- 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. -- You are receiving this mail because: You are the QA Contact for the bug.
Received on Wednesday, 5 February 2014 21:22:43 UTC