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Re: mapping (Was: Re: Re: . in for)

Subject: Re: mapping (Was: Re: Re: . in for)
From: Jeni Tennison <jeni@xxxxxxxxxxxxxxxx>
Date: Thu, 10 Jan 2002 12:32:16 +0000
Re: mapping (Was: Re:  Re: . in for)
Hi Joerg,

> Jeni Tennison <jeni@xxxxxxxxxxxxxxxx> wrote:
>> Function names are indistinguishable from element names,
> [...]
> Well, i realized it somewhat too late as you have seen in my
> follow-up...

Sure :) I'd just sent my message when yours came through.

> This means you want lambda expressions.

Yes please :)

>>  - define a map expression (rather than a map() function)
>
> Note that in you proposal of a mapping operator
>   $coordinates -> (. * 2)
> the second operand actually *is* already a lambda expression. Hint:
> the . is not bound to the value of the context node as it would
> have been in ordinary expressions.

Yes, I thought so, but I'm not an expert on this functional
programming stuff so I didn't want to claim too much. I think it gives
you lambda expressions in a limited set of circumstances, without
adding lambda expressions as objects in the data model (which I
personally can't see happening in XPath 2.0 - there's just too much
else to sort out).

Now you could see that as a step in the right direction, or you could
see it as potentially undermining future efforts to introduce lambdas
as full objects within XPath/XSLT, because it doesn't introduce a
syntax that discriminates between a lambda expression and a normal
expression.

One possibility would be to use a keyword to indicate a lambda
expression, so you'd do:

  LambdaExpr  ::=  "expression" "(" Expr ")"
  MapExpr     ::=  Expr "map" LambdaExpr

and then have:

  $coordinates map expression (. * 2)
               map expression (if (position() mod 2)
                               then . + 50 else .)

We could cover up the fact we were doing this by pretending that the
syntax was simply:

  MapExpr     ::= Expr "map" "expression" "(" Expr ")"

But it would mean that in XPath 3.0 then we could change Expr to
include LambdaExpr. Then you could do (with a properly-defined sum()
function):

  sum(item, expression (item * price))

It's a bit sneaky, but it might just work...

Cheers,

Jeni

---
Jeni Tennison
http://www.jenitennison.com/


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