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 XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
Current Thread |
---|
|
<- Previous | Index | Next -> |
---|---|---|
Re: [xsl] Can sets have order?, David Carlisle | Thread | Re: [xsl] Can sets have order?, David Carlisle |
RE: [xsl] XSLTUK conference/tutoria, Kevin Jones | Date | Re: [xsl] XSLTUK conference/tutoria, Michael Beddow |
Month |