Re: [xsl] The output of evaluating an XSLT transform is the same regardless of the order in which output elements are evaluated. Right?

[Liam R E Quin <liam@xxxxxx>]

On Tue, 2010-04-13 at 10:12 -0700, Dimitre Novatchev wrote:
> [...] The evaluation of
> 
> f(g(x))
> 
> requires that g(x) be evaluated, before f()  can be evaluated.

If f() takes a sequence as an argument, and g() produces a sequence,
and each item in the result of f() depends only on the corresponding
item in g(), then you can decompose f(g(x))
    for $i in g(x) return f($i)
in other words, evaluating f of g of each item in th sequence in turn.

And given
function f($input as item*) as item*
{
    return $nput[1]
}
it's not necessary to evaluate g(x) further than necessary to generate
one single item.

> There are datatypes that (by definition) impose a particular ordering
> on their processing. Take a list. in order to access the Nth item one
> needs to get the tail of the list and then acces its N-1st item. The
> Nth item is the head of the N-1st tail in this processing.

XSLT does not, however, mandate such data structures. It's perfectly
possible to implement an n-item sequence as an array, such that
accessing the nth item is O(1) in complexity.

> Also, a child cannot be born before its parents are born.

Given
   <a>
    <b>
     <c />
    </b>
   </a>
as a literal element constructor, an implementation is free to construct
the result in any order.

The oldest man in the Bible died before his father ;)

Best,

Liam

-- 
Liam Quin - XML Activity Lead, W3C, http://www.w3.org/People/Quin/
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