Does addTrack require the streams argument?

[jan-ivar]

In 5.1 the processing model for addTrack, there's no mention of the streams argument. Is streams required? E.g. Does this work or cause an error (which error)?

navigator.mediaDevices.getUserMedia({ video: true })
.then(stream => {
  var track = stream.getVideoTracks()[0];
  stream.removeTrack(track);
  new RTCPeerConnection().addTrack(track);
})

Interestingly, #processing-remote-mediastreamtracks tolerates reception of stream-less tracks:

1. If streams is an empty list, create a new MediaStream object and add it to streams.

But if the code I wrote above fails, then step 5 seems redundant.
If the code I wrote above does not fail, then I have a larger question.

The larger question is: Can a live track be consumed outside of a stream (assuming PeerConnection is a consumer)?

The MediaStream and Capture spec suggests no, defining “consumer” to be solely of a stream.

Yet the WebRTC spec seems to play fast and loose with this rule, talking about tracks with "inputs" and "outputs" with no mention of streams:

• "The concepts of an input and output to a given MediaStreamTrack ..."
• "The RTP media API lets a web application send and receive MediaStreamTracks over a peer-to-peer connection. Tracks, when added to a RTCPeerConnection, result in signaling; when this signaling is forwarded to a remote peer, it causes corresponding tracks to be created on the remote side."

I think some clarification in the spec here would help.

[stefhak]

My understanding is that your code should not fail (it should be OK to send just a track).
But I agree to that the spec should be clarified.

[stefhak]

List discussion initiated: https://lists.w3.org/Archives/Public/public-webrtc/2015Oct/0073.html

[stefhak]

Not much input given on the list, but the input given seems to agree that it should be OK to do addTrack without have a stream argument. Closing.