llvm dev - Sep 2014 - [LLVMdev] Bug 16257 - fmul of undef ConstantExpr not folded to undef

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.

—Owen
> On Sep 16, 2014, at 10:30 AM, Oleg Ranevskyy <llvm.mail.list at
gmail.com> wrote:
> 
> Hi Duncan,
> 
> 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.
> 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.
> 
> 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.
> 
> This also applies to "fadd undef, undef" and "fadd %x,
undef". We can't rely on getting an arbitrary NaN here on ARMs.
> 
> Would you be able to confirm this please?
> 
> Thank you in advance for your time!
> 
> Kind regards,
> Oleg
> 
> On 10.09.2014 22:50, Duncan Sands wrote:
>> Hi Oleg,
>> 
>> On 01/09/14 18:46, Oleg Ranevskyy wrote:
>>> Hi Duncan,
>>> 
>>> I looked through the IEEE standard and here is what I found:
>>> 
>>> *6.2 Operations with NaNs*
>>> /"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"/.
>>> 
>>> *6.2.3 NaN propagation*
>>> /"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"./
>> 
>> thanks for finding this out.
>> 
>>> 
>>> 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.
>> 
>> Yes, folding "fadd %x, -0.0" to "%x" is correct. 
This implies that "fadd undef, undef" can be folded to
"undef".
>> 
>>> 
>>> 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:
>>> /"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. *This standard does not specify which of the
input NaNs will
>>> provide the payload*"/.
>>> 
>>> Thus, this makes it possible to fold "fadd %x, undef" to
a NaN. Is this right?
>> 
>> Yes, I agree.
>> 
>> Ciao, Duncan.
>> 
>>> 
>>> Oleg
>>> 
>>> On 01.09.2014 10:04, Duncan Sands wrote:
>>>> Hi Oleg,
>>>> 
>>>> On 01/09/14 15:42, Oleg Ranevskyy wrote:
>>>>> Hi,
>>>>> 
>>>>> Thank you for your comment, Owen.
>>>>> My LLVM expertise is certainly not enough to make such
decisions yet.
>>>>> Duncan, do you have any comments on this or do you know
anyone else who can
>>>>> decide about preserving NaN payloads?
>>>> 
>>>> 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.
>>>> 
>>>> 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.
>>>> 
>>>> Ciao, Duncan.
>>> 
>> 
>

On 16/09/14 19:37, Owen Anderson wrote:
> 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.
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.

Ciao, Duncan.
>
> —Owen
>
>> On Sep 16, 2014, at 10:30 AM, Oleg Ranevskyy <llvm.mail.list at
gmail.com> wrote:
>>
>> Hi Duncan,
>>
>> 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.
>> 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.
>>
>> 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.
>>
>> This also applies to "fadd undef, undef" and "fadd %x,
undef". We can't rely on getting an arbitrary NaN here on ARMs.
>>
>> Would you be able to confirm this please?
>>
>> Thank you in advance for your time!
>>
>> Kind regards,
>> Oleg
>>
>> On 10.09.2014 22:50, Duncan Sands wrote:
>>> Hi Oleg,
>>>
>>> On 01/09/14 18:46, Oleg Ranevskyy wrote:
>>>> Hi Duncan,
>>>>
>>>> I looked through the IEEE standard and here is what I found:
>>>>
>>>> *6.2 Operations with NaNs*
>>>> /"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"/.
>>>>
>>>> *6.2.3 NaN propagation*
>>>> /"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"./
>>>
>>> thanks for finding this out.
>>>
>>>>
>>>> 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.
>>>
>>> Yes, folding "fadd %x, -0.0" to "%x" is
correct.  This implies that "fadd undef, undef" can be folded to
"undef".
>>>
>>>>
>>>> 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:
>>>> /"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. *This standard does not specify which of
the input NaNs will
>>>> provide the payload*"/.
>>>>
>>>> Thus, this makes it possible to fold "fadd %x, undef"
to a NaN. Is this right?
>>>
>>> Yes, I agree.
>>>
>>> Ciao, Duncan.
>>>
>>>>
>>>> Oleg
>>>>
>>>> On 01.09.2014 10:04, Duncan Sands wrote:
>>>>> Hi Oleg,
>>>>>
>>>>> On 01/09/14 15:42, Oleg Ranevskyy wrote:
>>>>>> Hi,
>>>>>>
>>>>>> Thank you for your comment, Owen.
>>>>>> My LLVM expertise is certainly not enough to make such
decisions yet.
>>>>>> Duncan, do you have any comments on this or do you know
anyone else who can
>>>>>> decide about preserving NaN payloads?
>>>>>
>>>>> 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.
>>>>>
>>>>> 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.
>>>>>
>>>>> Ciao, Duncan.
>>>>
>>>
>>
>

Hi,

Thank you for all your helpful comments.

