On Fri, 7 Aug 2026, Richard Biener wrote: > On Fri, Aug 7, 2026 at 9:31 AM Alexander Monakov <[email protected]> wrote: > > > > > > On Wed, 5 Aug 2026, Roger Sayle wrote: > > > > > Likewise when x is NaN, 0.0 - x > > > may change the payload, but -x is guaranteed not to. > > > > Not saying you should drop the condition in your patch, but do we care > > about NaN payload propagation anywhere else? The HONOR_NANS et al. only > > seem to cover presence/absence of NaNs, not observability of payloads. > > I think we don't. -x will still turn sNaN into qNaN?
It may not, negation can only change the sign bit. It's a good point, 0-x would quieten the input sNaN and raise FE_INVALID for a sNaN but not qNaN, negation wouldn't. As for payload propagation, IEEE754 recommends that computational operations propagate one of the input NaNs. Therefore 0-x should just propagate payload of x when it's a qNaN, just like -x. But, conversely, propagating the input NaN can result in 0-x having the same sign as x, not the opposite (this is what actually happens on amd64, subsd preserves the sign of a NaN). Alexander
