[Bug middle-end/52082] New: Memory loads not rematerialized

[Bug middle-end/52082] New: Memory loads not rematerialized

[jakub at gcc dot gnu.org]

http://gcc.gnu.org/bugzilla/show_bug.cgi?id=52082

             Bug #: 52082
           Summary: Memory loads not rematerialized
    Classification: Unclassified
           Product: gcc
           Version: 4.7.0
            Status: UNCONFIRMED
          Keywords: missed-optimization, ra
          Severity: normal
          Priority: P3
         Component: middle-end
        AssignedTo: unassigned@gcc.gnu.org
        ReportedBy: jakub@gcc.gnu.org
                CC: vmakarov@gcc.gnu.org
            Target: x86_64-linux


On the following testcase at -O2 (distilled from genautomata.c):

struct S { unsigned long *s1; struct S *s2; };
int v1 __attribute__((visibility ("hidden")));
struct T
{
  int a, b, c;
} *v2 __attribute__((visibility ("hidden")));
struct S **v3 __attribute__((visibility ("hidden")));
struct S **v4 __attribute__((visibility ("hidden")));

int __attribute__((noinline, noclone))
foo (unsigned long *x, unsigned long *y, int z)
{
  int j, k, l;
  unsigned int i;
  struct S *m;

  for (j = 0; j < v1; j++)
    if (y[j])
      for (i = 0; i < 8 * sizeof (unsigned long); i++)
  if ((y[j] >> i) & 1)
    {
      k = j * 8 * sizeof (unsigned long) + i;
      if (k >= v2->c)
        break;
      for (m = (z ? v4 [k] : v3 [k]); m != ((void *)0); m = m->s2)
        {
          for (l = 0; l < v1; l++)
            if ((x [l] & m->s1 [l]) != m->s1 [l] && m->s1 [l])
              break;
          if (l >= v1)
            return 0;
        }
    }
  return 1;
}

tree LIM moves the loads from v2/v3/v4 before the loop, but unfortunately the
register pressure is high and the pseudos holding the v3/v4 pointers don't get
a a hard register and are immediately spilled to the stack.  I wonder whether
we couldn't instead just rematerialize them and put the original MEM loads into
the loop (assuming they don't alias with anything on the way, but that must be
the case here when LIM moved them there first, after all this loop doesn't have
any MEM stores at all).

[Bug rtl-optimization/52082] Memory loads not rematerialized

[pinskia at gcc dot gnu.org]

http://gcc.gnu.org/bugzilla/show_bug.cgi?id=52082

Andrew Pinski <pinskia at gcc dot gnu.org> changed:

           What    |Removed                     |Added
----------------------------------------------------------------------------
             Status|UNCONFIRMED                 |NEW
   Last reconfirmed|                            |2012-02-03
          Component|middle-end                  |rtl-optimization
     Ever Confirmed|0                           |1

--- Comment #1 from Andrew Pinski <pinskia at gcc dot gnu.org> 2012-02-03 02:29:47 UTC ---
Confirmed.

[Bug rtl-optimization/52082] Memory loads not rematerialized

[pinskia at gcc dot gnu.org]

https://gcc.gnu.org/bugzilla/show_bug.cgi?id=52082

Andrew Pinski <pinskia at gcc dot gnu.org> changed:

           What    |Removed                     |Added
----------------------------------------------------------------------------
           Severity|normal                      |enhancement

--- Comment #3 from Andrew Pinski <pinskia at gcc dot gnu.org> ---
One thing I noticed that LLVM does to reduce the register pressure is:
(z ? v4 [k] : v3 [k])

Gets pulled out of the loop such that it is:
tmpaddr = z ? v4 : v3;

and then inside the loop it does:
(tempaddr)[k]

GCC still has (I changed the bb order just so it is easier to see what is going
on):
  if (z_39(D) != 0)
    goto <bb 9>; [50.00%]
  else
    goto <bb 11>; [50.00%]

  <bb 11> [local count: 5427362]:
  _21 = v3.3_18 + _157;
  iftmp.1_40 = *_21;
  goto <bb 10>; [100.00%]

  <bb 9> [local count: 5427362]:
  _17 = v4.2_14 + _157;
  iftmp.1_41 = *_17;

  <bb 10> [local count: 10854724]:
  # m_8 = PHI <iftmp.1_40(11), iftmp.1_41(9)>
  if (m_8 != 0B)
    goto <bb 12>; [94.50%]
  else
    goto <bb 19>; [5.50%]

we should able to do the similar it seems and need two less registers; one to
hold z and one to hold either v3 or v4.  This won't be enough for this testcase
but it will be something.