To sum up, below is the list of correct folding examples for fadd:
        (1)  fadd %x, -0.0            ->     %x
        (2)  fadd undef, undef    ->     undef
        (3)  fadd %x, undef         ->     NaN  (undef is a NaN which is 
propagated)

Looking through the code I found the "NoNaNs" flag accessed through an
instance of the FastMathFlags class.
(2) and (3) should probably depend on it.
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?

Other arithmetic FP operations (fsub, fmul, fdiv) also propagate NaNs. 
Thus, the same rules seem applicable to them as well:
---------------------------------------------------------------------
- fdiv:
        (4) "fdiv %x, undef" is now folded to undef.
        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.
        Moreover, this looks similar to (3) and can be folded to a NaN. 
/Is it worth doing?/

        (5) fdiv undef, undef    ->    undef
---------------------------------------------------------------------
- fmul:
        (6) fmul undef, undef    ->    undef
        (7) fmul %x, undef        ->    NaN or undef (undef is a NaN, 
which is propagated)
---------------------------------------------------------------------
- fsub:
        (8) fsub %x, -0.0           ->    %x  (if %x is not -0.0; works 
this way now)
        (9) fsub %x, undef        ->    NaN or undef (undef is a NaN, 
which is propagated)
      (10) fsub undef, undef   ->    undef
---------------------------------------------------------------------

I will be very thankful if you could review this final summary and share 
your thoughts.

Thank you.

P.S. Sorry for bothering you again and again.
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.

Kind regards,
Oleg

On 16.09.2014 21:42, Duncan Sands wrote:
> On 16/09/14 19:37, Owen Anderson wrote:
>> 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.
>
> 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.
>
> Ciao, Duncan.
>
>>
>> —Owen
>>
>>> On Sep 16, 2014, at 10:30 AM, Oleg Ranevskyy 
>>> <llvm.mail.list at gmail.com> wrote:
>>>
>>> Hi Duncan,
>>>
>>> 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.
>>> 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.
>>>
>>> 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.
>>>
>>> This also applies to "fadd undef, undef" and "fadd
%x, undef". We
>>> can't rely on getting an arbitrary NaN here on ARMs.
>>>
>>> Would you be able to confirm this please?
>>>
>>> Thank you in advance for your time!
>>>
>>> Kind regards,
>>> Oleg
>>>
>>> On 10.09.2014 22:50, Duncan Sands wrote:
>>>> Hi Oleg,
>>>>
>>>> On 01/09/14 18:46, Oleg Ranevskyy wrote:
>>>>> Hi Duncan,
>>>>>
>>>>> I looked through the IEEE standard and here is what I
found:
>>>>>
>>>>> *6.2 Operations with NaNs*
>>>>> /"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"/.
>>>>>
>>>>> *6.2.3 NaN propagation*
>>>>> /"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"./
>>>>
>>>> thanks for finding this out.
>>>>
>>>>>
>>>>> 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.
>>>>
>>>> Yes, folding "fadd %x, -0.0" to "%x" is
correct.  This implies that
>>>> "fadd undef, undef" can be folded to
"undef".
>>>>
>>>>>
>>>>> 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:
>>>>> /"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. *This standard does not specify which
of the
>>>>> input NaNs will
>>>>> provide the payload*"/.
>>>>>
>>>>> Thus, this makes it possible to fold "fadd %x,
undef" to a NaN. Is
>>>>> this right?
>>>>
>>>> Yes, I agree.
>>>>
>>>> Ciao, Duncan.
>>>>
>>>>>
>>>>> Oleg
>>>>>
>>>>> On 01.09.2014 10:04, Duncan Sands wrote:
>>>>>> Hi Oleg,
>>>>>>
>>>>>> On 01/09/14 15:42, Oleg Ranevskyy wrote:
>>>>>>> Hi,
>>>>>>>
>>>>>>> Thank you for your comment, Owen.
>>>>>>> My LLVM expertise is certainly not enough to make
such decisions
>>>>>>> yet.
>>>>>>> Duncan, do you have any comments on this or do you
know anyone
>>>>>>> else who can
>>>>>>> decide about preserving NaN payloads?
>>>>>>
>>>>>> 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.
>>>>>>
>>>>>> 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.
>>>>>>
>>>>>> Ciao, Duncan.
>>>>>
>>>>
>>>
>>
>
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