Subject: Re: [xsl] XSLT repetition constructs From: "Michael Kay mike@xxxxxxxxxxxx" <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx> Date: Fri, 8 Mar 2019 08:40:13 -0000 |
Because if $a is an empty sequence (which is the situation on the first call), then $a[last()] is an empty sequence, and ($x + ()) returns an empty sequence, not $x. Michael Kay Saxonica > On 8 Mar 2019, at 05:35, Mukul Gandhi gandhi.mukul@xxxxxxxxx <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx> wrote: > > Hi Mike, > > On Thu, Mar 7, 2019 at 8:49 PM Michael Kay mike@xxxxxxxxxxxx <mailto:mike@xxxxxxxxxxxx> <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx <mailto:xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx>> wrote: > Your solution could be rewritten > > fold-left($x, (), function($a, $b) {$a, ($a[last()], 0)[1] + $b}) > > I can confirm that, above expression evaluates correctly. > > But why does the following doesn't give the correct answer ? > > <xsl:variable name="inpValues" select="(1,2,3,4)" as="xs:integer*"/> > fold-left($inpValues, (), function($a, $b) {$a, $a[last()] + $b}) > > i.e why can't $a[last()] work instead your's ($a[last()], 0)[1] ? > > > > > -- > Regards, > Mukul Gandhi > XSL-List info and archive <http://www.mulberrytech.com/xsl/xsl-list> > EasyUnsubscribe <http://lists.mulberrytech.com/unsub/xsl-list/293509> (by email <>)
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