5 Oct
2014
5 Oct
'14
8:27 p.m.
Lets say we have a synth and a sequencer connected via jack.
The sequencer spits out a command timed at position 4 (i'm deliberately keeping
the numbers generic), and the synth then knows that on the 4th sample it has to
do whatever the command says - again, I know it's a bit more 'interesting'
than
that. All fine and dandy, the synth bundles up it's data an chucks the buffer
full of audio to jack.
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.
Say now, I have this brilliant counter melody idea - well, it could happen :)
However, I don't have a keyboard with me. Still I can just fire up jack
keyboard and link it to the sequencer. Only now. the sequencer is pushing out
its MIDI data before it gets anything from the keyboard.
I know this sounds contrived but I have frequently done exactly this, using
alsa MIDI, and I confess I've no idea what actually happens in that situation
either!
--
Will J Godfrey
http://www.musically.me.uk
Say you have a poem and I have a tune.
Exchange them and we can both have a poem, a tune, and a song.
5 Oct
5 Oct
8:58 p.m.
On Sun, Oct 5, 2014 at 2:27 PM, Will Godfrey <willgodfrey(a)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.
9:38 p.m.
On Sun, 5 Oct 2014 14:58:36 -0400
Paul Davis <paul(a)linuxaudiosystems.com> wrote:
On Sun, Oct 5, 2014 at 2:27 PM, Will Godfrey
<willgodfrey(a)musically.me.uk>
wrote:
Thanks. That's a relief. I'd got the impression they were executed in the order
they were first registered.
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.
Never even thought of trying to do that!
--
It wasn't me! (Well actually, it probably was)
... the hard part is not dodging what life throws at you,
but trying to catch the good bits.
10:31 p.m.
On Sun, Oct 05, 2014 at 02:58:36PM -0400, Paul Davis wrote:
if you're silly enough to create a feedback loop,
then the order becomes
undefined.
You don't need to be very silly to do this. If you have
sequencer -> synth -> audio recorder
then if the sequencer and recorder are the same Jack client
you always get a loop. But the final result (provided there
are no other connections to the synth) is still defined: one
period of extra latency.
Ciao,
--
FA
A world of exhaustive, reliable metadata would be an utopia.
It's also a pipe-dream, founded on self-delusion, nerd hubris
and hysterically inflated market opportunities. (Cory Doctorow)
11:02 p.m.
On 5 Oct 2014, at 19:58, Paul Davis <paul(a)linuxaudiosystems.com> wrote:
On Sun, Oct 5, 2014 at 2:27 PM, Will Godfrey
<willgodfrey(a)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
3694
days inactive
3694
days old
linux-audio-dev@lists.linuxaudio.org
4 comments
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participants (5)
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Will J Godfrey