<html><head><meta http-equiv="Content-Type" content="text/html charset=iso-2022-jp"></head><body style="word-wrap: break-word; -webkit-nbsp-mode: space; -webkit-line-break: after-white-space; ">Ok. Looks reasonable.<div>-Jim<br><div><br><div><div>
<!--EndFragment--><div><div>On Jun 17, 2013, at 1:50 AM, Mike Cowlishaw <<a href="mailto:mfc@speleotrove.com">mfc@speleotrove.com</a>> wrote:</div><br class="Apple-interchange-newline"><blockquote type="cite">
<meta http-equiv="Content-Type" content="text/html; charset=iso-2022-jp">
<title></title>
<meta name="GENERATOR" content="MSHTML 10.00.9200.16576">
<div><!-- Converted from text/plain format --><p><font size="2"><br></font><font size="3" face="Arial">As promised, I've checked
754. The final sentence of 5.2 (Decimal exponent calculation)
has:<br><br> If x is infinite, Q(x) is +∞.<br><br>Therefore the
result of quantum(x) where x is infinite should be +∞ (not x).
This is consistent in that if x is finite the result is also not affected by the
sign of x.<br><br>For when x is qNaN or sNaN I can think of no reason why the
results would be different from any other single-argument function or
general operation.<br><br>Mike </font></p></div>
</blockquote></div><br></div></div></div></div></body></html>