re PR middle-end/18005 (ICE with simple loop with VLA)

	* tree-data-ref.c (analyze_subscript_affine_affine): Correctly
	compute the first overlapping iterations.

	PR middle-end/18005
	* tree-data-ref.c (estimate_niter_from_size_of_data): Ensure
	that arguments of EXACT_DIV_EXPR are INTEGER_CST.

From-SVN: r90510

gcc/ChangeLog

+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

gcc/tree-data-ref.c

@@ -513,11 +513,12 @@ estimate_niter_from_size_of_data (struct loop *loop,
   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
@@ -1435,12 +1436,21 @@ analyze_subscript_affine_affine (tree chrec_a,
 
 		  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;