- From: <bugzilla@jessica.w3.org>
- Date: Wed, 26 Dec 2012 17:10:32 +0000
- To: public-audio@w3.org
https://www.w3.org/Bugs/Public/show_bug.cgi?id=20505
--- Comment #6 from Marcos Caceres <w3c@marcosc.com> ---
(In reply to comment #5)
> (In reply to comment #4)
> > Just wanted to record a random thought here:
> >
> > getPort(DOMString id, optional MIDIPortType type);
> >
> > As in:
> > var port = midi.getPort("12e23f3", "input");
>
> But why would you do that? For any given id, the type is predetermined (and
> fixed).
Jussi said that the fingerprint might not be reliable (i.e., an input and and
output would have the same fingerprint): "It would be especially annoying if
faced with a UA that doesn't have enough data to give reliable fingerprints,
an application would've stored a fingerprint and assumed that the method would
return an output port, but was given an input port instead, resulting in an
error if it tried to send anything to it."
> If you were going down this path (of collapsing Input and Output
> together), I would expect:
>
> interface MIDIAccess {
> sequence<MIDIPort> listInputs ();
> sequence<MIDIPort> listOutputs ();
> MIDIPort getPort (MIDIPort or DOMString or short target);
> };
Sorry to again be a dumbass, but I really don't understand why you send a
MIDIPort to get a MIDIPort? Can you please explain the logic there?
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Received on Wednesday, 26 December 2012 17:10:34 UTC