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Re: Looking for a shorter mapping expression

Subject: Re: Looking for a shorter mapping expression
From: Dimtre Novatchev <dnovatchev@xxxxxxxxx>
Date: Mon, 6 Dec 2004 20:21:44 +1100
Re:  Looking for a shorter mapping expression
Thanks, Mike.

Now this is clear.

Cheers,
Dimitre.


On Mon, 6 Dec 2004 09:14:04 -0000, Michael Kay <mike@xxxxxxxxxxxx> wrote:
> > It seemed to me that both constraints would be in force for any LHS
> > operand, because it happens that at the same time this LHS operand is
> > also the RHS operand of the preceding from left  "/".
> 
> I think it's a mistake to think of a path expression as a sequence of steps
> separated by "/". (It's therefore also a mistake for the XPath spec to speak
> of "the last step in a path expression".) It's better to think of "/" as a
> binary higher-order operator. So A/B/C/D is simply ((A/B)/C)/D. The
> left-hand operand of "/" is not necessarily sorted and deduplicated; it will
> only be sorted and deduplicated if it is itself one of the kinds of
> expression that returns its results sorted and deduplicated.
> 
> >
> > Because every E1 is also some E2 for another "/", then the above
> > applies fully for this E1, too -- in fact it has been first evaluated
> > as an E2 and only then it serves as an E1 for the next "/".
> 
> It's not true that every E1 is the right-hand operand of a "/" operator. For
> example, E1 might be the expression (author, title). Or it might be the
> expression reverse(ancestor::*).
> 
> 
> 
> Michael Kay
> http://www.saxonica.com/

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