209 lines
6.8 KiB
C
209 lines
6.8 KiB
C
/*
|
|
* Elastic Binary Trees - exported functions for operations on pointer nodes.
|
|
* Version 6.0.6
|
|
* (C) 2002-2011 - Willy Tarreau <w@1wt.eu>
|
|
*
|
|
* This library is free software; you can redistribute it and/or
|
|
* modify it under the terms of the GNU Lesser General Public
|
|
* License as published by the Free Software Foundation, version 2.1
|
|
* exclusively.
|
|
*
|
|
* This library 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
|
|
* Lesser General Public License for more details.
|
|
*
|
|
* You should have received a copy of the GNU Lesser General Public
|
|
* License along with this library; if not, write to the Free Software
|
|
* Foundation, Inc., 51 Franklin Street, Fifth Floor, Boston, MA 02110-1301 USA
|
|
*/
|
|
|
|
/* Consult ebpttree.h for more details about those functions */
|
|
|
|
#include "ebpttree.h"
|
|
|
|
REGPRM2 struct ebpt_node *ebpt_insert(struct eb_root *root, struct ebpt_node *new)
|
|
{
|
|
return __ebpt_insert(root, new);
|
|
}
|
|
|
|
REGPRM2 struct ebpt_node *ebpt_lookup(struct eb_root *root, void *x)
|
|
{
|
|
return __ebpt_lookup(root, x);
|
|
}
|
|
|
|
/*
|
|
* Find the last occurrence of the highest key in the tree <root>, which is
|
|
* equal to or less than <x>. NULL is returned is no key matches.
|
|
*/
|
|
REGPRM2 struct ebpt_node *ebpt_lookup_le(struct eb_root *root, void *x)
|
|
{
|
|
struct ebpt_node *node;
|
|
eb_troot_t *troot;
|
|
|
|
troot = root->b[EB_LEFT];
|
|
if (unlikely(troot == NULL))
|
|
return NULL;
|
|
|
|
while (1) {
|
|
if ((eb_gettag(troot) == EB_LEAF)) {
|
|
/* We reached a leaf, which means that the whole upper
|
|
* parts were common. We will return either the current
|
|
* node or its next one if the former is too small.
|
|
*/
|
|
node = container_of(eb_untag(troot, EB_LEAF),
|
|
struct ebpt_node, node.branches);
|
|
if (node->key <= x)
|
|
return node;
|
|
/* return prev */
|
|
troot = node->node.leaf_p;
|
|
break;
|
|
}
|
|
node = container_of(eb_untag(troot, EB_NODE),
|
|
struct ebpt_node, node.branches);
|
|
|
|
if (node->node.bit < 0) {
|
|
/* We're at the top of a dup tree. Either we got a
|
|
* matching value and we return the rightmost node, or
|
|
* we don't and we skip the whole subtree to return the
|
|
* prev node before the subtree. Note that since we're
|
|
* at the top of the dup tree, we can simply return the
|
|
* prev node without first trying to escape from the
|
|
* tree.
|
|
*/
|
|
if (node->key <= x) {
|
|
troot = node->node.branches.b[EB_RGHT];
|
|
while (eb_gettag(troot) != EB_LEAF)
|
|
troot = (eb_untag(troot, EB_NODE))->b[EB_RGHT];
|
|
return container_of(eb_untag(troot, EB_LEAF),
|
|
struct ebpt_node, node.branches);
|
|
}
|
|
/* return prev */
|
|
troot = node->node.node_p;
|
|
break;
|
|
}
|
|
|
|
if ((((ptr_t)x ^ (ptr_t)node->key) >> node->node.bit) >= EB_NODE_BRANCHES) {
|
|
/* No more common bits at all. Either this node is too
|
|
* small and we need to get its highest value, or it is
|
|
* too large, and we need to get the prev value.
|
|
*/
|
|
if (((ptr_t)node->key >> node->node.bit) < ((ptr_t)x >> node->node.bit)) {
|
|
troot = node->node.branches.b[EB_RGHT];
|
|
return ebpt_entry(eb_walk_down(troot, EB_RGHT), struct ebpt_node, node);
|
|
}
|
|
|
|
/* Further values will be too high here, so return the prev
|
|
* unique node (if it exists).
|
|
*/
|
|
troot = node->node.node_p;
|
|
break;
|
|
}
|
|
troot = node->node.branches.b[((ptr_t)x >> node->node.bit) & EB_NODE_BRANCH_MASK];
|
|
}
|
|
|
|
/* If we get here, it means we want to report previous node before the
|
|
* current one which is not above. <troot> is already initialised to
|
|
* the parent's branches.
