MIDI player laggy

[daniloprates]

I'm trying to trigger some MIDI players on noteOn, but the behaviour is a bit strange.

The MIDI player starts immediately, but there's a delay until I hear the sounds.
Sometimes it also triggers the wrong player, and sometimes the stop doesn't work properly.

Is there anything I'm doing wrong?

Synth.deferCallbacks(false);

const mp0 = Synth.getMidiPlayer("mp0");
const mp1 = Synth.getMidiPlayer("mp1");

const players = [mp0, mp1];
const noteTriggers = [60, 62];
const triggers = [];

for (var i=0;i<players.length;i++) {
	triggers[noteTriggers[i]] = players[i];
}
function onNoteOn()
{
	const note = Message.getNoteNumber();
	if (!!triggers[note]) {
		Message.ignoreEvent(true);
		triggers[note].play(0);
	}
}
 function onNoteOff()
{
	if (!!triggers[note]) {
		Message.ignoreEvent(true);
        triggers[note].stop(0);
	}
}

[d.healey]

@daniloprates said in MIDI player laggy:

for (var i=0;i<players.length;i++) {

No need for the var

@daniloprates said in MIDI player laggy:

const note = Message.getNoteNumber();

Use a local here.

@daniloprates said in MIDI player laggy:

!!triggers[note]

Doesn't the first ! cancel out the second?

[daniloprates]

@d-healey said in MIDI player laggy:

note = Message.getNoteNumber();

!! returns a true instead of the value. Just a more secure way of checking if something exists.

I've just realized that there's a latency even when pressing play on the MIDI player directly, so it's not related to the code. The player starts immediately, but the sound takes a while to start.

[d.healey]

@daniloprates said in MIDI player laggy:

!! returns a true instead of the value. Just a more secure way of checking if something exists.

isDefined() is probably a better way, because !! might return the wrong value for invalid values.

[daniloprates]

@d-healey thanks for the heads up!

[dannytaurus]

@d-healey I think isDefined() is for different cases. Similar in many ways but different enough to throw you off at times.

The double quote !! turns any truthy/falsey value into a boolean. But isDefined() will consider a value of false as defined, and therefore return true.

[d.healey]

@dannytaurus said in MIDI player laggy:

But isDefined() will consider a value of false as defined,

Yeah that's what I meant by he might get invalid values, because he wants to check "if something exists" not if it's true or false - or at least that's what I got from the statement.

[dannytaurus]

@d-healey said in MIDI player laggy:

Yeah that's what I meant by he might get invalid values, because he wants to check "if something exists" not if it's true or false - or at least that's what I got from the statement.

Yeah, true (no pun intended). To check if something is defined I'd always use isDefined() because it's more readable and explicit.

But I do use !! a lot in other places because it's so consistent in what it returns.

And !! does return false for undefined, for reference.

Console.print(isDefined(somethingUndefined)); // 0
Console.print(!!somethingUndefined); // 0