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Hi Duncan,<br>
<br>
I looked through the IEEE standard and here is what I found:<br>
<br>
<b>6.2 Operations with NaNs</b><br>
<i>"For an operation with quiet NaN inputs, other than maximum and
minimum operations, if a floating-point result is to be delivered
the result shall be a quiet NaN which should be one of the input
NaNs"</i>.<br>
<br>
<b>6.2.3 NaN propagation</b><br>
<i>"An operation that propagates a NaN operand to its result and has
a single NaN as an input should produce a NaN with the payload of
the input NaN if representable in the destination format".</i><br>
<br>
Floating point add propagates a NaN. There is no conversion in the
context of LLVM's fadd. So, if %x in "fadd %x, -0.0" is a NaN, the
result is also a NaN with the same payload.<br>
<br>
As regards "fadd %x, undef", where %x might be a NaN and undef might
be chosen to be (probably some different) NaN, and a possibility to
fold this to a constant (NaN), the standard says:<br>
<i>"If two or more inputs are NaN, then the payload of the resulting
NaN should be identical to the payload of one of the input NaNs if
representable in the destination format. <b>This standard does
not specify which of the input NaNs will provide the payload</b>"</i>.<br>
<br>
Thus, this makes it possible to fold "fadd %x, undef" to a NaN. Is
this right?<br>
<br>
Oleg<br>
<br>
<div class="moz-cite-prefix">On 01.09.2014 10:04, Duncan Sands
wrote:<br>
</div>
<blockquote cite="mid:540444A8.8090501@deepbluecap.com" type="cite">Hi
Oleg,
<br>
<br>
On 01/09/14 15:42, Oleg Ranevskyy wrote:
<br>
<blockquote type="cite">Hi,
<br>
<br>
Thank you for your comment, Owen.
<br>
My LLVM expertise is certainly not enough to make such decisions
yet.
<br>
Duncan, do you have any comments on this or do you know anyone
else who can
<br>
decide about preserving NaN payloads?
<br>
</blockquote>
<br>
my take is that the first thing to do is to see what the IEEE
standard says about NaNs. Consider for example "fadd x, -0.0".
Does the standard specify the exact NaN bit pattern produced as
output when a particular NaN x is input? Or does it just say that
the output is a NaN? If the standard doesn't care exactly which
NaN is output, I think it is reasonable for LLVM to assume it is
whatever NaN is most convenient for LLVM; in this case that means
using x itself as the output.
<br>
<br>
However this approach does implicitly mean that we may end up not
folding floating point operations completely deterministically:
depending on the optimization that kicks in, in one case we might
fold to NaN A, and in some different optimization we might fold
the same expression to NaN B. I think this is pretty reasonable,
but it is something to be aware of.
<br>
<br>
Ciao, Duncan.
<br>
</blockquote>
<br>
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