or to versions of the operating system earlier than Windows XP with SP2.
Is it that TCP data cannot be sent over raw sockets? However, if yes, is there a way to establish the TCP connection anyway? Maybe some WinAPI to add the stream identifier?
I'd love to hear any solution for establishing TCP connections with a raw socket. Thanks!
I have around 20 clients communicating together with a central server in the same LAN. The clients can make transaction simultaneously with the server. The server forward each transaction to external appliance in the network. Sometimes it works, sometimes my application shows a "time out" message in a client screen (randomly)
I mirrored all traffic and found TCP Retransmission after TCP Reset packets for the first TCP Sequence. I immediately thought about packet loss but all my cables/NIC are fine, and I do not see DUP ACK in the capture.
It seems that RST packets may have different significations.
What causes those TCP Reset?
Where should I focus my investigation: network or application design ?
I would appreciate any help. Thanks in advance.
Judging by the capture, I assume your central server is 137.56.64.31. What's happening is the clients are initiating a connection to the server with a SYN packet and the server responds with a RST. This is typical if the server has no application listening on that particular port e.g. the webserver application isn't running and a client tries to connect to port 80.
The clients are all connecting to different ports on the server, which is unusual for an central server, but not unheard of. The destination ports the clients are connecting to on the server are: 11007, 11012, 11014, 11108, and 11115. Is that normal for the application? If not, the clients should be connecting to whatever port the application server is listening on.
The reason for the retransmits is that instead of giving up on the connection upon receiving a RST from the server, the client tries to initiate the connection again so Wireshark considers it a retransmission.
I started to learn Linux Networking and packets filtering. In the iptables documentation it is stated that:
If a packet is destined for this box, the packet passes downwards in the diagram, to the INPUT chain. If it passes this, any processes waiting for that packet will receive it.
So, suppose there're 3 server apps on a host. Servers A and B are TCP servers, and C is UDP server.
Is it true, that if we receive an UDP packet, at IP level this packet is to be delivered for apps A, B, C? Or sockets of apps A & B wouldn't receive this packet at all?
TCP servers and UDP servers operate in very different ways.
At most one TCP server will listen on a given TCP port (corner cases ignored for the sake of simplicity). Connection requests (encapsulated in IP packets) destined for that port are "accepted" by exactly one process (more accurately, accepted by a process that has a file descriptor corresponding to exactly one listening endpoint). The combination of [remote_address,remote_port] and [local_address,local_port] is unique. A TCP server doesn't really receive "packets", it receives a stream of data that doesn't have any specific relationship to the underlying packets that carry the data (packet "boundaries" are not directly visible to the receiving process). And a TCP packet that is neither a connection request nor associated with any existing connection would simply be discarded.
With UDP, each UDP datagram is logically independent and may be received by multiple listening processes. That is, more than one process can bind to the same UDP endpoint and receive datagrams sent to it. Typically, each datagram corresponds to a single IP packet though it is possible for a datagram to be broken into multiple packets for transmission.
So, in your example: no, a server that is listening for TCP requests (a "TCP server") will never receive a UDP packet. The port namespaces for TCP and UDP are completely separate.
The delivery of the packet will depend on its destination port.
Lets assume that the servers A, B and C are listening on port 1111, 2222 and 3333 respectively, so when a packet with destination port 2222 is arrived, it will be delivered to server B.
My question wasn't well formulated, unfortunatelly. I understood it when I had seen the answers. Here is an explanation which I was looking for, it's from http://www.cs.unh.edu/cnrg/people/gherrin/linux-net.html#tth_chAp6: > When the process scheduler sees that there are networking tasks to do it runs the network bottom-half. This function pops packets off of the backlog queue, matches them to a known protocol (typically IP), and passes them to that protocol's receive function. The IP layer examines the packet for errors and routes it; the packet will go into an outgoing queue (if it is for another host) or up to the transport layer (such as TCP or UDP). This layer again checks for errors, looks up the socket associated with the port specified in the packet, and puts the packet at the end of that socket's receive queue.
The SYN packet has the same source dest IP address & port with the established connection, so what will happen in this case?
The server will silently drop the packet since it already has a connection in the ESTABLISHED state, one of the four values from (client-ip, src-port, server-ip, dest-port) must be different for the new SYN to be accepted.
The server will attempt a new connection.
in tech terms it will send a syn,ack packet and wait for the client to finish the tcp handshake
and open the connection.
http://en.wikipedia.org/wiki/Transmission_Control_Protocol
will explain the process alot better than me.
the server will send some information to identify the connection in its syn,ack packet.
and that information is used to keep that connection seperate from others.
Most the time, the ports will not be the same
but when it is, it can cause problems with low grade nat routers,
They try to rewrite that ports that are used, and can get the connections confused.