- From: <bugzilla@jessica.w3.org>
- Date: Wed, 05 Feb 2014 21:22:41 +0000
- To: public-qt-comments@w3.org
https://www.w3.org/Bugs/Public/show_bug.cgi?id=24532
Bug ID: 24532
Summary: Streamability of NodeComp, before/after operator
Product: XPath / XQuery / XSLT
Version: Last Call drafts
Hardware: PC
OS: Windows NT
Status: NEW
Severity: normal
Priority: P2
Component: XSLT 3.0
Assignee: mike@saxonica.com
Reporter: abel.braaksma@xs4all.nl
QA Contact: public-qt-comments@w3.org
This is somewhat similar as the previous bug 24531. The same that was written
there might apply here. In addition:
How can a processor possibly check using forward-only streaming whether one
node is before the other? There is only two situations using streaming nodes on
both operands, that is currently streamable and that is when one of the
operands is motionless and the other is consuming, or when both are motionless.
a) foo >> bar
b) foo[@bar << @zzz]
c) @foo << bar
d) foo << .
e) ancestor-or-self::foo >> bar
Hmm, while writing this down, I am finding out the obvious logic here: one of
the nodes will always be in the current stack, which is always before any
consuming expression. And if both are motionless, their relative position is
known.
Is this true always? Is there an easy proof for this? Common sense tells me
this is correct, but still, I find operators << and >> to "feel" very
free-ranging.
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Received on Wednesday, 5 February 2014 21:22:43 UTC