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/*
* Copyright (c) 2008 Google Inc. All rights reserved.
*
* $Header: $
*/
#include "config.h"
#include "system.h"
#include "coretypes.h"
#include "tm.h"
#include "rtl.h"
#include "function.h"
#include "basic-block.h"
#include "tree-pass.h"
#include "icg.h"
/*
* TODO: the perm vector is only defined for subject nodes that are matched
* by the root of a pattern tree, where the pattern tree has >= 2 non-terminals.
* This lets us avoid overwriting legitimate permutations with the 0:0 permutation
* induced by chain rules.
*/
static
void suOrder2(icg_node *p, icg_node *x, icg_node *y,
icg_node **kid, int kids) {
int i, j;
int cost0, cost1;
icg_node *left;
icg_node *right;
p->perm_kids = 2;
j = 0;
for (i = 0; i < kids; i++)
if (kid[i] == x ||
kid[i] == y)
p->perm[j++] = i;
/* j should equal 2 */
if (j != 2)
fprintf(stderr, "\n" "j != 2\n");
left = kid[p->perm[0]];
right = kid[p->perm[1]];
cost0 = left->extra + left->freed > right->extra ?
left->extra : right->extra - left->freed + 1;
cost1 = right->extra + right->freed > left->extra ?
right->extra : left->extra - right->freed + 1;
if (cost0 > cost1) {
p->extra = cost1;
i = p->perm[0];
p->perm[0] = p->perm[1];
p->perm[1] = i;
}
else
p->extra = cost0;
p->freed = x->freed + y->freed;
}
/* see Knuth, Volume 4, Section 7.2.1.2, Algorithm L */
static
int next_perm(short *a, int n) {
int j, k, l, t;
j = n - 1;
while (a[j] >= a[j + 1])
j--;
if (j == 0) return 0;
l = n;
while (a[j] >= a[l])
l--;
t = a[j]; a[j] = a[l]; a[l] = t;
k = j + 1;
l = n;
while (k < l) {
t = a[k]; a[k] = a[l]; a[l] = t;
k++;
l--;
}
return 1;
}
/* Is this correct? Check closely. */
static
int cost_of_visit(icg_node *p, int n, icg_node **kid) {
int i;
short *perm = p->perm;
int cost = 0;
int spare = 0;
for (i = 0; i < n; i++) {
int freed = kid[perm[i]]->freed;
int extra = kid[perm[i]]->extra;
if (spare < extra) {
cost += extra - spare;
spare = freed + extra - 1;
}
else
spare += freed - 1;
}
return cost;
}
/* This is pretty gross. Can I find a better approach? */
static
void suOrder3(icg_node *p,
icg_node *x, icg_node *y, icg_node *z,
icg_node **kid, int kids) {
int i, j;
int best_index, best_cost;
int cost[MAX_KIDS];
p->perm_kids = 3;
j = 0;
for (i = 0; i < kids; i++)
if (kid[i] == x ||
kid[i] == y ||
kid[i] == z)
p->perm[j++] = i;
/* j should equal 3 */
if (j != 3)
fprintf(stderr, "\n" "j != 3\n");
i = 0;
do {
cost[i++] = cost_of_visit(p, 3, kid);
} while (next_perm(p->perm, 3));
best_cost = cost[0];
best_index = 0;
for (j = 1; j < i; j++)
if (cost[j] < best_cost) {
best_cost = cost[j];
best_index = j;
}
i = p->perm[0];
p->perm[0] = p->perm[2];
p->perm[2] = i;
for (i = 0; i < best_index; i++)
(void) next_perm(p->perm, 3);
p->extra = best_cost;
p->freed = x->freed + y->freed + z->freed;
}
static
void reduce_supairs(icg_node *p, NT goalNT) {
icg_node *kid[MAX_KIDS];
const RuleNumber rule = icg_burm_rule(p->state_label, goalNT);
const NT *nts = icg_burm_nts[rule];
int i, kids;
icg_burm_kids(p, rule, kid);
for (i = 0; nts[i]; i++)
reduce_supairs(kid[i], nts[i]);
kids = i;
switch (rule) {
#include "icg-supairs.cases"
}
}
void icg_supairs(void)
{
basic_block bb;
(void)suOrder3; /* this function is not currently used; shut up gcc */
FOR_EACH_BB(bb)
{
rtx insn;
FOR_BB_INSNS(bb, insn)
{
int id = INSN_UID(insn);
if (icg_insn2dirty[id])
{
icg_node *tree = icg_insn2tree[id];
if (tree) {
reduce_supairs(tree, burm_goal_NT);
}
}
}
}
}
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