I want to convert the decimal integers returned by the XPath 2
function fn:string-to-codepoints into a sequence of strings that use
the conventional "U+..." hexadecimal notation of the Unicode standard.

Neither the XPath 2 spec nor the XSLT FAQ offered a complete
solution to this problem, so I developed my own. It is working, but
I wonder how to integrate these two functions, which convert a
decimal integer into its hexadecimal equivalent, into a single
function which still returns just one string:

<xsl:function name="my:decimal-to-hex" as="xs:string">
 <xsl:param name="decimalNumber"/>
 <xsl:sequence select="string-join(my:decimal-to-hex-
internal($decimalNumber),'')"/>
</xsl:function>

<xsl:function name="my:decimal-to-hex-internal" as="xs:string+">
 <xsl:param name="decimalNumber"/>
 <xsl:variable name="hexDigits" select="'0123456789ABCDEF'"/>
 <xsl:if test="$decimalNumber &gt;= 16">
   <xsl:sequence
     select="my:decimal-to-hex-internal(floor($decimalNumber div
16))"/>
 </xsl:if>
 <xsl:sequence
   select="substring($hexDigits, ($decimalNumber mod 16) + 1, 1)"/>
</xsl:function>

I don't think it is elegant at all, but you can collect the recursive
digits in a variable and then do this:

<xsl:function name="my:decimal-to-hex" as="xs:string">
  <xsl:param name="decimalNumber"/>
  <xsl:variable name="hexDigits" select="'0123456789ABCDEF'"/>
  <xsl:variable name="upperDigits">
    <xsl:if test="$decimalNumber &gt;= 16">
      <xsl:sequence select="string-join(my:decimal-to-
hex(floor($decimalNumber div 16)), '')"/>
    </xsl:if>
  </xsl:variable>
  <xsl:sequence select="string-join(($upperDigits,
substring($hexDigits, ($decimalNumber mod 16) + 1, 1)), '')"/>
</xsl:function>

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_______________________________________________________________
Michael M|ller-Hillebrand: Dokumentations-Technologie
Adobe Certified Expert, FrameMaker
Lvsungen und Training, FrameScript, XML/XSL, Unicode
Blog: http://cap-studio.de/ - Tel. +49 (9131) 28747