On Sat, 23 Sept 2023, 16:45 Chris Gary via Std-Proposals, <std-proposals@lists.isocpp.org> wrote:

All computable data can be made isomorphic to integers.It is always possible to define a total ordering.Even N-d points have a total ordering in the form of space-filling curves.N-d rational points are the same: Reduce the coordinates so they are unique, then either use zig-zag or interleave the bits numerator then denominator to get a Morton index for some weird N+1 or 2*N dimensional space (not the same in the spatial sense, but this at least allows them to be used in an ordered container).For all scalar types, spaceship could become int-valued and return the sign of the difference in the form of an int.

What does that mean for NaN <=> 1.0 ?

Let the ordering in the case of an 'int'-valued operator <=> be assumed the same as std::strong_order where it is not otherwise defined explicitly.In the context of deducing the result type of operator <=>, it could be allowed either for backwards compatibility. That, or implicit "cast" of "int" to a "std::strong_order" where it is required.

Frankly, I've avoided using operator <=> due to header issues and the strangeness of a formal operator returning a privileged type in namespace std. I wouldn't be surprised if I've missed a detail...

That seems like a rather silly self-imposed limitation.