Re: [xsl] [XPath 3.0] How to do partial function application on the min() function?

[Dimitre Novatchev <dnovatchev@xxxxxxxxx>]

On Mon, Oct 29, 2012 at 2:50 PM, Costello, Roger L. <costello@xxxxxxxxx> wrote:
> Hi Folks,
>
> I would like to fix one argument of the min() function to 2 and leave the other argument undefined, thus producing a partially applied function. Then, I would like to give the partially applied function a value. Here's what I tried:
>
>             let $min2 := min((?,2))
>             return $min2(3)
>
> I get this error: Unexpected token "?" in path expression

Min has a single argument, which is a sequence.  Therefore:

>             let $min2 := min((?,2))

isn't correct, indeed, because the argument placeholder doesn't
substitute the whole and only argument of the function.

It seems to me the what you want is:

          let $min2 := $compose(min#1, $makeSequence(?,2))
            return
               $min2(3)

where $compose should be similar to what was already discussed in this
list a few days ago and $makeSequence() must be essentially the same
as LISP's cons() funxction.


Cheers,

Dimitre



>
> What's the correct way to accomplish this?
>
> /Roger
>



-- 
Cheers,
Dimitre Novatchev
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