RE: [xsl] Can sets have order?

[Dimitre Novatchev <dnovatchev at yahoo dot com>]

Wolfgang May wrote:

> Not the same node, yes, but nodes which have the same values for all
> their attributes and attributes of subelements (i.e., which are
> deep-equal).  Such nodes may be exchanged in the node set without
> changing the behavior of any XPath query.


Not quite so...

Some XPath expressions (defined on the nodes of a nodeset), in which 
position() is combined with some of the following:

count(preceding::node()) 
 or 
count(following::node()) 
 or 
count(ancestor::node())
 or
count(descendant::node())

... etc.

will be affected by even a single swap of two so called "deep equal"
nodes.

As David pointed out generate-id() returns (two) distinct values for
any pair of such "deep equal" nodes.

Any Xpath expression that evaluates to exactly one of the nodes in a
"deep equal" pair -- this expression will be different from any Xpath
expression that evaluates to the other node.

Dimitre Novatchev.

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