struct depend {
/* We can have more than one */
- struct list_node list;
- unsigned int file;
- unsigned int op;
-};
-
-struct depend_xmit {
- unsigned int dst_op;
- unsigned int src_file, src_op;
+ struct list_node pre_list;
+ struct list_node post_list;
+ unsigned int needs_file;
+ unsigned int needs_opnum;
+ unsigned int satisfies_file;
+ unsigned int satisfies_opnum;
};
-static void remove_matching_dep(struct list_head *deps,
- unsigned int file, unsigned int op)
-{
- struct depend *dep;
-
- list_for_each(deps, dep, list) {
- if (dep->file == file && dep->op == op) {
- list_del(&dep->list);
- return;
- }
- }
- errx(1, "Failed to find depend on file %u line %u\n", file, op+1);
-}
-
static void check_deps(const char *filename, struct op op[], unsigned int num)
{
#ifdef DEBUG_DEPS
struct depend *dep;
printf("%s:%u still waiting for:\n", filename[file], i+1);
- list_for_each(&op[i].pre, dep, list)
- printf(" %s:%u\n", filename[dep->file], dep->op+1);
+ list_for_each(&op[i].pre, dep, pre_list)
+ printf(" %s:%u\n",
+ filename[dep->satisfies_file], dep->satisfies_opnum+1);
check_deps(filename[file], op, i);
}
+/* We simply read/write pointers, since we all are children. */
static void do_pre(char *filename[], unsigned int file, int pre_fd,
struct op op[], unsigned int i)
{
while (!list_empty(&op[i].pre)) {
- struct depend_xmit dep;
+ struct depend *dep;
#if DEBUG_DEPS
printf("%s:%u:waiting for pre\n", filename[file], i+1);
#if DEBUG_DEPS
printf("%s:%u:got pre %u from %s:%u\n", filename[file], i+1,
- dep.dst_op+1, filename[dep.src_file], dep.src_op+1);
+ dep->needs_opnum+1, filename[dep->satisfies_file],
+ dep->satisfies_opnum+1);
fflush(stdout);
#endif
/* This could be any op, not just this one. */
- remove_matching_dep(&op[dep.dst_op].pre,
- dep.src_file, dep.src_op);
+ talloc_free(dep);
}
}
{
struct depend *dep;
- list_for_each(&op[i].post, dep, list) {
- struct depend_xmit dx;
-
- dx.src_file = file;
- dx.src_op = i;
- dx.dst_op = dep->op;
+ list_for_each(&op[i].post, dep, post_list) {
#if DEBUG_DEPS
printf("%s:%u:sending to file %s:%u\n", filename[file], i+1,
- filename[dep->file], dep->op+1);
+ filename[dep->needs_file], dep->needs_opnum+1);
#endif
- if (write(pipes[dep->file].fd[1], &dx, sizeof(dx))
- != sizeof(dx))
+ if (write(pipes[dep->needs_file].fd[1], &dep, sizeof(dep))
+ != sizeof(dep))
err(1, "%s:%u failed to tell file %s",
- filename[file], i+1, filename[dep->file]);
+ filename[file], i+1, filename[dep->needs_file]);
}
}
return key_eq(*data, *need);
}
+static void move_to_front(struct key_user res[], unsigned int elem)
+{
+ if (elem != 0) {
+ struct key_user tmp = res[elem];
+ memmove(res + 1, res, elem*sizeof(res[0]));
+ res[0] = tmp;
+ }
+}
+
+static void restore_to_pos(struct key_user res[], unsigned int elem)
+{
+ if (elem != 0) {
+ struct key_user tmp = res[0];
+ memmove(res, res + 1, elem*sizeof(res[0]));
+ res[elem] = tmp;
+ }
+}
+
static bool sort_deps(char *filename[], struct op *op[],
struct key_user res[], unsigned num,
- const TDB_DATA *data)
+ const TDB_DATA *data, unsigned num_files)
{
- unsigned int i;
+ unsigned int i, files_done;
+ struct op *this_op;
+ bool done[num_files];
/* Nothing left? We're sorted. */
if (num == 0)
return true;
- for (i = 0; i < num; i++) {
- struct op *this_op = &op[res[i].file][res[i].op_num];
+ memset(done, 0, sizeof(done));
+
+ /* Since ops within a trace file are ordered, we just need to figure
+ * out which file to try next. Since we don't take into account
+ * inter-key relationships (which exist by virtue of trace file order),
+ * we minimize the chance of harm by trying to keep in serial order. */
+ for (files_done = 0, i = 0; i < num && files_done < num_files; i++) {
+ if (done[res[i].file])
+ continue;
+ this_op = &op[res[i].file][res[i].op_num];
/* Is what we have good enough for this op? */
if (satisfies(data, needs(this_op))) {
- /* Try this one next. */
- struct key_user tmp = res[0];
- res[0] = res[i];
- res[i] = tmp;
+ move_to_front(res, i);
if (sort_deps(filename, op, res+1, num-1,
- gives(this_op, data)))
+ gives(this_op, data), num_files))
return true;
+ restore_to_pos(res, i);
}
+ done[res[i].file] = true;
+ files_done++;
}
+
/* No combination worked. */
return false;
}
+static void check_dep_sorting(struct key_user user[], unsigned num_users,
+ unsigned num_files)
+{
+#if DEBUG_DEPS
+ unsigned int i;
+ unsigned minima[num_files];
+
+ memset(minima, 0, sizeof(minima));
+ for (i = 0; i < num_users; i++) {
+ assert(minima[user[i].file] < user[i].op_num);
+ minima[user[i].file] = user[i].op_num;
+ }
+#endif
+}
+
/* All these ops have the same serial number. Which comes first?
