265 lines
8.5 KiB
C
265 lines
8.5 KiB
C
/*
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* Elastic Binary Trees - macros to manipulate String data nodes.
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* Version 5.1
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* (C) 2002-2009 - Willy Tarreau <w@1wt.eu>
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*
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* This program is free software; you can redistribute it and/or modify
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* it under the terms of the GNU General Public License as published by
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* the Free Software Foundation; either version 2 of the License, or
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* (at your option) any later version.
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*
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* This program is distributed in the hope that it will be useful,
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* but WITHOUT ANY WARRANTY; without even the implied warranty of
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* MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
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* GNU General Public License for more details.
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*
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* You should have received a copy of the GNU General Public License
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* along with this program; if not, write to the Free Software
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* Foundation, Inc., 51 Franklin St, Fifth Floor, Boston, MA 02110-1301 USA
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*/
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/* These functions and macros rely on Multi-Byte nodes */
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#ifndef _EBSTTREE_H
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#define _EBSTTREE_H
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#include "ebtree.h"
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#include "ebmbtree.h"
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/* The following functions are not inlined by default. They are declared
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* in ebsttree.c, which simply relies on their inline version.
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*/
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REGPRM2 struct ebmb_node *ebst_lookup(struct eb_root *root, const char *x);
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REGPRM3 struct ebmb_node *ebst_lookup_len(struct eb_root *root, const char *x, unsigned int len);
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REGPRM2 struct ebmb_node *ebst_insert(struct eb_root *root, struct ebmb_node *new);
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/* Find the first occurence of a zero-terminated string <x> in the tree <root>.
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* It's the caller's reponsibility to use this function only on trees which
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* only contain zero-terminated strings. If none can be found, return NULL.
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*/
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static forceinline struct ebmb_node *__ebst_lookup(struct eb_root *root, const void *x)
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{
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struct ebmb_node *node;
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eb_troot_t *troot;
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unsigned int bit;
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troot = root->b[EB_LEFT];
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if (unlikely(troot == NULL))
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return NULL;
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bit = 0;
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while (1) {
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if ((eb_gettag(troot) == EB_LEAF)) {
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node = container_of(eb_untag(troot, EB_LEAF),
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struct ebmb_node, node.branches);
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if (strcmp((char *)node->key, x) == 0)
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return node;
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else
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return NULL;
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}
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node = container_of(eb_untag(troot, EB_NODE),
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struct ebmb_node, node.branches);
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if (node->node.bit < 0) {
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/* We have a dup tree now. Either it's for the same
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* value, and we walk down left, or it's a different
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* one and we don't have our key.
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*/
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if (strcmp((char *)node->key, x) != 0)
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return NULL;
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troot = node->node.branches.b[EB_LEFT];
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while (eb_gettag(troot) != EB_LEAF)
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troot = (eb_untag(troot, EB_NODE))->b[EB_LEFT];
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node = container_of(eb_untag(troot, EB_LEAF),
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struct ebmb_node, node.branches);
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return node;
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}
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/* OK, normal data node, let's walk down */
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bit = string_equal_bits(x, node->key, bit);
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if (bit < node->node.bit)
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return NULL; /* no more common bits */
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troot = node->node.branches.b[(((unsigned char*)x)[node->node.bit >> 3] >>
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(~node->node.bit & 7)) & 1];
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}
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}
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/* Insert ebmb_node <new> into subtree starting at node root <root>. Only
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* new->key needs be set with the zero-terminated string key. The ebmb_node is
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* returned. If root->b[EB_RGHT]==1, the tree may only contain unique keys. The
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* caller is responsible for properly terminating the key with a zero.
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*/
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static forceinline struct ebmb_node *
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__ebst_insert(struct eb_root *root, struct ebmb_node *new)
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{
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struct ebmb_node *old;
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unsigned int side;
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eb_troot_t *troot;
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eb_troot_t *root_right = root;
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int diff;
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int bit;
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side = EB_LEFT;
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troot = root->b[EB_LEFT];
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root_right = root->b[EB_RGHT];
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if (unlikely(troot == NULL)) {
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/* Tree is empty, insert the leaf part below the left branch */
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root->b[EB_LEFT] = eb_dotag(&new->node.branches, EB_LEAF);
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new->node.leaf_p = eb_dotag(root, EB_LEFT);
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new->node.node_p = NULL; /* node part unused */
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return new;
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}
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/* The tree descent is fairly easy :
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* - first, check if we have reached a leaf node
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* - second, check if we have gone too far
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* - third, reiterate
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* Everywhere, we use <new> for the node node we are inserting, <root>
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* for the node we attach it to, and <old> for the node we are
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* displacing below <new>. <troot> will always point to the future node
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* (tagged with its type). <side> carries the side the node <new> is
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* attached to below its parent, which is also where previous node
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* was attached.
