XSLT Bitwise Logic

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I have an existing data set that utilizes an integer to store multiple values; the legacy front end did a simple bitwise check (e.g. in C#: iValues & 16 == 16) to see if

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执笔经年

2020-12-03 16:34

I haven't seen anything like this in XSLT / XPath. But I've found someone implementing this kind of operations manually. Maybe you could use the same approach, if you really need to.

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既然无缘

2020-12-03 16:36
XSLT does not define bitwise operations. If you want them, you have to roll your own.

If you're using XSLT specifically in .NET 2.0 context - that is, XslCompiledTransform class - then the simplest solution is to use a scripting block to introduce a C# function that does it, and then just call that:
```
<xsl:stylesheet xmlns:bitwise="urn:bitwise">

  <msxsl:script language="CSharp" implements-prefix="bitwise">
  <![CDATA[
    public int and(int x, int y) { return x & y; }
    public int or(int x, int y) { return x | y; }
    ...
  ]]>
  </msxsl:script>

  ...

  <xsl:value-of select="bitwise:and(@foo, @bar)" />
  <xsl:value-of select="bitwise:or(@foo, @bar)" />
  ...

</xsl:stylesheet>
```
Or you can define the more high-level primitives in a scripting block, such as HasFlag, and then use those.

When loading such a stylesheet, you will need to explicitly enable scripting in it:
```
XslCompiledTransform xslt = new XslCompiledTransform();
xslt.Load("foo.xsl",
  new XsltSettings { EnableScript = true },
  new XmlUrlResolver());
```
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面向向阳花

2020-12-03 16:41
```
<xsl:value-of select="for $n in (128, 64, 32, 16, 8, 4, 2, 1) return if ((floor($var div $n) mod 2) = 1) then 1 else 0"/>
```
this will return the binary array of your variable (stored in $var)

by the way I used XPath 2.0 to do this
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南笙

2020-12-03 16:42

XSLT is Turing-complete, see for example here or here, hence it can be done. But I have used XSLT only one or two times and can give no solution.

UPDATE

I just read a tutorial again and found a solution using the following fact. bitset(x, n) returns true, if the n-th bit of x is set, false otherwise.

bitset(x, n) := floor(x / 2^n) mod 2 == 1

The following XSLT

<?xml version="1.0" encoding="ISO-8859-1"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">

  <xsl:template match="/">
    <html>
      <body>
        <table border="1" style="text-align:center;">
          <tr bgcolor="#9acd32">
            <th>Number</th>
            <th>Bit 3</th>
            <th>Bit 2</th>
            <th>Bit 1</th>
            <th>Bit 0</th>
          </tr>
          <xsl:for-each select="numbers/number">
            <tr>
              <td>
                <xsl:value-of select="."/>
              </td>
              <td>
                <xsl:choose>
                  <xsl:when test="floor(. div 8) mod 2 = 1">1</xsl:when>
                  <xsl:otherwise>0</xsl:otherwise>
                </xsl:choose>
              </td>
              <td>
                <xsl:choose>
                  <xsl:when test="floor(. div 4) mod 2 = 1">1</xsl:when>
                  <xsl:otherwise>0</xsl:otherwise>
                </xsl:choose>
              </td>
              <td>
                <xsl:choose>
                  <xsl:when test="floor(. div 2) mod 2 = 1">1</xsl:when>
                  <xsl:otherwise>0</xsl:otherwise>
                </xsl:choose>
              </td>
              <td>
                <xsl:choose>
                  <xsl:when test="floor(. div 1) mod 2 = 1">1</xsl:when>
                  <xsl:otherwise>0</xsl:otherwise>
                </xsl:choose>
              </td>
            </tr>
          </xsl:for-each>
        </table>
      </body>
    </html>
  </xsl:template>
</xsl:stylesheet>

will turn this XML

<?xml version="1.0" encoding="ISO-8859-1"?>
<numbers>
  <number>0</number>
  <number>1</number>
  <number>2</number>
  <number>3</number>
  <number>4</number>
  <number>5</number>
  <number>6</number>
  <number>7</number>
  <number>8</number>
  <number>9</number>
  <number>10</number>
  <number>11</number>
  <number>12</number>
  <number>13</number>
  <number>14</number>
  <number>15</number>
</numbers>

into a HTML document with a table showing the bits of the numbers.

Number | Bit 3 | Bit 2 | Bit 1 | Bit 0 
---------------------------------------
   0   |   0   |   0   |   0   |   0 
   1   |   0   |   0   |   0   |   1 
   2   |   0   |   0   |   1   |   0 
   3   |   0   |   0   |   1   |   1 
   4   |   0   |   1   |   0   |   0 
   5   |   0   |   1   |   0   |   1 
   6   |   0   |   1   |   1   |   0 
   7   |   0   |   1   |   1   |   1 
   8   |   1   |   0   |   0   |   0 
   9   |   1   |   0   |   0   |   1 
  10   |   1   |   0   |   1   |   0 
  11   |   1   |   0   |   1   |   1 
  12   |   1   |   1   |   0   |   0 
  13   |   1   |   1   |   0   |   1 
  14   |   1   |   1   |   1   |   0 
  15   |   1   |   1   |   1   |   1

This is neither elegant nor nice in any way and there are probably much simpler solution, but it works. And given that it is my first contact with XSLT, I am quite satisfied.

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