Re: [xsl] XPath 3.0 How to implement the function composition operator?

[Michael Kay <mike@xxxxxxxxxxxx>]

compose is a function that takes two functions as input and produces a 
third function as output, so it looks like this:

$compose   :=  function($a as function(item()*) as item()*,
                        $b as function(item()*) as item()*)
                     as (function(item()*) as item()*)
{  function($c as item()*) as item()* { $b($a($c)) } }

(Or the other way around. I don't know which way Haskell does it.)

Michael Kay
Saxonica


On 15/10/2012 23:08, Costello, Roger L. wrote:
> Hi Folks,
>
> How is function composition implemented in XPath 3.0?
>
> Example: Suppose I want to compose these two function:
>
>      1. increment: this function increases its argument by 1.
>
>      2. double: this function multiplies its argument by 2.
>
> In Haskell I can compose the two functions like so:
>
>      f = double . increment
>
> And then I can apply the composed functions to an argument:
>
>      f 2
>
> The result is 6.
>
> How is f implemented in XPath 3.0?
>
> Here is my attempt, which is not correct:
>
>              let $increment :=  function($x as xs:integer) {$x + 1},
>                   $double        :=  function($y as xs:integer) {$y * 2},
>                   $compose   :=  function(
>                                                                 $a as function(item()*) as item()*,
>                                                                 $b as function(item()*) as item()*
>                                                              )
>                                                             as item()*
>                                                 {$b($a)},
>                  $f  :=  $compose($double, $increment)
>              return $f(2)
>
> What is the correct way?
>
> /Roger