I've seen many answers similar to this one in regards to serversockets in java: "Let's say you have a server with a serversocket on port 5000. Client A and Client B will be connecting to our server.
Client A sends out a request to the Server on port 5000. The port on Client A's side is chosen by the Operating System. Usually, the OS picks the next port that is available. The starting point for this search is the previously-used port number + 1 (so for instance if the OS happened to us port 45546 recently, the OS would then try 45547).
Assuming there are no connection problems, the Server receives Client A's request to connect on port 5000. The Server then opens up its own next available port, and sends that to the client. Here, Client A connects to the new port, and the server now has port 5000 available again."
I've seen answers like this in multiple questions on stackoverflow about how a different port is used in the returned socket of the accept() than the port that the ServerSocket is listening on. I was always under the impression that TCP is identified by the quartet of information:
Client IP : Client Port and Server IP : Server Port ->protocol too (to distinguish TCP from UDP)
So why would the accept() need to return a socket bound to a different port? Doesn't the quartet of information sent in every header distinguish multiple connections to the same server port from different machines enough where it would not need to use different ports on the server machine for communication?
You are correct in the TCP packet header's information. It contains:
Client IP | Client Port | Server IP | Server Port | Protocol
Or, more appropriately (since client/server become confusing when you think about bi-directional transfer):
Source IP | Source Port | Destination IP | Destination Port | Protocol
Multiple connections to the same server port will come from different ports on the client. An example may be:
0.0.0.0:45000 -> 220.127.116.11:80 0.0.0.0:45001 -> 18.104.22.168:80
The difference in client ports is enough to disambiguate the two sockets, and thus have two separate connections. There is no need for the server to open another socket on another port. It does receive a socket from the
accept method, but it's assigned to the same port and is now a route back to the newly accepted client.
FTP, on the other hand, does have a model where the server will open a new unprivileged port (> 1023) and send that back to the client for the client to connect to (this is referred to as "Passive FTP"). This is to resolve issues where the client is behind a firewall and can't accept incoming data connections from the server. However, this is not the case in a typical HTTP server (or any other standard socket implementation). It's functionality that is layered on top of FTP.