I don't think that's enough; I think it's much better to explicitly say that a server MUST provide a full NM resource for each NM.
If we just say that a filtered NM must have a "uses" attribute, then a server might define a filtered NM that uses a nonexistent resource ID. The associated cost maps would use that nonexistent resource as well. Yes, most people will assume that the dependent resource must exist -- but I don't think the protocol actually requires that. BTW, a sleazy implementor might even declare the filtered NM as "using" itself. In any case, we should say that a filtered NM MUST have a "uses" clause for a *full* NM. Otherwise a filtered NM is not very useful -- it would be in limbo, and a client could not connect it with any cost maps. Also, we should say that cost maps must "use" a full network map resource, not a filtered one. - Wendy From: "Y. Richard Yang" <[email protected]> Date: Sun, August 11, 2013 13:24 To: Wendy Roome <[email protected]> Cc: IETF ALTO <[email protected]> Subject: Re: [alto] required resources On Aug 9, 2013 11:16 AM, "Wendy Roome" <[email protected]> wrote: > > I have a few questions about required resources that need to be clarified, > especially in the case of multiple network maps. > > > 1. {10.1.1} says that a server MUST provide at least one full Network Map > resource. If a server has multiple network maps, it can satisfy that > requirement by providing a full NM resource for one map -- and it doesn't > have to be the default map -- and filtered network map resources for the > others. > > I'm not a big fan of requirements, but I do think a network map isn't very > useful unless the server is willing to give the full map. So I suggest > changing {10.1.1} to say, "An ALTO Server MUST provide a Network Map > resource for each Network Map." In -17, there is no "uses" for a filtered network map. Hence, one may not know from which base NM a filtered NM is derived. So with your proposal, if we require that each filtered NM MUST have a "uses" to declare the base NM, the problem is solved. Do I understand it correct?
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