On Sun, Oct 5, 2014 at 2:27 PM, Will Godfrey <willgodfrey@musically.me.uk> 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