*
* This can happen both because read ops or failed write ops don't
* in which case we'll deadlock and report: fix manually in that case).
*/
static void figure_deps(char *filename[], struct op *op[],
- struct key_user user[], unsigned start, unsigned end)
+ struct key_user user[], unsigned num_users,
+ unsigned num_files)
{
- unsigned int i;
/* We assume database starts empty. */
const struct TDB_DATA *data = &tdb_null;
- /* What do we have to start with? */
- for (i = 0; i < start; i++)
- data = gives(&op[user[i].file][user[i].op_num], data);
-
- if (!sort_deps(filename, op, user + start, end - start, data))
- fail(filename[user[start].file], user[start].op_num+1,
+ if (!sort_deps(filename, op, user, num_users, data, num_files))
+ fail(filename[user[0].file], user[0].op_num+1,
"Could not resolve inter-dependencies");
+
+ check_dep_sorting(user, num_users, num_files);
}
-static void sort_ops(struct keyinfo hash[], char *filename[], struct op *op[])
+static void sort_ops(struct keyinfo hash[], char *filename[], struct op *op[],
+ unsigned int num)
{
unsigned int h;
int compare_serial(const void *_a, const void *_b)
{
const struct key_user *a = _a, *b = _b;
+
+ /* First, maintain order within any trace file. */
+ if (a->file == b->file)
+ return a->op_num - b->op_num;
+
+ /* Otherwise, arrange by serial order. */
return op[a->file][a->op_num].serial
- op[b->file][b->op_num].serial;
}
- /* Now sort into seqnum order. */
+ /* Now sort into serial order. */
for (h = 0; h < total_keys * 2; h++) {
- unsigned int i, same;
struct key_user *user = hash[h].user;
qsort(user, hash[h].num_users, sizeof(user[0]), compare_serial);
-
- /* Try to deal with same serial numbers. */
- for (i = 1, same = 0; i < hash[h].num_users; i++) {
- if (op[user[i].file][user[i].op_num].serial
- == op[user[i-1].file][user[i-1].op_num].serial) {
- same++;
- continue;
- }
-
- if (same) {
- figure_deps(filename, op, user, i-same-1, i);
- same = 0;
- }
- }
- if (same)
- figure_deps(filename, op, user, i-same-1, i);
+ figure_deps(filename, op, user, hash[h].num_users, num);
}
}
+static int destroy_depend(struct depend *dep)
+{
+ list_del(&dep->pre_list);
+ list_del(&dep->post_list);
+ return 0;
+}
+
static void add_dependency(void *ctx,
struct op *op[],
char *filename[],
unsigned int satisfies_file,
unsigned int satisfies_opnum)
{
- struct depend *post, *pre;
+ struct depend *dep;
unsigned int needs_start, sat_start;
/* We don't depend on ourselves. */
- if (needs_file == satisfies_file)
+ if (needs_file == satisfies_file) {
+ assert(satisfies_opnum < needs_opnum);
return;
+ }
#if DEBUG_DEPS
printf("%s:%u: depends on %s:%u\n",
}
}
- post = talloc(ctx, struct depend);
- post->file = needs_file;
- post->op = needs_opnum;
- list_add(&op[satisfies_file][satisfies_opnum].post, &post->list);
-
- pre = talloc(ctx, struct depend);
- pre->file = satisfies_file;
- pre->op = satisfies_opnum;
- list_add(&op[needs_file][needs_opnum].pre, &pre->list);
+ dep = talloc(ctx, struct depend);
+ dep->needs_file = needs_file;
+ dep->needs_opnum = needs_opnum;
+ dep->satisfies_file = satisfies_file;
+ dep->satisfies_opnum = satisfies_opnum;
+ list_add(&op[satisfies_file][satisfies_opnum].post, &dep->post_list);
+ list_add(&op[needs_file][needs_opnum].pre, &dep->pre_list);
+ talloc_set_destructor(dep, destroy_depend);
}
#if TRAVERSALS_TAKE_TRANSACTION_LOCK
}
#endif /* TRAVERSALS_TAKE_TRANSACTION_LOCK */
+static bool changes_db(const struct op *op)
+{
+ return gives(op, NULL) != NULL;
+}
+
+static void depend_on_previous(struct op *op[],
+ char *filename[],
+ unsigned int num,
+ struct key_user user[],
+ unsigned int i,
