On Mon, Aug 18, 2025 at 03:49 PM -07, Eduard Zingerman wrote:
> 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.
Makes sense if I think about it like that. Thanks.
