RE: [xsl] copy vs. copy-of performance in xsltproc

Subject: RE: [xsl] copy vs. copy-of performance in xsltproc
From: "Michael Kay" <mike@xxxxxxxxxxxx>
Date: Tue, 5 Oct 2004 11:38:11 +0100
Intrinsically one would expect copy-of to be a bit faster than a recursive
descent applying template rules to every node, but of course the actual
answer for a particular product (and source document) can only be obtained
by measurement. It may depend on how many other template rules there are.

If you have a performance problem, I think that any difference here is
unlikely to be the solution to it. (And if you don't, why trouble yourself?)

Michael Kay
http://www.saxonica.com/ 


> -----Original Message-----
> From: Werner, Wolfgang [mailto:mail@xxxxxxxxxxxxxxxxxxx] 
> Sent: 05 October 2004 09:45
> To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
> Subject: [xsl] copy vs. copy-of performance in xsltproc
> 
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> Hash: SHA1
> 
> Hi,
> 
> has any of you experience regarding the performance of copy 
> compared to
> copy-of in xsltproc?
> Right now I'm copying some parts of the source tree completly while
> filtering some other parts.
> I use the following template:
> 
> ~  <xsl:template match="/ | @* | node()">
> ~    <xsl:copy>
> ~      <xsl:apply-templates select="@* | node()"/>
> ~    </xsl:copy>
> ~  </xsl:template>
> 
> My question is, if I want to copy a complete element, is it 
> faster to use
> 
> ~  <xsl:template match='dontfilterme'>
> ~    <xsl:copy-of select='.'/>
> ~  </xsl:template>
> 
> or the default template above?
> How much does the complexity of the copied element influence 
> the speed?
> 
> Any pointers appreciated,
> Wolfgang
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