Im trying to write a server client program, where client sends request through UDP socket to a server, then server responds back to a client through TCP socket.
My question is, how can server establish a TCP connection back to a client after getting the request through UDP?
I'll add code parts on Monday, but I more interested in pseudocode for that. Does that mean that the client should listen on tcp port after sending udp request? So confused
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By definition, HTTP pipelining is a technique in which multiple HTTP requests are sent on a single TCP connection without waiting for the corresponding responses, according to Wikipedia, and some other resources. What is meant by TCP connection, is it the connection that starts when the client and server first communicate doing the 3 way handshake? Is it the same TCP stream? TCP stream meaning that the client and server communicate with each other using the same combination of ports.
Let say I have a server XYZ that listens on port 50000 for TCP clients and port 80 for HTTP clients. And on the other side, I have a client that uses a WebSocket to establish a socket connection to port 50000 and will use HTTP port 80 for the handshake (of course).
Now, when the client begins, it will first send a request to server XYZ via the HTTP port 80, and the server will receive its request on port 80 for the handshake and will send a response for welcome. So, in that case, both parties are using port 80 (of course).
Now, when the handshake is done, the standard documentation says that the same TCP connection that is used for HTTP request/response for handshake purposes is then converted to the TCP socket connection. Ok right.
But, but if this whole handshake process and TCP connection for the HTTP request/response uses port 80 the first time, and that the same TCP connection is converted to the TCP socket connection, and this whole process is done via port 80, then how does the same TCP connection get converted to port 50000 for the TCP socket on both parties? Does the client initialize another TCP connection internally for changing to port 50000?
So, can anyone tell how the port conversion is performed and works in the WebSocket from port 80 to a different port in both parties? How does a complete single socket connection get established on the different ports? How does the same TCP connection change/flip its ports?
A TCP socket connection cannot change ports at all. Once a connection has been established, its ports are locked in and cannot be changed. If you have a TCP socket connection on port 80, the only way to have a connection on port 50000 is to make a completely separate TCP socket connection.
A WebSocket cannot connect to port 80 and then switch to port 50000. However, an HTML page that is served to a browser from port 80 can contain client-side scripting that allows the browser to make a WebSocket object and connect it to port 50000. The two TCP connections (HTTP and WebSocket) are completely separate from each other (in fact, the HTTP socket connection does not even need to stay open once the HTML is served, since HTTP is a stateless protocol).
My friend told me that TCP doesn't need port forward.
What exactly he said is if the server is port forwarded the client can request something and the server will respond without port forward.
And I agreed with that even though I'm not sure it is true.
Later he said it is the same with UDP which I do not believe.
MAINLY THE QUESTION IS
If a client requests something on a server with TCP, does it need to be port forwarded to receive the response?
Also is it the same for UDP?
If the request from the client is a SYN for connect call then only a SYN-ACK response will be allowed through NAT. If the NAT supports simultaneous open connection then a SYN response from server will also be allowed through NAT. After the connection is established then client and server can communicate freely without any restriction. Port forwarding is not needed.
For UDP after a packet from client to server is sent then anything from server can be received through exact same public port of the NAT from which the first packet was sent. No port forwarding needed.
I don't fully understand when a TCP connection ends. That is, when a client sends a request to a server and the server responds, is that response part of the same TCP connection? Or is that response made through a brand new TCP connection?
A TCP connection ends when both sides have closed it. A response is sent over the same connection as the request, and there can be many request/response pairs over a single connection. Or none, just a download for example.
when a TCP client wants to establish a tcp connection with a tcp server
it needs to send SYN and then ACK
while tcp server only sends SYN/ACK
so they are different
but , after the 3_way handshaking,
is this connection symmetric, namely, are TCP client and server in equal status
for example, after the 3-way handshake, usually the client send packet first,
can TCP server send packet first?
No, the procedure is not different at all, but instead of sending a SYN then an ACK in two different packets, the servers concatenate them by sending them via a single packet!
In the other hand, remember always that the client/server nomenclature is relative. The server is the party that remains in listening mode, while the client is the party that initiates the connection ...
After the establishment of the connection, both parties are equivalent (same status as you said: ESTABLISHED). For that reason, both can send the FIN statement to close the connection ...
After the connection is established, both ends are indeed "symmetric". Who sends first is decided by the underlying protocol and differes amongst them.
For example, HTTP starts with the GET <path> HTTP/1.0 command, while other protocols let the server give a greeting line first, and only then the client sends its request.
So in general, both ends are free to send their stuff first.