s_roundPackToF32.c rounding to odd

[Kou66666]

Hi,

s_roundPackToF32.c, line 96 implement rounding to odd as following:

96 if ( roundingMode == softfloat_round_odd ) {
97 sig |= 1;

Does this means rod always or the fraction part bit0 to 1'b1?

The correct behavior of rod suppose to be frac round up only:

1. round > 0.5
2. if 0.5 & (guard bit == 0)

Kou

[aswaterman]

I think you're describing a different rounding mode. Round to odd always makes the result odd if the amount rounded off is nonzero.

[Kou66666]

Hi Waterman,

Thanks for your reply. u are right with rounding to odd.

Another question is how to deal with overflow with rounding to odd.

s_roundPackToF32.c, line 58 ~ 64, It seems roundIncrement take 0 in case of rod.
then line 86, rod overflow will output infinite minus 1(! roundIncrement) e.g. the largest finite value.

IEEE754 defines the overflow value with other rounding mode, but without rod. How to know rod will take the largest finite value for overflow instead of infinite?

Thanks,
Kou

86 uiZ = packToF32UI( sign, 0xFF, 0 ) - ! roundIncrement;
87 goto uiZ;

58 if ( ! roundNearEven && (roundingMode != softfloat_round_near_maxMag) ) {
roundIncrement =
(roundingMode
== (sign ? softfloat_round_min : softfloat_round_max))
? 0x7F
: 0;
64 }

IEEE754-2019, page 53:
a) roundTiesToEven and roundTiesToAway carry all overflows to ∞ with the sign of the
intermediate result.
b) roundTowardZero carries all overflows to the format’s largest finite number with the sign of the
intermediate result.
c) roundTowardNegative carries positive overflows to the format’s largest finite number, and carries
negative overflows to −∞.
d) roundTowardPositive carries negative overflows to the format’s most negative finite number, and
carries positive overflows to +∞.