Re: What happened to onaddstream?

On 24/06/15 12:17, Silvia Pfeiffer wrote:
>
> On 24 Jun 2015 5:09 pm, "Stefan Håkansson LK"
> <stefan.lk.hakansson@ericsson.com
> <mailto:stefan.lk.hakansson@ericsson.com>> wrote:
>  >
>  > On 23/06/15 13:07, Silvia Pfeiffer wrote:
>  >
>  > >
>  > > I did some more digging.
>  > >
>  > > Reading the spec, I stumble over both an ontrack and an onaddtrack
>  > > event. The ontrack event is used in Example 4 to replace the
>  > > onaddstream event from earlier on the RTCPeerConnection. The
>  > > onaddtrack fires on MediaStream directly.
>  > > So, we're really talking about the ontrack event here.
>  >
>  > Right.
>  >
>  > >
>  > > It seems to me that ontrack has the exact same definition as
>  > > onaddstream, except it is a RTCTrackEvent (and not a MediaStreamEvent)
>  > > and can report the addition of multiple MediaStream objects in one go.
>  > >
>  > > http://w3c.github.io/webrtc-pc/#idl-def-RTCTrackEvent
>  > > I'm actually a bit confused about this description: it seems that a
>  > > track can be part of multiple streams there. That's not how I
>  > > understand MediaStreams to work - rather, a track would always be part
>  > > of one stream, but a stream could have multiple tracks.
>  >
>  > I think both ways are OK. A MediaStream is basically a collection of
>  > MediaStreamTrack's, but a MediaStreamTrack can be part of more than one
>  > MediaStream.
>  >
>  > Just assume you have MediaStream's A and B, and a MediastreamTrack X,
>  > AFAIK nothing forbids
>  >
>  > A.addTrack(X);
>  > B.addTrack(X);
>  >
>
> Oh and that doesn't clone the track?

Not the way I read http://w3c.github.io/mediacapture-main/#methods

> Feels wrong...
>
> I'm particularly thinking about e.g. A being a stream between two
> endpoints X and Y and B being a stream between Y and Z. How can it still
> be the same track?

I don't think there is anything stopping you from forwarding a track 
received over one PeerConnection to another endpoint (over another 
PeerConnection). If the track is identity bound then nothing would come 
out at Z, but that's another story.

>
> Cheers,
> Silvia.
>

Received on Wednesday, 24 June 2015 10:32:46 UTC