[Asterisk-Dev] Asterisk / OpenH323 Bearer-Capability

[Michael Procter]

> We weren't attempting to change that particular aspect of the output,
> so it didn't directly affect our code, only the auxiliary
> understanding of it.  Thanks very much for pointing out the detail.

I realised that it wouldn't affect your problem, but thought it might
clear up your confusion.  I also wanted to reiterate that ITU specs
can be obtained for free (3 free ones when you register your email
address), since many people seem to be oblivious to this.

> Since you have a far firmer grasp on 
> the art of standard-reading than I do, perhaps you can shed 
> some light 
> on what that first (8th) bit in the 5th octet is supposed to 
> indicate?  
> The only reference to it I've seen is in the overall diagram, as "0/1 
> ext".  Is that supposed to indicate an "extended" interpretation or 
> something?

Yes.  When the bit is set to zero, the octet 'extends' to the next one.
In this case, coding 23 instead of A3 in octet 5 would indicate that octet
5a is present.  Similarly, coding the top bit of octet 5a as a zero would
indicate that 5b is present.  The last octet of what is laughingly called
octet 5 must have the top bit set, to indicate that it isn't extended
further.

This is described in Q.931 at the end of section 4.5.1:  Clause c of the
rules for coding variable length information elements starts by saying:

	An octet group is formed by using some extension mechanism.  The
	preferred extension mechanism is to extend an octet (N) through
	the next octet(s) (Na, Nb, etc.) by using bit 8 in each octet as
	an extension bit.  The bit value "0" indicates that the octet
	continues through the next octet.  The bit value "1" indicates
	that this octet is the last octet.  If one octet (Nb) is present,
	also the preceding octets (N and Na) must be present.

I hope this helps your understanding - spec reading can be an aquired taste.
I don't think it will directly help to solve your problem though.  Sorry.

Regards,

Michael