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RE: [xsl] Best way to count siblings

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> count(../*) -1 is shorter, you;d have to ask an implementer which is
> more efficient. Often the siblings are the current node list in which
> case last()-1 is likely to be the most efficient.

last() can be very efficient or very inefficient depending on the
circumstances. XSLT processors often use lazy evaluation, in which the nodes
are processed one-at-a-time without ever reading the whole list into memory.
Using last() can break this pipelined execution model, forcing the processor
to allocate memory for the entire list of say a million nodes - or to
evaluate the whole expression twice, if it decides that's better. Saxon uses
both these techniques depending on the circumstances. But if the node list
is already in memory for some other reason, e.g. because it is being sorted,
then last() should be very fast.

The only way to tell is by measuring.

Michael Kay
http://www.saxonica.com/

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