Re: XSLT repetition constructs

The best I could come up with was

tail(fold-left($x, 0, function($a, $b) {$a, $a[last()] + $b}))

Your solution could be rewritten

fold-left($x, (), function($a, $b) {$a, ($a[last()], 0)[1] + $b})

Another option is

fold-left(tail($x), $x[1], function($a, $b) {$a, $a[last()] + $b})

xsl:iterate is more intuitive though of course more verbose:

<xsl:iterate select="1 to 4">
  <xsl:param name="total" select="0"/>
  <xsl:variable name="new-total" select="$total + ."/>
  <xsl:sequence select="$new-total"/>
  <xsl:next-iteration>
    <xsl:with-param name="total" select="$new-total"/>
  </xsl:next-iteration>
</xsl:iterate>

Michael Kay
Saxonica

> On 7 Mar 2019, at 13:28, Martin Honnen martin.honnen@xxxxxx
<xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx> wrote:
>
> Am 07.03.2019 um 12:55 schrieb Michael Kay mike@xxxxxxxxxxxx:
>> A good simple use case for fold-left() is to accumulate a running total,
i.e. turn (1,2,3,4) into (1,3,6,10).
>>
>
> The example to simply compute the running total (e.g. map (1,2,3,4) to 10)
is in the spec with
>
>    fold-left((1 to 4), 0, function($a, $b) { $a + $b})
>
>
> But to map the whole sequence (1,2,3,4) with fold-left to a new sequence of
(1,3,6,10) I am already struggling to express that in a compact way, is
>
> fold-left(
>   (1 to 4),
>   (),
>   function ($a, $b) {  $a, if (empty($a)) then $b else $b + $a[last()] }
> )
>
> a good way? Or can the third argument, the function be expressed in a more
compact way?