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