Paolo<br><br><div class="gmail_quote">On Dec 3, 2007 7:59 AM, Paolo Bonzini <<a href="mailto:paolo.bonzini@lu.unisi.ch">paolo.bonzini@lu.unisi.ch</a>> wrote:<br><div> </div><blockquote class="gmail_quote" style="border-left: 1px solid rgb(204, 204, 204); margin: 0pt 0pt 0pt 0.8ex; padding-left: 1ex;">
But we're on violent agreement on this. emax is, with no possible way<br>to disagree, the highest exponent that can be seen for a valid float, in<br>the sign/mantissa/exponent representation of the float itself. This is
<br>unambiguous.</blockquote><div><br><br>Ok... then we can leave poor emax alone... for a while. Later, we can decide to agree violently on something else :)...<br><br> </div><blockquote class="gmail_quote" style="border-left: 1px solid rgb(204, 204, 204); margin: 0pt 0pt 0pt 0.8ex; padding-left: 1ex;">
It is just that the definition *of the mantissa* is not what you're<br>accustomed to. You believe that emax should be the exponent of a number<br>that is realizable with doubles, while LIA is clear that it is not so<br>
because it defines the mantissa to be <1.<br><div><div></div><div class="Wj3C7c"><br></div></div></blockquote><div><br><br></div></div>Hmmm... well, but even then, we have that the integer 2^emax is a double, but 2^(emax + 1) is not a double. Is perhaps the convenience of coincidence the confusing part here? I didn't mean to write such amounts in any particular mantissa notation.
<br><br>Andres.<br>