[xslt] "|" operator

[Phil Shafer <phil juniper net>]

I have another odd xslt question:

What's the difference between 'select=". | one"'
and 'select="(.) | one"'?  The latter does what
I thought the former should have done.

Here's is a script that does three selects and calls a
template that for-each'es thru the selected nodes and
prints their name():

<?xml version="1.0" encoding="utf-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform";
                version="1.0">

  <xsl:template match="top">
    <out>
      <test1>
        <xsl:variable name="test1" select=". | one"/>
        <xsl:call-template name="out">
          <xsl:with-param name="test" select="$test1"/>
        </xsl:call-template>
      </test1>

      <test2>
        <xsl:variable name="test2" select="(.) | one"/>
        <xsl:call-template name="out">
          <xsl:with-param name="test" select="$test2"/>
        </xsl:call-template>
      </test2>

      <test3>
        <xsl:variable name="top" select="."/>
        <xsl:variable name="test2" select="$top | one"/>
        <xsl:call-template name="out">
          <xsl:with-param name="test" select="$test2"/>
        </xsl:call-template>
      </test3>

    </out>
  </xsl:template>

  <xsl:template name="out">
    <xsl:param name="test"/>

    <xsl:for-each select="$test">
      <found>
        <name><xsl:value-of select="name(.)"/></name>
      </found>
    </xsl:for-each>
  </xsl:template>

</xsl:stylesheet>

Here is my test input:

<top>
  <one/>
  <two/>
  <three/>
</top>

The output of xsltproc (formatted) is:

<?xml version="1.0"?>
<out>
  <test1>
    <found>
      <name>top</name>
    </found>
  </test1>
  <test2>
    <found>
      <name>top</name>
    </found>
    <found>
      <name>one</name>
    </found>
  </test2>
  <test3>
    <found>
      <name>top</name>
    </found>
    <found>
      <name>one</name>
    </found>
  </test3>
</out>

The problem is that ". | one" selects only <top>, not <one>.  But
if I wrap it in parens (test2) or use a variable (test3), it works
correctly.  Any insight appreciated.

Thanks,
 Phil