[Bug tree-optimization/72443] VRP derives poor range for "y = (int)x + C;" when x has an anti-range

[rguenth at gcc dot gnu.org]

https://gcc.gnu.org/bugzilla/show_bug.cgi?id=72443

Richard Biener <rguenth at gcc dot gnu.org> changed:

           What    |Removed                     |Added
----------------------------------------------------------------------------
             Status|UNCONFIRMED                 |NEW
   Last reconfirmed|                            |2016-07-26
                 CC|                            |rguenth at gcc dot gnu.org
     Ever confirmed|0                           |1

--- Comment #2 from Richard Biener <rguenth at gcc dot gnu.org> ---
It works just fine for unsigned arithmetic - the ~INT_MIN is just from
undefined overflow behavior...

But yes, union_ranges at the moment doesn't try to do anything fancy but
has the "trivial" implementation:

      else if (*vr0type == VR_RANGE
               && vr1type == VR_RANGE)
        {
          /* The result is the convex hull of both ranges.  */
          if (operand_less_p (*vr0max, vr1min) == 1)
            {
              /* If the result can be an anti-range, create one.  */
              if (TREE_CODE (*vr0max) == INTEGER_CST
                  && TREE_CODE (vr1min) == INTEGER_CST
                  && vrp_val_is_min (*vr0min)
                  && vrp_val_is_max (vr1max))
                {

so we restrict the anti-range case to when both ends of the ranges are
MIN/MAX.  Enhancing this to minimize the number of elements in the result
would be indeed good (though generally we prefer ranges over anti-ranges
due to the weak anti-range handling).

Sth that would greatly simplify VRP (IMHO) is to dump anti-ranges from the
range representation and instead have the lattice value be composed of
the union of two ranges (that's more powerful for ranges _and_ it can
represent all anti-ranges) [just make it the union of N ranges with N
being configurable - it might be possible to ditch equivalences on the
way though they compose the range as intersections of N ranges]

Maybe just split union_ranges into two parts, one where it handles
all the overlap cases and anti-ranges and one where it chooses the
best representation from sth that is the union of N ranges (N <= 4).

Well.  Confirmed.