[testcase] IA-32 ICE due to not splitted insn

[Jakub Jelinek <jakub at redhat dot com>]

Hi!

The following testcase causes ICE on gcc-3_3-branch (as of today) and
gcc-3_3-rhl-branch.
The problem is that at -O scheduling is not done after reload and thus also
split_all_insns is not called after reload.
If reload creates a splittable instruction and HAVE_attr_length, nothing
until final_scan_insns splits it and final_scan_insns aborts:
        /* If the template is the string "#", it means that this insn must
           be split.  */
        if (template[0] == '#' && template[1] == '\0')
          {
            rtx new = try_split (body, insn, 0);

            /* If we didn't split the insn, go away.  */
            if (new == insn && PATTERN (new) == body)
              fatal_insn ("could not split insn", insn);

#ifdef HAVE_ATTR_length
            /* This instruction should have been split in shorten_branches,
               to ensure that we would have valid length info for the
               splitees.  */
            abort ();
#endif

Should shorten_branches call split_all_insns, or call it only when toplev.c
did not do it:
#ifdef INSN_SCHEDULING
  if (optimize == 0 || !flag_schedule_insns_after_reload)
#else
  if (1)
#endif
    split_all_insns (0)
or look at insn templates and try_split only if template[0] == '#'?

2003-08-25  Jakub Jelinek  <jakub@redhat.com>

	* gcc.dg/20030825-1.c: New test.

--- gcc/testsuite/gcc.dg/20030825-1.c.jj	2003-01-30 05:24:37.000000000 -0500
+++ gcc/testsuite/gcc.dg/20030825-1.c	2003-08-25 13:34:39.000000000 -0400
@@ -0,0 +1,97 @@
+/* { dg-do compile } */
+/* { dg-options "-O" } */
+
+struct A
+{
+  long a1;
+  double *a2;
+};
+
+struct B
+{
+  void *b1;
+  double b2, b3;
+  struct
+  {
+    int d1;
+    double d2;
+  } b4;
+};
+
+struct C
+{
+  struct A *c1;
+  void *c2;
+};
+
+long fn1 (struct A *, double);
+void fn2 (void *, const char *);
+double fn3 (double);
+double fn4 (double);
+int fn5 (void *, double, double);
+
+int
+foo (struct B *x)
+{
+  struct C *e = x->b1;
+  struct A *f = e->c1;
+  long g, h, i;
+  double *j, k;
+  g = fn1 (f, 0.5 * (x->b2 + x->b3)), h = g + 1, i = f->a1;
+  j = f->a2, k = x->b4.d2;
+  fn2 (x, "something");
+  if (g <= 0)
+    {
+      double l = j[2] - j[1];
+      if (l > 0.0 && l <= 0.02)
+        k = (x->b4.d1 == 1
+             ? ((1.0 / l) < 25 ? 25 : (1.0 / l))
+             : fn3 ((1.0 / l) < 25 ? 25 : (1.0 / l)));
+    }
+  else
+    {
+      double m = j[h] - j[g], n = 0.0, l = 0.0;
+      if (g > 1)
+        n = j[g] - j[g - 1];
+      if (h < i)
+        l = j[h + 1] - j[h];
+      if (n > 0.02)
+        n = 0;
+      if (m > 0.02)
+        m = 0;
+      if (l > 0.02)
+        l = 0;
+      if (m < n)
+        {
+          double o = m;
+          m = n;
+          n = o;
+        }
+      if (l < n)
+        {
+          double o = l;
+          l = n;
+          n = o;
+        }
+      if (l < m)
+        {
+          double o = l;
+          l = m;
+          m = o;
+        }
+      if (n != 0.0)
+        k = (x->b4.d1 == 1
+             ? ((1 / m) < 25 ? 25 : (1 / m))
+             : fn3 ((1 / m) < 25 ? 25 : (1 / m)));
+      else if (m != 0.0)
+        k = (x->b4.d1 == 1
+             ? ((2 / (m + l)) < 25 ? 25 : (2 / (m + l)))
+             : fn3 ((2 / (m + l)) < 25 ? 25 : (2 / (m + l))));
+      else if (l != 0.0)
+        k = (x->b4.d1 == 1
+             ? ((1 / l) < 25 ? 25 : (1 / l))
+             : fn3 ((1 / l) < 25 ? 25 : (1 / l)));
+    }
+  fn5 (e->c2, 0.5 * (x->b2 + x->b3), (x->b4.d1 == 1 ? k : fn4 (k)));
+  return 1;
+}

	Jakub