[Bug tree-optimization/59594] [4.9 Regression] wrong code (by tree vectorizer) at -O3 on x86_64-linux-gnu

[jakub at gcc dot gnu.org]

http://gcc.gnu.org/bugzilla/show_bug.cgi?id=59594

--- Comment #6 from Jakub Jelinek <jakub at gcc dot gnu.org> ---
On:
#define N 1024
int ia[N + 1];
int ib[N + 1];

void
f1 (void)
{
  int i;
  for (i = 0; i < N; i++)
    {
      ia[i + 1] = 1;
      ib[i] = ia[i];
    }
}

void
f2 (void)
{
  int i;
  for (i = 0; i < N; i++)
    {
      ia[i] = 1;
      ib[i] = ia[i + 1];
    }
}

void
f3 (void)
{
  int i;
  for (i = N - 1; i >= 0; i--)
    {
      ia[i + 1] = 1;
      ib[i] = ia[i];
    }
}

void
f4 (void)
{
  int i;
  for (i = N - 1; i >= 0; i--)
    {
      ia[i] = 1;
      ib[i] = ia[i + 1];
    }
}

we properly vectorize f2 and f3 where the write/read DDR is DDR_REVERSED_P and
not f1/f4.

On
#define N 1024
int ia[N + 1];
int ib[N + 1];

void
f1 (void)
{
  int i;
  for (i = 0; i < N; i++)
    {
      ia[i + 1] = 1;
      ia[i] = 2;
    }
}

void
f2 (void)
{
  int i;
  for (i = 0; i < N; i++)
    {
      ia[i] = 1;
      ia[i + 1] = 2;
    }
}

void
f3 (void)
{
  int i;
  for (i = N - 1; i >= 0; i--)
    {
      ia[i + 1] = 1;
      ia[i] = 2;
    }
}

void
f4 (void)
{
  int i;
  for (i = N - 1; i >= 0; i--)
    {
      ia[i] = 1;
      ia[i + 1] = 2;
    }
}

we don't vectorize f1 and f2, in both cases for the write/write DDR
DDR_REVERSED_P is false, and vectorize f3/f4, where DDR_REVERSED_P is true in
both cases.  f2 and f3 shouldn't be vectorizable (at least not as is, when we'd
be trying to vectorize the two stores just by putting a vector store at that
position), f1 and f4 can.  So, this leads me to believe that for write/write we
don't have a way to differentiate between the bad and good cases using dist > 0
&& DDR_REVERSED_P test.  In that case, I'd think best would be to not ignore
dist > 0 && DDR_REVERSED_P (ddr)
ddrs if (!DR_IS_READ (dra) && !DR_IS_READ (drb)).