|
|
*/
|
|
while (eb_gettag(troot) == EB_LEFT) {
|
|
/* Walking up from left branch. We must ensure that we never
|
|
* walk beyond root.
|
|
*/
|
|
if (unlikely(eb_clrtag((eb_untag(troot, EB_LEFT))->b[EB_RGHT]) == NULL))
|
|
return NULL;
|
|
troot = (eb_root_to_node(eb_untag(troot, EB_LEFT)))->node_p;
|
|
}
|
|
/* Note that <troot> cannot be NULL at this stage */
|
|
troot = (eb_untag(troot, EB_RGHT))->b[EB_LEFT];
|
|
node = ebpt_entry(eb_walk_down(troot, EB_RGHT), struct ebpt_node, node);
|
|
return node;
|
|
}
|
|
|
|
/*
|
|
* Find the first occurrence of the lowest key in the tree <root>, which is
|
|
* equal to or greater than <x>. NULL is returned is no key matches.
|
|
*/
|
|
REGPRM2 struct ebpt_node *ebpt_lookup_ge(struct eb_root *root, void *x)
|
|
{
|
|
struct ebpt_node *node;
|
|
eb_troot_t *troot;
|
|
|
|
troot = root->b[EB_LEFT];
|
|
if (unlikely(troot == NULL))
|
|
return NULL;
|
|
|
|
while (1) {
|
|
if ((eb_gettag(troot) == EB_LEAF)) {
|
|
/* We reached a leaf, which means that the whole upper
|
|
* parts were common. We will return either the current
|
|
* node or its next one if the former is too small.
|
|
*/
|
|
node = container_of(eb_untag(troot, EB_LEAF),
|
|
struct ebpt_node, node.branches);
|
|
if (node->key >= x)
|
|
return node;
|
|
/* return next */
|
|
troot = node->node.leaf_p;
|
|
break;
|
|
}
|
|
node = container_of(eb_untag(troot, EB_NODE),
|
|
struct ebpt_node, node.branches);
|
|
|
|
if (node->node.bit < 0) {
|
|
/* We're at the top of a dup tree. Either we got a
|
|
* matching value and we return the leftmost node, or
|
|
* we don't and we skip the whole subtree to return the
|
|
* next node after the subtree. Note that since we're
|
|
* at the top of the dup tree, we can simply return the
|
|
* next node without first trying to escape from the
|
|
* tree.
|
|
*/
|
|
if (node->key >= x) {
|
|
troot = node->node.branches.b[EB_LEFT];
|
|
while (eb_gettag(troot) != EB_LEAF)
|
|
troot = (eb_untag(troot, EB_NODE))->b[EB_LEFT];
|
|
return container_of(eb_untag(troot, EB_LEAF),
|
|
struct ebpt_node, node.branches);
|
|
}
|
|
/* return next */
|
|
troot = node->node.node_p;
|
|
break;
|
|
}
|
|
|
|
if ((((ptr_t)x ^ (ptr_t)node->key) >> node->node.bit) >= EB_NODE_BRANCHES) {
|
|
/* No more common bits at all. Either this node is too
|
|
* large and we need to get its lowest value, or it is too
|
|
* small, and we need to get the next value.
|
|
*/
|
|
if (((ptr_t)node->key >> node->node.bit) > ((ptr_t)x >> node->node.bit)) {
|
|
troot = node->node.branches.b[EB_LEFT];
|
|
return ebpt_entry(eb_walk_down(troot, EB_LEFT), struct ebpt_node, node);
|
|
}
|
|
|
|
/* Further values will be too low here, so return the next
|
|
* unique node (if it exists).
|
|
*/
|
|
troot = node->node.node_p;
|
|
break;
|
|
}
|
|
troot = node->node.branches.b[((ptr_t)x >> node->node.bit) & EB_NODE_BRANCH_MASK];
|
|
}
|
|
|
|
/* If we get here, it means we want to report next node after the
|
|
* current one which is not below. <troot> is already initialised
|
|
* to the parent's branches.
|
|
*/
|
|
while (eb_gettag(troot) != EB_LEFT)
|
|
/* Walking up from right branch, so we cannot be below root */
|
|
troot = (eb_root_to_node(eb_untag(troot, EB_RGHT)))->node_p;
|
|
|
|
/* Note that <troot> cannot be NULL at this stage */
|
|
troot = (eb_untag(troot, EB_LEFT))->b[EB_RGHT];
|
|
if (eb_clrtag(troot) == NULL)
|
|
return NULL;
|
|
|
|
node = ebpt_entry(eb_walk_down(troot, EB_LEFT), struct ebpt_node, node);
|
|
return node;
|
|
}
|