ConnectionOpen -> ConnectionStart - question about ports in use

[timprepscius]

Am I allowed to:
(pseudo)

var connection = connectionOpen();
connectionStart(connection, ..., callbackForServerA, serverA_IP, serverAPort);
connectionStart(connection, ..., callbackForServerB, serverB_IP, serverBPort);

Is this allowed? If I want to make connections to let's say, 10000 computers,
do I need to use more than one port ?

When is the port actually created (I assumed in connectionOpen)?

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In a nutshell, I am looking for a solution where a client uses only one UDP port to connect to multiple servers.
Just as the server would use one port to listen for connections.

Is this possible with QUIC?

Thanks,

-tim

[nibanks]

You can use a single port to practically start unlimited client connections. You do have to open separate connection objects/handles for each. The one extra step you need to set QUIC_PARAM_CONN_SHARE_UDP_BINDING to TRUE before calling start.

[timprepscius]

Thanks for this.

[timprepscius]

I have probably a super obvious question:

So, I'm reading through RpsClient.cpp

So let's say I:

1. connectionOpen
   with my connectionHandler
2. and then set the flag of the QUIC_PARAM_CONN_SHARE_UDP_BINDING

and then I do a
3. connectionStart to serverA
4. and connectionStart to serverB

Let's say serverB initiates a shutdown
SHUTDOWN_INITIATED_BY_PEER

I assume my connectionHandler will receive that event.
But how do I know whether it pertains to serverA or serverB?
I'm looking at msquic.h: QUIC_CONNECTION_EVENT
and I'm not seeing how to differentiate events for different connectionStarts

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Or am I getting this completely wrong?

Thanks,

-tim

[ThadHouse]

The HQUIC will be different, and each connection can also be passed its own context pointer that you can use to differentiate between the connections.

[timprepscius]

Oh, I reread nibanks comment, and it says, "You do have to open separate connection objects/handles for each."

Ah, so I am doing another open before each connectionStart. Sorry, brain skipped a few words.