On Wed, Jan 24, 2007 at 03:30:07PM +0100, Bjoern Hoehrmann wrote:
> * Bjoern Hoehrmann wrote:
> >Your test case somewhat simplified:
> >
> > a = element a { empty }
> > b = element b { empty }
> > c = element c { empty }
> > d = element d { empty }
> >
> > e = a & b
> > f = c & d
> >
> > start = element root { e & f }
>
> As Daniel asked, this would be
>
> <grammar xmlns="http://relaxng.org/ns/structure/1.0";>
> <define name="a">
> <element name="a">
> <empty/>
> </element>
> </define>
> <define name="b">
> <element name="b">
> <empty/>
> </element>
> </define>
> <define name="c">
> <element name="c">
> <empty/>
> </element>
> </define>
> <define name="d">
> <element name="d">
> <empty/>
> </element>
> </define>
> <define name="e">
> <interleave>
> <ref name="a"/>
> <ref name="b"/>
> </interleave>
> </define>
> <define name="f">
> <interleave>
> <ref name="c"/>
> <ref name="d"/>
> </interleave>
> </define>
> <start>
> <element name="root">
> <interleave>
> <ref name="e"/>
> <ref name="f"/>
> </interleave>
> </element>
> </start>
> </grammar>
It doesn't look obvious to me that
<root><a/><c/><b/><d/></root>
should be accepted. This is equivalent to
<element name="root">
<interleave>
<interleave>
<element name="a">
<empty/>
</element>
<element name="b">
<empty/>
</element>
</interleave>
<interleave>
<element name="c">
<empty/>
</element>
<element name="d">
<empty/>
</element>
</interleave>
</interleave>
</element>
I don't see the reduction rule which would be equivalent to flattening out
to a single interleave. The interleaves are at that point in a binary form,
explain because I don't see how your draw that conclusion.
Daniel
--
Red Hat Virtualization group http://redhat.com/virtualization/
Daniel Veillard | virtualization library http://libvirt.org/
[EMAIL PROTECTED] | libxml GNOME XML XSLT toolkit http://xmlsoft.org/
http://veillard.com/ | Rpmfind RPM search engine http://rpmfind.net/
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