Re: [xslt] number() floated?

[Vincent Lefevre <vincent+gnome vinc17 org>]

On 2012-01-03 20:53:40 +0000, Laurence Rowe wrote:
> On 16 December 2011 17:06, Konrad Korzeniowski <konrad pandur net> wrote:
> > I got following simple xslt file:
> >
> > <?xml version="1.0" encoding="UTF-8"?>
> >
> > <xsl:stylesheet version="1.0"
> > xmlns:xsl='http://www.w3.org/1999/XSL/Transform'>
> >     <xsl:template match="/">
> >     <xsl:value-of select="number('8.95')"/>
> >     </xsl:template>
> > </xsl:stylesheet>
> >
> >
> > running xsltproc with this xsl against any xml gives:
> > 8.949999999999999
> >
> > Is this expected? Shouldn't this be 8.85?
> >
> > Thanks for any explanation.
> 
> Not all decimals may be exactly represented in binary floating point.
> See: http://en.wikipedia.org/wiki/Floating_point#Representable_numbers.2C_conversion_and_rounding

Except that this seems to be a bug here:

$ /usr/bin/printf "%a\n" 8.95
0x8.f33333333333333p+0

This is the binary value represented internally.

$ /usr/bin/printf "%a\n" 8.949999999999999
0x8.f33333333332eb2p+0

This is quite different!

Now, if the XSLT spec says that the value should be rounded *downward*
to decimal with a fixed number of decimal digits, this would be correct.
But this would be a quite surprising choice.

-- 
Vincent Lefèvre <vincent vinc17 net> - Web: <http://www.vinc17.net/>
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