+ int prev)
+{
+ bool deps[num];
+ int j;
+
+ if (i == 0)
+ return;
+
+ if (prev == i - 1) {
+ /* Just depend on previous. */
+ add_dependency(NULL, op, filename,
+ user[i].file, user[i].op_num,
+ user[prev].file, user[prev].op_num);
+ return;
+ }
+
+ /* We have to wait for the readers. Find last one in *each* file. */
+ memset(deps, 0, sizeof(deps));
+ deps[user[i].file] = true;
+ for (j = i - 1; j > prev; j--) {
+ if (!deps[user[j].file]) {
+ add_dependency(NULL, op, filename,
+ user[i].file, user[i].op_num,
+ user[j].file, user[j].op_num);
+ deps[user[j].file] = true;
+ }
+ }
+}
+
+/* This is simple, but not complete. We don't take into account
+ * indirect dependencies. */
+static void optimize_dependencies(struct op *op[], unsigned int num_ops[],
+ unsigned int num)
+{
+ unsigned int i, j;
+
+ /* There can only be one real dependency on each file */
+ for (i = 0; i < num; i++) {
+ for (j = 1; j < num_ops[i]; j++) {
+ struct depend *dep, *next;
+ struct depend *prev[num];
+
+ memset(prev, 0, sizeof(prev));
+
+ list_for_each_safe(&op[i][j].pre, dep, next, pre_list) {
+ if (!prev[dep->satisfies_file]) {
+ prev[dep->satisfies_file] = dep;
+ continue;
+ }
+ if (prev[dep->satisfies_file]->satisfies_opnum
+ < dep->satisfies_opnum) {
+ talloc_free(prev[dep->satisfies_file]);
+ prev[dep->satisfies_file] = dep;
+ } else
+ talloc_free(dep);
+ }
+ }
+ }
+
+ for (i = 0; i < num; i++) {
+ int deps[num];
+
+ for (j = 0; j < num; j++)
+ deps[j] = -1;
+
+ for (j = 1; j < num_ops[i]; j++) {
+ struct depend *dep, *next;
+
+ list_for_each_safe(&op[i][j].pre, dep, next, pre_list) {
+ if (deps[dep->satisfies_file]
+ >= (int)dep->satisfies_opnum)
+ talloc_free(dep);
+ else
+ deps[dep->satisfies_file]
+ = dep->satisfies_opnum;
+ }
+ }
+ }
+}
+
static void derive_dependencies(char *filename[],
struct op *op[], unsigned int num_ops[],
unsigned int num)
{
struct keyinfo *hash;
- unsigned int i, j;
+ unsigned int h, i;
/* Create hash table for faster key lookup. */
hash = hash_ops(op, num_ops, num);
- /* Now handle the hard cases: same serial number. */
- sort_ops(hash, filename, op);
-
- /* We make the naive assumption that two ops on the same key
- * have to be ordered; it's overkill. */
- for (i = 0; i < total_keys * 2; i++) {
- for (j = 1; j < hash[i].num_users; j++) {
- add_dependency(hash, op, filename,
- hash[i].user[j].file,
- hash[i].user[j].op_num,
- hash[i].user[j-1].file,
- hash[i].user[j-1].op_num);
+ /* Sort them by serial number. */
+ sort_ops(hash, filename, op, num);
+
+ /* Create dependencies back to the last change, rather than
+ * creating false dependencies by naively making each one
+ * depend on the previous. This has two purposes: it makes
+ * later optimization simpler, and it also avoids deadlock with
+ * same sequence number ops inside traversals (if one
+ * traversal doesn't write anything, two ops can have the same
+ * sequence number yet we can create a traversal dependency
+ * the other way). */
+ for (h = 0; h < total_keys * 2; h++) {
+ int prev = -1;
+
+ if (hash[h].num_users < 2)
+ continue;
+
+ for (i = 0; i < hash[h].num_users; i++) {
+ if (changes_db(&op[hash[h].user[i].file]
+ [hash[h].user[i].op_num])) {
+ depend_on_previous(op, filename, num,
+ hash[h].user, i, prev);
+ prev = i;
+ } else if (prev >= 0)
+ add_dependency(hash, op, filename,
+ hash[h].user[i].file,
+ hash[h].user[i].op_num,
+ hash[h].user[prev].file,
+ hash[h].user[prev].op_num);
}
}
#if TRAVERSALS_TAKE_TRANSACTION_LOCK
make_traverse_depends(filename, op, num_ops, num);
#endif
+
+ optimize_dependencies(op, num_ops, num);
}
int main(int argc, char *argv[])
projects / ccan / blobdiff