On Mon, 2025-08-18 at 20:23 +0200, Jakub Sitnicki wrote:
> On Fri, Aug 15, 2025 at 07:35 PM +0530, Nandakumar Edamana wrote:
>
> [...]
>
> > @@ -155,6 +163,14 @@ struct tnum tnum_intersect(struct tnum a, struct tnum b)
> > return TNUM(v & ~mu, mu);
> > }
> >
> > +struct tnum tnum_union(struct tnum a, struct tnum b)
> > +{
> > + u64 v = a.value & b.value;
> > + u64 mu = (a.value ^ b.value) | a.mask | b.mask;
> > +
> > + return TNUM(v & ~mu, mu);
> > +}
> > +
>
> Not sure I follow. So if I have two tnums that represent known contants,
> say a=(v=0b1010, m=0) and b=(v=0b0101, m=0), then their union is an
> unknown u=(v=0b0000, m=0b1111)?
Yes, because a and b have no bits in common.
As far as I understand, tnum_union() computes a tnum that is a
superset of both `a` and `b`. Maybe `union` is not the best name.
