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Hi,<br>
<br>
Thank you for all your helpful comments.<br>
<br>
To sum up, below is the list of correct folding examples for fadd:<br>
(1) fadd %x, -0.0 -> %x<br>
(2) fadd undef, undef -> undef<br>
(3) fadd %x, undef -> NaN (undef is a NaN
which is propagated)<br>
<br>
Looking through the code I found the "NoNaNs" flag accessed through
an instance of the FastMathFlags class.<br>
(2) and (3) should probably depend on it.<br>
If the flag is set, (2) and (3) cannot be folded as there are no
NaNs and we are not guaranteed to get an arbitrary bit pattern from
fadd, right?<br>
<br>
Other arithmetic FP operations (fsub, fmul, fdiv) also propagate
NaNs. Thus, the same rules seem applicable to them as well:<br>
---------------------------------------------------------------------<br>
- fdiv:<br>
(4) "fdiv %x, undef" is now folded to undef. <br>
The code comment states this is done because undef might be a
sNaN. We can't rely on sNaNs as they can either be masked or the
platform might not have FP exceptions at all. Nevertheless, such
folding is still correct due to the NaN propagation rules we found
in the Standard - undef might be chosen to be a NaN and its payload
will be propagated.<br>
Moreover, this looks similar to (3) and can be folded to a
NaN. <i>Is it worth doing?</i><br>
<br>
(5) fdiv undef, undef -> undef<br>
---------------------------------------------------------------------<br>
- fmul:<br>
(6) fmul undef, undef -> undef <br>
(7) fmul %x, undef -> NaN or undef (undef is a
NaN, which is propagated)<br>
---------------------------------------------------------------------<br>
- fsub:<br>
(8) fsub %x, -0.0 -> %x (if %x is not -0.0;
works this way now)<br>
(9) fsub %x, undef -> NaN or undef (undef is a
NaN, which is propagated)<br>
(10) fsub undef, undef -> undef<br>
---------------------------------------------------------------------<br>
<br>
I will be very thankful if you could review this final summary and
share your thoughts.<br>
<br>
Thank you.<br>
<br>
P.S. Sorry for bothering you again and again. <br>
Just want to make sure I clearly understand the subject in order to
make correct code changes and to be able to help others with this in
the future.<br>
<br>
Kind regards,<br>
Oleg<br>
<br>
<div class="moz-cite-prefix">On 16.09.2014 21:42, Duncan Sands
wrote:<br>
</div>
<blockquote cite="mid:54187677.40708@deepbluecap.com" type="cite">On
16/09/14 19:37, Owen Anderson wrote:
<br>
<blockquote type="cite">As far as I know, LLVM does not try very
hard to guarantee constant folded NaN payloads that match
exactly what the target would generate.
<br>
</blockquote>
<br>
I'm with Owen here. Unless ARM people object, I think it is
reasonable to say that at the LLVM IR level we may assume that the
IEEE rules are followed.
<br>
<br>
Ciao, Duncan.
<br>
<br>
<blockquote type="cite">
<br>
—Owen
<br>
<br>
<blockquote type="cite">On Sep 16, 2014, at 10:30 AM, Oleg
Ranevskyy <a class="moz-txt-link-rfc2396E" href="mailto:llvm.mail.list@gmail.com"><llvm.mail.list@gmail.com></a> wrote:
<br>
<br>
Hi Duncan,
<br>
<br>
I reread everything we've discussed so far and would like to
pay closer attention to the the ARM's FPSCR register mentioned
by Stephen.
<br>
It's really possible on ARM systems that floating point
operations on one or more qNaN operands return a NaN different
from the operands. I.e. operand NaN is not propagated. This
happens when the "default NaN" flag is set in the FPSCR
(floating point status and control register). The result in
this case is some default NaN value.
<br>
<br>
This means "fadd %x, -0.0", which is currently folded to %x by
InstructionSimplify, might produce a different result if %x is
a NaN. This breaks the NaN propagation rules the IEEE standard
establishes and significantly reduces folding capabilities for
the FP operations.
<br>
<br>
This also applies to "fadd undef, undef" and "fadd %x, undef".
We can't rely on getting an arbitrary NaN here on ARMs.
<br>
<br>
Would you be able to confirm this please?
<br>
<br>
Thank you in advance for your time!
<br>
<br>
Kind regards,
<br>
Oleg
<br>
<br>
On 10.09.2014 22:50, Duncan Sands wrote:
<br>
<blockquote type="cite">Hi Oleg,
<br>
<br>
On 01/09/14 18:46, Oleg Ranevskyy wrote:
<br>
<blockquote type="cite">Hi Duncan,
<br>
<br>
I looked through the IEEE standard and here is what I
found:
<br>
<br>
*6.2 Operations with NaNs*
<br>
/"For an operation with quiet NaN inputs, other than
maximum and minimum
<br>
operations, if a floating-point result is to be delivered
the result shall be a
<br>
quiet NaN which should be one of the input NaNs"/.
<br>
<br>
*6.2.3 NaN propagation*
<br>
/"An operation that propagates a NaN operand to its result
and has a single NaN
<br>
as an input should produce a NaN with the payload of the
input NaN if
<br>
representable in the destination format"./
<br>
</blockquote>
<br>
thanks for finding this out.
<br>
<br>
<blockquote type="cite">
<br>
Floating point add propagates a NaN. There is no
conversion in the context of
<br>
LLVM's fadd. So, if %x in "fadd %x, -0.0" is a NaN, the
result is also a NaN
<br>
with the same payload.
<br>
</blockquote>
<br>
Yes, folding "fadd %x, -0.0" to "%x" is correct. This
implies that "fadd undef, undef" can be folded to "undef".
<br>
<br>
<blockquote type="cite">
<br>
As regards "fadd %x, undef", where %x might be a NaN and
undef might be chosen
<br>
to be (probably some different) NaN, and a possibility to
fold this to a
<br>
constant (NaN), the standard says:
<br>
/"If two or more inputs are NaN, then the payload of the
resulting NaN should be
<br>
identical to the payload of one of the input NaNs if
representable in the
<br>
destination format. *This standard does not specify which
of the input NaNs will
<br>
provide the payload*"/.
<br>
<br>
Thus, this makes it possible to fold "fadd %x, undef" to a
NaN. Is this right?
<br>
</blockquote>
<br>
Yes, I agree.
<br>
<br>
Ciao, Duncan.
<br>
<br>
<blockquote type="cite">
<br>
Oleg
<br>
<br>
On 01.09.2014 10:04, Duncan Sands wrote:
<br>
<blockquote 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
<br>
about NaNs. Consider for example "fadd x, -0.0". Does
the standard specify
<br>
the exact NaN bit pattern produced as output when a
particular NaN x is
<br>
input? Or does it just say that the output is a NaN?
If the standard doesn't
<br>
care exactly which NaN is output, I think it is
reasonable for LLVM to assume
<br>
it is whatever NaN is most convenient for LLVM; in this
case that means using
<br>
x itself as the output.
<br>
<br>
However this approach does implicitly mean that we may
end up not folding
<br>
floating point operations completely deterministically:
depending on the
<br>
optimization that kicks in, in one case we might fold to
NaN A, and in some
<br>
different optimization we might fold the same expression
to NaN B. I think
<br>
this is pretty reasonable, but it is something to be
aware of.
<br>
<br>
Ciao, Duncan.
<br>
</blockquote>
<br>
</blockquote>
<br>
</blockquote>
<br>
</blockquote>
<br>
</blockquote>
<br>
</blockquote>
<br>
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