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*/
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bit = 0;
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while (1) {
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if (unlikely(eb_gettag(troot) == EB_LEAF)) {
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eb_troot_t *new_left, *new_rght;
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eb_troot_t *new_leaf, *old_leaf;
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old = container_of(eb_untag(troot, EB_LEAF),
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struct ebmb_node, node.branches);
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new_left = eb_dotag(&new->node.branches, EB_LEFT);
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new_rght = eb_dotag(&new->node.branches, EB_RGHT);
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new_leaf = eb_dotag(&new->node.branches, EB_LEAF);
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old_leaf = eb_dotag(&old->node.branches, EB_LEAF);
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new->node.node_p = old->node.leaf_p;
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/* Right here, we have 3 possibilities :
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* - the tree does not contain the key, and we have
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* new->key < old->key. We insert new above old, on
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* the left ;
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*
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* - the tree does not contain the key, and we have
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* new->key > old->key. We insert new above old, on
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* the right ;
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*
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* - the tree does contain the key, which implies it
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* is alone. We add the new key next to it as a
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* first duplicate.
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*
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* The last two cases can easily be partially merged.
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*/
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bit = string_equal_bits(new->key, old->key, bit);
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diff = cmp_bits(new->key, old->key, bit);
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if (diff < 0) {
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new->node.leaf_p = new_left;
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old->node.leaf_p = new_rght;
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new->node.branches.b[EB_LEFT] = new_leaf;
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new->node.branches.b[EB_RGHT] = old_leaf;
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} else {
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/* we may refuse to duplicate this key if the tree is
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* tagged as containing only unique keys.
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*/
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if (diff == 0 && eb_gettag(root_right))
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return old;
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/* new->key >= old->key, new goes the right */
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old->node.leaf_p = new_left;
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new->node.leaf_p = new_rght;
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new->node.branches.b[EB_LEFT] = old_leaf;
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new->node.branches.b[EB_RGHT] = new_leaf;
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if (diff == 0) {
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new->node.bit = -1;
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root->b[side] = eb_dotag(&new->node.branches, EB_NODE);
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return new;
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}
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}
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break;
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}
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/* OK we're walking down this link */
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old = container_of(eb_untag(troot, EB_NODE),
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struct ebmb_node, node.branches);
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/* Stop going down when we don't have common bits anymore. We
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* also stop in front of a duplicates tree because it means we
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* have to insert above. Note: we can compare more bits than
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* the current node's because as long as they are identical, we
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* know we descend along the correct side.
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*/
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if (old->node.bit < 0) {
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/* we're above a duplicate tree, we must compare till the end */
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bit = string_equal_bits(new->key, old->key, bit);
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goto dup_tree;
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}
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else if (bit < old->node.bit) {
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bit = string_equal_bits(new->key, old->key, bit);
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}
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if (bit < old->node.bit) { /* we don't have all bits in common */
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/* The tree did not contain the key, so we insert <new> before the node
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* <old>, and set ->bit to designate the lowest bit position in <new>
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* which applies to ->branches.b[].
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*/
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eb_troot_t *new_left, *new_rght;
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eb_troot_t *new_leaf, *old_node;
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dup_tree:
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new_left = eb_dotag(&new->node.branches, EB_LEFT);
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new_rght = eb_dotag(&new->node.branches, EB_RGHT);
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new_leaf = eb_dotag(&new->node.branches, EB_LEAF);
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old_node = eb_dotag(&old->node.branches, EB_NODE);
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new->node.node_p = old->node.node_p;
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diff = cmp_bits(new->key, old->key, bit);
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if (diff < 0) {
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new->node.leaf_p = new_left;
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old->node.node_p = new_rght;
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new->node.branches.b[EB_LEFT] = new_leaf;
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new->node.branches.b[EB_RGHT] = old_node;
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}
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else if (diff > 0) {
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old->node.node_p = new_left;
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new->node.leaf_p = new_rght;
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new->node.branches.b[EB_LEFT] = old_node;
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new->node.branches.b[EB_RGHT] = new_leaf;
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}
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else {
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struct eb_node *ret;
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ret = eb_insert_dup(&old->node, &new->node);
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return container_of(ret, struct ebmb_node, node);
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}
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break;
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}
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/* walk down */
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root = &old->node.branches;
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side = (new->key[old->node.bit >> 3] >> (~old->node.bit & 7)) & 1;
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troot = root->b[side];
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}
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/* Ok, now we are inserting <new> between <root> and <old>. <old>'s
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* parent is already set to <new>, and the <root>'s branch is still in
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* <side>. Update the root's leaf till we have it. Note that we can also
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* find the side by checking the side of new->node.node_p.
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*/
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/* We need the common higher bits between new->key and old->key.
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* This number of bits is already in <bit>.
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*/
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new->node.bit = bit;
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root->b[side] = eb_dotag(&new->node.branches, EB_NODE);
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return new;
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}
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#endif /* _EBSTTREE_H */
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