[LLVMdev] Bug 16257 - fmul of undef ConstantExpr not folded to undef

[Oleg Ranevskyy]

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"./

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.

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?

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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