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<p><font face="Arial">Dear co-Scilabers,</font></p>
<p>I met a strange Scilab's bug this week-end. But today, i tried
with Octave, Matlab2016 and R, and i found the same strange
behavior. So, either i am missing something, or the bug affects
all these languages in the same way. It is reported @
<a class="moz-txt-link-freetext" href="http://bugzilla.scilab.org/15276">http://bugzilla.scilab.org/15276</a><br>
</p>
<p>In a few words, here it is:</p>
<p>The mantissa of any decimal number is recorded on 53 bits
(numbered #0 to #52) + 1 bit for the sign.<br>
This relative absolute accuracy sets the value of the
epsilon-machine :<br>
<tt>--> %eps == 2^0 / 2^52</tt><tt><br>
</tt><tt> ans =</tt><tt><br>
</tt><tt> T</tt><br>
From here, it comes, as expected:</p>
<p><tt>--> 2^52 + 1 == 2^52</tt><tt><br>
</tt><tt> ans =</tt><tt><br>
</tt><tt> F</tt><tt><br>
</tt><tt><br>
</tt><tt>--> 2^53 - 1 == 2^53<br>
ans =<br>
F<br>
<br>
--> 2^53 + 1 == 2^53</tt><tt> // (A)<br>
</tt><tt> ans =</tt><tt><br>
</tt><tt> T</tt><tt><br>
</tt><br>
Now comes the issue:<br>
In (A), the relative difference 1/2^53 is too small (< %eps) to
be recorded and to change the number. OK.<br>
Since 1 / (2^53 +2) is even smaller than 1 / (2^53), it should nor
make a difference. Yet, it does:</p>
<p><tt>--> (2^53 + 2^1) + 1 == (2^53 + 2^1)</tt><tt><br>
</tt><tt> ans =</tt><tt><br>
</tt><tt> F</tt><tt><br>
</tt><br>
How is this possible ??!<br>
But this occurs only when THIS bit #0 is set. For higher bits
(hebelow the #1 one), we get back to the expected behavior:<br>
<tt>--> (2^53 + 2^2) + 1 == (2^53 + 2^2)</tt><tt><br>
</tt><tt> ans =</tt><tt><br>
</tt><tt> T</tt><tt><br>
</tt><br>
So, when the bit #0 is set and we add a value on the bit "#-1",
the language somewhat behaves as if there was a "withholding"
value on the bit #0, and seems to toogle it.<br>
Is is a part of any IEEE floating point convention, or am i
missing anything, or is it a true bug?<br>
Again, R, Octave and Matlab behave exactly in the same way...</p>
<span id="result_box" class="short_text" lang="en"><span class="">Looking
forward to reading your thoughts,<br>
<br>
</span></span>Samuel<br>
<br>
<p><br>
</p>
<p><br>
</p>
<p><br>
</p>
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