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Re: Proposal for dealing with root

From: Martin Gudgin <martin.gudgin@btconnect.com>
Date: Tue, 23 Apr 2002 14:51:15 +0100
Message-ID: <001201c1eacd$f243cf10$b47ba8c0@zerogravitas>
To: "Jacek Kopecky" <jacek@systinet.com>
Cc: "XML Protocol Discussion" <xml-dist-app@w3.org>


----- Original Message -----
From: "Jacek Kopecky" <jacek@systinet.com>
To: "Martin Gudgin" <martin.gudgin@btconnect.com>
Cc: "XML Protocol Discussion" <xml-dist-app@w3.org>
Sent: Tuesday, April 23, 2002 12:53 PM
Subject: Re: Proposal for dealing with root

> Gudge, generally I like it, with just one remark inline.
>                    Jacek Kopecky
>                    Senior Architect, Systinet (formerly Idoox)
>                    http://www.systinet.com/
> On Tue, 23 Apr 2002, Martin Gudgin wrote:
>  >   'A graph node that has no inbound edges is a root of the graph.'
> What we needed is not a graph root but a serialization root. You
> define the former, I don't know at the moment how one could
> easily define the latter. Consider the following case:
>   <env:Body>
>     <m:foo encodingStyle="{soap-encoding}" id="id1">
>       <m:value>42</m:value>
>       <m:next ref="id1"/>
>     </m:foo>
>   </env:Body>
> It's a circular graph and the serialization root is not a graph
> root (this graph has no root).
> I'd propose that if we don't come up with a definition of a
> seriailzation root (as I cannot at the moment), we can just skip
> this as I don't feel a strong need for this definition anyway.

I think you just did define serialization root. It's m:foo in your example
above, that is, it is always the outermost element of the serialization (
given no independent elements )

> Below, I think you also mean to talk about serialization roots as
> opposed to graph roots.

Well, it's actually hard to seperate the two, but the graph has two roots
( and the serialization has two roots too )


Received on Tuesday, 23 April 2002 09:51:18 UTC

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