+2004-11-12 Sebastian Pop <pop@cri.ensmp.fr>
+
+ * tree-data-ref.c (analyze_subscript_affine_affine): Correctly
+ compute the first overlapping iterations.
+
+2004-11-12 Sebastian Pop <pop@cri.ensmp.fr>
+
+ PR middle-end/18005
+ * tree-data-ref.c (estimate_niter_from_size_of_data): Ensure
+ that arguments of EXACT_DIV_EXPR are INTEGER_CST.
+
2004-11-12 Steven Bosscher <stevenb@suse.de>
PR tree-optimization/18419
array_size = TYPE_SIZE (TREE_TYPE (opnd0));
element_size = TYPE_SIZE (TREE_TYPE (TREE_TYPE (opnd0)));
if (array_size == NULL_TREE
- || element_size == NULL_TREE)
+ || TREE_CODE (array_size) != INTEGER_CST
+ || TREE_CODE (element_size) != INTEGER_CST)
return;
data_size = fold (build2 (EXACT_DIV_EXPR, integer_type_node,
- array_size, element_size));
+ array_size, element_size));
if (init != NULL_TREE
&& step != NULL_TREE
if (j1 > 0)
{
- int last_conflict;
+ int last_conflict, min_multiple;
tau1 = MAX (tau1, CEIL (-j0, j1));
tau2 = MIN (tau2, FLOOR_DIV (niter - j0, j1));
- x0 = (i1 * tau1 + i0) % i1;
- y0 = (j1 * tau1 + j0) % j1;
+ x0 = i1 * tau1 + i0;
+ y0 = j1 * tau1 + j0;
+
+ /* At this point (x0, y0) is one of the
+ solutions to the Diophantine equation. The
+ next step has to compute the smallest
+ positive solution: the first conflicts. */
+ min_multiple = MIN (x0 / i1, y0 / j1);
+ x0 -= i1 * min_multiple;
+ y0 -= j1 * min_multiple;
+
tau1 = (x0 - i0)/i1;
last_conflict = tau2 - tau1;