On Tue, Nov 08, 2022 at 02:06:58PM +0100, Aldy Hernandez wrote:
> + gcc_checking_assert (!r.nan_signbit_p (sign1));
> + if (op1_nan && op2_nan)
> + {
> + // If boths signs agree, we could use that sign, but IEEE754
> + // does not guarantee this for a binary operator. The x86_64
> + // architure does keep the common known sign, but further tests
> + // are needed to see if other architectures do the same (i387
> + // long double, ARM/aarch64, PowerPC, s390,{,x}, RSICV, etc).
s/RSICV/RISCV/
> + // In the meantime, keep sign VARYING.
> + ;
> + }
> + else if (op1_nan)
> + {
> + if (op1.nan_signbit_p (sign1))
> + r.update_nan (sign1);
> + }
> + else if (op2_nan)
> + {
> + if (op2.nan_signbit_p (sign2))
> + r.update_nan (sign2);
> + }
Well, these cases also aren't guaranteed for binary operator.
I think a conforming implementation can say copy the NaN operand
to result and toggle the sign. Or, if the operand would be a sNaN,
it must turn it into a qNaN (don't remember right now if there are
requirements on what it can do with the mantissa which needs to change
for the sNaN -> qNaN difference at least, but whether it can just
generate a canonical qNaN or needs to preserve at least some bits),
but could e.g. clear or toggle the sign of the NaN as well.
Whether there are any such implementations or not is a question.
For the single qNaN operand case, it would surprise me if anybody
bothered to tweak the sign bit in any way, just copying the input
seems simpler to me, but for the sNaN -> qNaN conversion it wouldn't
surprise me that much.
Jakub
