haproxy/ebtree/ebistree.h

264 lines
8.6 KiB
C

/*
* Elastic Binary Trees - macros to manipulate Indirect String data nodes.
* Version 5.0
* (C) 2002-2009 - Willy Tarreau <w@1wt.eu>
*
* This program is free software; you can redistribute it and/or modify
* it under the terms of the GNU General Public License as published by
* the Free Software Foundation; either version 2 of the License, or
* (at your option) any later version.
*
* This program is distributed in the hope that it will be useful,
* but WITHOUT ANY WARRANTY; without even the implied warranty of
* MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
* GNU General Public License for more details.
*
* You should have received a copy of the GNU General Public License
* along with this program; if not, write to the Free Software
* Foundation, Inc., 51 Franklin St, Fifth Floor, Boston, MA 02110-1301 USA
*/
/* These functions and macros rely on Multi-Byte nodes */
#include <string.h>
#include "ebtree.h"
#include "ebpttree.h"
/* These functions and macros rely on Pointer nodes and use the <key> entry as
* a pointer to an indirect key. Most operations are performed using ebpt_*.
*/
/* The following functions are not inlined by default. They are declared
* in ebistree.c, which simply relies on their inline version.
*/
REGPRM2 struct ebpt_node *ebis_lookup(struct eb_root *root, const char *x);
REGPRM2 struct ebpt_node *ebis_insert(struct eb_root *root, struct ebpt_node *new);
/* Find the first occurence of a zero-terminated string <x> in the tree <root>.
* It's the caller's reponsibility to use this function only on trees which
* only contain zero-terminated strings. If none can be found, return NULL.
*/
static forceinline struct ebpt_node *__ebis_lookup(struct eb_root *root, const void *x)
{
struct ebpt_node *node;
eb_troot_t *troot;
unsigned int bit;
troot = root->b[EB_LEFT];
if (unlikely(troot == NULL))
return NULL;
bit = 0;
while (1) {
if ((eb_gettag(troot) == EB_LEAF)) {
node = container_of(eb_untag(troot, EB_LEAF),
struct ebpt_node, node.branches);
if (strcmp(node->key, x) == 0)
return node;
else
return NULL;
}
node = container_of(eb_untag(troot, EB_NODE),
struct ebpt_node, node.branches);
if (node->node.bit < 0) {
/* We have a dup tree now. Either it's for the same
* value, and we walk down left, or it's a different
* one and we don't have our key.
*/
if (strcmp(node->key, x) != 0)
return NULL;
troot = node->node.branches.b[EB_LEFT];
while (eb_gettag(troot) != EB_LEAF)
troot = (eb_untag(troot, EB_NODE))->b[EB_LEFT];
node = container_of(eb_untag(troot, EB_LEAF),
struct ebpt_node, node.branches);
return node;
}
/* OK, normal data node, let's walk down */
bit = string_equal_bits(x, node->key, bit);
if (bit < node->node.bit)
return NULL; /* no more common bits */
troot = node->node.branches.b[(((unsigned char*)x)[node->node.bit >> 3] >>
(~node->node.bit & 7)) & 1];
}
}
/* Insert ebpt_node <new> into subtree starting at node root <root>. Only
* new->key needs be set with the zero-terminated string key. The ebpt_node is
* returned. If root->b[EB_RGHT]==1, the tree may only contain unique keys. The
* caller is responsible for properly terminating the key with a zero.
*/
static forceinline struct ebpt_node *
__ebis_insert(struct eb_root *root, struct ebpt_node *new)
{
struct ebpt_node *old;
unsigned int side;
eb_troot_t *troot;
eb_troot_t *root_right = root;
int diff;
int bit;
side = EB_LEFT;
troot = root->b[EB_LEFT];
root_right = root->b[EB_RGHT];
if (unlikely(troot == NULL)) {
/* Tree is empty, insert the leaf part below the left branch */
root->b[EB_LEFT] = eb_dotag(&new->node.branches, EB_LEAF);
new->node.leaf_p = eb_dotag(root, EB_LEFT);
new->node.node_p = NULL; /* node part unused */
return new;
}
/* The tree descent is fairly easy :
* - first, check if we have reached a leaf node
* - second, check if we have gone too far
* - third, reiterate
* Everywhere, we use <new> for the node node we are inserting, <root>
* for the node we attach it to, and <old> for the node we are
* displacing below <new>. <troot> will always point to the future node
* (tagged with its type). <side> carries the side the node <new> is
* attached to below its parent, which is also where previous node
* was attached.
