Re: [xsl] Can sets have order?

Subject: Re: [xsl] Can sets have order?
From: Wolfgang May <may@xxxxxxxxxxxxxxxxxxxxxxxxxx>
Date: Fri, 2 Feb 2001 14:00:52 +0100 (MET)
David Carlisle writes:
 > 
 > > Then if x' is obtained by exchanging X1 and x2 in X,
 > 
 > what does exchanging two elements of a set mean?
 > (This is a real question, I don't understand your point.)

What I meant is the following:

The node set is exported, e.g., as ASCII representation, and, e.g.,
put as a file on the Web or sent to somebody who should use it.
Then, in this file, the two nodes which are deep-equal are exchanged.

This does not effect anything the recipient does with the XML instance.

 > Every node in a note set is uniquely identifiable, so for example
 > 
 > <xsl:value-of select="//*[generate-id(.)=generate-id(current())]"/>
 > 
 > always returns the current node (if we are currently on an element node)

Yes, but only due to the use of an additional function.

Thus, I argued that XML node sets - without any additional information 
due to internal representation - are multisets.  When using the
internal representation, the set is not only not a multiset, but it is 
also ordered (which was the initial topic of the thread).

Wolfgang

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