Hi,
  Any feedback on this point... for connection_driver, must I call tick() if 
I'm not interested in Heartbeats ?

Thanks,
- Tom

From: Tom Crowe
Sent: Wednesday, October 25, 2023 5:44 PM
To: users@qpid.apache.org
Subject: RE: external event loop

Thanks Cliff for your helpful answer to this topic way back on June 30th.

"If you must manage the AMQP connection's IO yourself, then the
connection_driver class is a solid choice. The gotchas are to make
sure to call the tick() function regularly to preserve heartbeats and
to note that both input to and output from the driver can generate new
events to be dispatched."

I am curious about the tick() function.
Is preserving heartbeats the ONLY purpose of tick() ?

I don't need to use the heartbeat feature in AMQP because there is a heartbeat 
at application level.
So perhaps I don't need to call the tick() function at all ?

Or does tick() service other needs e.g. message aging in queues OR timeouts 
related to settlement OR something else ?

Thanks.

From: Tom Crowe
Sent: Friday, June 30, 2023 11:09 AM
To: users@qpid.apache.org<mailto:users@qpid.apache.org>
Subject: RE: external event loop

Discovered that search function on mail lists is returning hits, albeit hidden 
from view.
Found enough info to work from.
Thanks,
- Tom

On 2023/06/28 14:12:14 Tom Crowe wrote:
> Hi,
>   I have a working prototype with qpid proton c++ v0.37 consuming an amqps 
> 1.0 data feed.
>
> Next, I would like to incorporate the prototype into existing software. That 
> software has other networking interfaces and has a select() based event loop.

>
> How can I incorporate my qpid proton prototype into this event loop?
>
> I am encouraged when I see the 'connection_driver' class which is 'designed 
> to work with a classic reactor'.
> I am discouraged by the fact that I see 'support for third party event loop' 
> as a nice-to-have feature in the README.
>
> Is it possible to incorporate my qpid proton c++ client into the existing 
> event loop?
> If yes, how?
> Are there any examples of this?
>
> Thanks for your help.
> - Tom.
>

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