On 5 Oct 2014, at 19:58, Paul Davis <[email protected]> wrote:
> On Sun, Oct 5, 2014 at 2:27 PM, Will Godfrey <[email protected]>
> wrote:
>
> But, what happens if the synth was registered with jack before the sequencer?
> Presumably it is now going to get it's MIDI data *after* it has already
> processed that callback.
>
> JACK clients are executed in the order required by their interconnections. If
> client A *sends* data to client B, then client A will always execute before
> client B.
>
> if you're silly enough to create a feedback loop, then the order becomes
> undefined.
Actually the order did get defined in the end: If you make a connection that
closes a feedback loop then all existing connections stay in the same order and
the new connection therefore becomes a ‘backwards’ connection. All new
connections either impose a new forward ordering constraint between clients
or, if that is impossible, become backwards connections to preserve the
existing order. Prior to this change adding a feedback connection or any
subsequent connection to a graph could alter the order that connected clients
were executing in but that doesn’t happen any more.
There’s an edge case though where you add a feedback connection and then remove
any forward connections. Last I looked jack 1 re-orders things so the feedback
connection becomes a forwards one (better for low latency) but jack 2 leaves
the connection backwards (better for “clickless connection”).
Simon Jenkins
Bristol, UK
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