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[Bug 24508] xsl:fork streamability rule #2

From: <bugzilla@jessica.w3.org>
Date: Tue, 04 Mar 2014 11:12:28 +0000
To: public-qt-comments@w3.org
Message-ID: <bug-24508-523-SoMlScXUkI@http.www.w3.org/Bugs/Public/>
https://www.w3.org/Bugs/Public/show_bug.cgi?id=24508

Abel Braaksma <abel.braaksma@xs4all.nl> changed:

           What    |Removed                     |Added
----------------------------------------------------------------------------
             Status|CLOSED                      |REOPENED
         Resolution|FIXED                       |---

--- Comment #3 from Abel Braaksma <abel.braaksma@xs4all.nl> ---
The bottom line of streamaiblity of xsl:fork is that xsl:sequences must be
grounded, and then the widest sweep counts. Hence, I suggest to simplify this
even further, as follows:

1. If the posture of any of the child xsl:sequence instructions is not
grounded, then roaming and free-ranging.

2. Otherwise, the posture is grounded and the sweep is the widest sweep of any
of the xsl:sequence children (or motionless if there are no children).

This then gets more in line with the xsl:map streamability rules, (which is
only slightly different because xsl:map-entry has different rules than
xsl:sequence).

----
If we decide to stick to the current rules, note:
The current rule #4 has a typo: "then the grounded..." => "then grounded...".

----
As a result of the new rules, the last paragraph has become redundant. The rule
about "exactly one child", for which I filed this bug, is now gone, which I
think is the right thing to do: why allow an exceptional case for which
xsl:fork is pointless anyway?

The last para, should probably go:

"The only case where xsl:fork is permitted to return streamed nodes is in the
case where only one of the xsl:sequence instructions is consuming (in which
case the xsl:fork instruction is pointless)." 

(I took the liberty to reopen the bug report, it was closed)

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