Re: [xsl] Two versions of sum over node list by recursion--why and how does second one work?

["andrew welch" <andrew.j.welch@xxxxxxxxx>]

On 9/5/06, hanged.man@xxxxxxxxx <hanged.man@xxxxxxxxx> wrote:
> SECOND EXAMPLE:
>
> <xsl:stylesheet
>    xmlns:xsl="http://www.w3.org/1999/XSL/Transform";
>    version="1.0">
>
> <xsl:template name="total-sales-value">
>    <xsl:param name="list"/>
>    <xsl:choose>
>       <xsl:when test="$list">
>          <xsl:variable name="first" select="$list[1]"/>
>          <xsl:variable name="total-of-rest">
>             <xsl:call-template name="total-sales-value">
>                <xsl:with-param name="list" select="$list[position()!=1]"/>
>             </xsl:call-template>
>          </xsl:variable>
>          <xsl:value-of select="$first/sales * $first/price +
> $total-of-rest"/>
>       </xsl:when>
>       <xsl:otherwise>0</xsl:otherwise>
>    </xsl:choose>
> </xsl:template>
> QUESTION:
>
> In Mr. Kay's example, why does the variable $total-of-rest indeed contain
> the total
> and not just the result of the addition in the last recursion?

It's as the recursion unravels - it essentially adds it from last item
to the first.  It's a great example of a mind-bending recursive
programming :)

cheers
andrew