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

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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@xxxxxxxxxxxx> 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
>



-- 
Cheers,
Dimitre Novatchev
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