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

2012-10-15 19:57:26
I thought that using the argument placeholder "?" could be used to
specify a more readable implementation.

However it seems tht Saxon EE 9.3.05 (coming with oXygen) doesn't
support argument place holders.

For this query:

         let $f := function($m as xs:integer, $n as xs:integer) as xs:integer
                         {$m + $n}
           return
               $f(5, ?)(3)

an error message is raised:

Unexpected token "?" in path expression
Start location: 24:0
URL: http://www.w3.org/TR/xpath20/#ERRXPST0003

Could someone, please, explain what is the issue with this expression?


Cheers,
Dimitre



On Mon, Oct 15, 2012 at 4:02 PM, Michael Kay <mike(_at_)saxonica(_dot_)com> 
wrote:
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

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-- 
Cheers,
Dimitre Novatchev
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