+/**
+ * This test case should provoke incompatible heuristical decision of the
+ * PBQP solver, which lead to infinity cost, i.e. no solution of the PBQP could
+ * found.
+ *
+ * It is necessary to manipulation some cost vectors to let the PBQP solver
+ * "crash". More details are shown below.
+ *
+ * This also seems more a missing feature of the PBQP solver than a counter
+ * example against the PBQP approach for instruction selection.
+ */
+
unsigned *block;
-unsigned *block1, *block2, *block3, *block4, *block5, *block6, *block7;
+unsigned *block1, *block2, *block3, *block4, *block5, *block6;
+unsigned *block7, *block8,*block9,*block10,*block11;
volatile unsigned arr[100];
-unsigned ca,cb,cc;
unsigned b = 3008;
+unsigned g1,g2,g3;
+unsigned g4,g5,g6;
+unsigned g7,g8;
+unsigned h1,h2,h3;
+unsigned h4,h5,h6;
+unsigned h7,h8,h9;
+unsigned k1,k2,k3;
+unsigned k4,k5,k6;
+unsigned kb1,kb2,kb3;
+unsigned kc1,kc2,kc3;
+unsigned kd1,kd2,kd3;
+unsigned ke1,ke2,ke3;
-// TODO what if we use unsigned* instead of char*?
unsigned k3_3(char* base, unsigned i1, unsigned i2, unsigned i3, unsigned k1, unsigned k2, unsigned k3)
{
char a1, a2, a3;
return 0;
}
-unsigned g1,g2,g3;
-unsigned g4,g5,g6;
-unsigned g7,g8,g9;
-unsigned g10,g11,g12;
-unsigned h1,h2,h3;
-unsigned h4,h5,h6;
-unsigned h7,h8,h9;
-unsigned k1,k2,k3;
-unsigned k4,k5,k6;
-unsigned k7,k8,k9;
-unsigned k10,k11,k12;
-
-void full_am(unsigned base, int index)
+void full_am(unsigned base, int index, unsigned base2, int index2, unsigned base3, int index3, unsigned base4, int index4, unsigned base5, int index5)
{
+ /*
+ * This is the core of this example.
+ * The following line can be done with one instruction
+ * (add with address mode) on x86.
+ * We provoke four heuristical decision on nodes of this address mode
+ * pattern. The mean idea is that the root (the add node) choose this
+ * pattern, but the decision on the shift const forbid these.
+ * To achieve this you have to manipulate the cost of the shift const by
+ * hand!
+ * The other two heuristical decision ensure, that the two heuristical
+ * decision above are separated, i.e. there are at least two irreducible
+ * nodes between them.
+ */
unsigned ca = arr[index] + base;
- /* user for shift const */
+ /*
+ * The following function ensure irreducible users of given nodes.
+ * All of these function have to be inlined.
+ */
+
+ /* users for shift const */
b = k3_3_am(block, h1, h2, h3, 2, 3, 4);
b = k3_3_am(block, h4, h5, h6, 2, 5, 6);
b = k3_3_am(block, h7, h8, h9, 2, 7, 8);
- b = k3_3_2(block1, 4 * index, g2, g3, 31, 32, 33);
- b = k3_3_2(block2, 4 * index, g5, g6, 34, 35, 36);
- b = k3_3_2(block6, 4 * index, g7, g8, 37, 38, 39);
- b = k3_3_2(block7, 4 * index, g9, g10, 40, 41, 42);
- //b = k3_3(base + 4 * index, base + 4 * index, g8, g9, 37, 38, 39);
-
- /* user for computed value */
- //b = k3_3(block1, ca, k2, k3, 7, 8, 9);
- //b = k3_3(block2, ca, k5, k6, 10, 11, 12);
- b = k3_3(block3, ca, k8, k9, 13, 14, 15);
- b = k3_3(block4, ca, k11, k12, 16, 17, 18);
- b = k3_3(block5, ca, k7, k10, 19, 20, 21);
+ /* users for symconst */
+ unsigned cb = arr[index2] + base2;
+ b = k3_3(block1, cb, kb1, kb2, 101, 102, 103);
+ unsigned cc = arr[index3] + base3;
+ b = k3_3(block2, cc, kc1, kc2, 111, 112, 113);
+ unsigned cd = arr[index4] + base4;
+ b = k3_3(block3, cd, kd1, kd2, 121, 122, 123);
+ unsigned ce = arr[index5] + base5;
+ b = k3_3(block4, ce, ke1, ke2, 131, 132, 133);
+
+ /* users for offset */
+ b = k3_3_2(block5, 4 * index, g1, g2, 31, 32, 33);
+ b = k3_3_2(block6, 4 * index, g3, g4, 34, 35, 36);
+ b = k3_3_2(block7, 4 * index, g5, g6, 37, 38, 39);
+ b = k3_3_2(block8, 4 * index, g7, g8, 40, 41, 42);
+
+ /* users for computed value */
+ b = k3_3(block9, ca, k1, k2, 13, 14, 15);
+ b = k3_3(block10, ca, k3, k4, 16, 17, 18);
+ b = k3_3(block11, ca, k5, k6, 19, 20, 21);
}
int main(int argc, char **argv) {
projects / libfirm / blobdiff