*/
bit = 0;
while (1) {
if (unlikely(eb_gettag(troot) == EB_LEAF)) {
eb_troot_t *new_left, *new_rght;
eb_troot_t *new_leaf, *old_leaf;
old = container_of(eb_untag(troot, EB_LEAF),
struct ebpt_node, node.branches);
new_left = eb_dotag(&new->node.branches, EB_LEFT);
new_rght = eb_dotag(&new->node.branches, EB_RGHT);
new_leaf = eb_dotag(&new->node.branches, EB_LEAF);
old_leaf = eb_dotag(&old->node.branches, EB_LEAF);
new->node.node_p = old->node.leaf_p;
/* Right here, we have 3 possibilities :
* - the tree does not contain the key, and we have
* new->key < old->key. We insert new above old, on
* the left ;
*
* - the tree does not contain the key, and we have
* new->key > old->key. We insert new above old, on
* the right ;
*
* - the tree does contain the key, which implies it
* is alone. We add the new key next to it as a
* first duplicate.
*
* The last two cases can easily be partially merged.
*/
bit = string_equal_bits(new->key, old->key, bit);
diff = cmp_bits(new->key, old->key, bit);
if (diff < 0) {
new->node.leaf_p = new_left;
old->node.leaf_p = new_rght;
new->node.branches.b[EB_LEFT] = new_leaf;
new->node.branches.b[EB_RGHT] = old_leaf;
} else {
/* we may refuse to duplicate this key if the tree is
* tagged as containing only unique keys.
*/
if (diff == 0 && eb_gettag(root_right))
return old;
/* new->key >= old->key, new goes the right */
old->node.leaf_p = new_left;
new->node.leaf_p = new_rght;
new->node.branches.b[EB_LEFT] = old_leaf;
new->node.branches.b[EB_RGHT] = new_leaf;
if (diff == 0) {
new->node.bit = -1;
root->b[side] = eb_dotag(&new->node.branches, EB_NODE);
return new;
}
}
break;
}
/* OK we're walking down this link */
old = container_of(eb_untag(troot, EB_NODE),
struct ebpt_node, node.branches);
/* Stop going down when we don't have common bits anymore. We
* also stop in front of a duplicates tree because it means we
* have to insert above. Note: we can compare more bits than
* the current node's because as long as they are identical, we
* know we descend along the correct side.
*/
if (old->node.bit < 0) {
/* we're above a duplicate tree, we must compare till the end */
bit = string_equal_bits(new->key, old->key, bit);
goto dup_tree;
}
else if (bit < old->node.bit) {
bit = string_equal_bits(new->key, old->key, bit);
}
if (bit < old->node.bit) { /* we don't have all bits in common */
/* The tree did not contain the key, so we insert <new> before the node
* <old>, and set ->bit to designate the lowest bit position in <new>
* which applies to ->branches.b[].
*/
eb_troot_t *new_left, *new_rght;
eb_troot_t *new_leaf, *old_node;
dup_tree:
new_left = eb_dotag(&new->node.branches, EB_LEFT);
new_rght = eb_dotag(&new->node.branches, EB_RGHT);
new_leaf = eb_dotag(&new->node.branches, EB_LEAF);
old_node = eb_dotag(&old->node.branches, EB_NODE);
new->node.node_p = old->node.node_p;
diff = cmp_bits(new->key, old->key, bit);
if (diff < 0) {
new->node.leaf_p = new_left;
old->node.node_p = new_rght;
new->node.branches.b[EB_LEFT] = new_leaf;
new->node.branches.b[EB_RGHT] = old_node;
}
else if (diff > 0) {
old->node.node_p = new_left;
new->node.leaf_p = new_rght;
new->node.branches.b[EB_LEFT] = old_node;
new->node.branches.b[EB_RGHT] = new_leaf;
}
else {
struct eb_node *ret;
ret = eb_insert_dup(&old->node, &new->node);
return container_of(ret, struct ebpt_node, node);
}
break;
}
/* walk down */
root = &old->node.branches;
side = (((unsigned char *)new->key)[old->node.bit >> 3] >> (~old->node.bit & 7)) & 1;
troot = root->b[side];
}
/* Ok, now we are inserting <new> between <root> and <old>. <old>'s
* parent is already set to <new>, and the <root>'s branch is still in
* <side>. Update the root's leaf till we have it. Note that we can also
* find the side by checking the side of new->node.node_p.
*/
/* We need the common higher bits between new->key and old->key.
* This number of bits is already in <bit>.
*/
new->node.bit = bit;
root->b[side] = eb_dotag(&new->node.branches, EB_NODE);
return new;
}