Re: [MMUSIC] 10 BUNDLE questions

[Justin Uberti <[e-mail]>]

You're looking at the handling of a=mid, not a=group.

Section 9.2 indicates the handling of a=group, which seems to indicate a
subset is allowed:


   A SIP entity that receives an offer that contains an "a=group" line
   with semantics that are understood MUST return an answer that
   contains an "a=group" line with the same semantics.  The
   identification-tags contained in this "a=group" line MUST be the same
   as those received in the offer, or a subset of them (zero
   identification-tags is a valid subset).  When the identification-tags
   in the answer are a subset, the "group" value to be used in the
   session MUST be the one present in the answer.




On Mon, Mar 18, 2013 at 12:08 AM, Harald Alvestrand <[e-mail]>wrote:

>  On 03/18/2013 04:19 AM, Justin Uberti wrote:
>
>
>
>>
>>    1. If the BUNDLE mids in the answer doesn't match the BUNDLE mids in
>>    the offer, what happens? (assume fail)
>>
>>   I think this violates RFC 5888 section 9.1, so it should be a FAIL.
>>
>
>  According to 9.2, the only restriction is the answer mids must be a
> subset of the offer, so we would have to specify an exact match directly.
> But I think we do want to impose this restriction, or else we can't BUNDLE
> a new m-line until we know the other side wants to BUNDLE it as well.
>
>
> The important phrase from 9.1 is
>
>    If a media stream that contained a particular "mid" identifier in the
>    offer contains a different identifier in the answer, the application
>    ignores all of the "mid" and "group" lines that might appear in the
>    session description.  The following example illustrates this
>    scenario.
>
> So if there's any mismatch, we don't get BUNDLE negotiated. Negotiation
> might not fail, but we don't get BUNDLE (or any other form of grouping).
>
>
>