325 lines
10 KiB
C
325 lines
10 KiB
C
/*
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* Elastic Binary Trees - macros to manipulate String data nodes.
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* Version 6.0.6
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* (C) 2002-2011 - Willy Tarreau <w@1wt.eu>
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*
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* This library is free software; you can redistribute it and/or
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* modify it under the terms of the GNU Lesser General Public
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* License as published by the Free Software Foundation, version 2.1
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* exclusively.
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*
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* This library 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 GNU
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* Lesser General Public License for more details.
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*
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* You should have received a copy of the GNU Lesser General Public
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* License along with this library; if not, write to the Free Software
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* Foundation, Inc., 51 Franklin Street, 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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struct ebmb_node *ebst_lookup(struct eb_root *root, const char *x);
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struct ebmb_node *ebst_insert(struct eb_root *root, struct ebmb_node *new);
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/* Find the first occurrence of a length <len> string <x> in the tree <root>.
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* It's the caller's responsibility to use this function only on trees which
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* only contain zero-terminated strings, and that no null character is present
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* in string <x> in the first <len> chars. If none can be found, return NULL.
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*/
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static forceinline struct ebmb_node *
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ebst_lookup_len(struct eb_root *root, const char *x, unsigned int len)
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{
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struct ebmb_node *node;
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node = ebmb_lookup(root, x, len);
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if (!node || node->key[len] != 0)
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return NULL;
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return node;
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}
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/* Find the first occurrence of a zero-terminated string <x> in the tree <root>.
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* It's the caller's responsibility 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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int bit;
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int node_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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node_bit = node->node.bit;
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if (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 but don't compare data
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* if we already reached the end of the key.
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*/
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if (likely(bit >= 0)) {
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bit = string_equal_bits(x, node->key, bit);
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if (likely(bit < node_bit)) {
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if (bit >= 0)
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return NULL; /* no more common bits */
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/* bit < 0 : we reached the end of the key. If we
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* are in a tree with unique keys, we can return
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* this node. Otherwise we have to walk it down
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* and stop comparing bits.
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*/
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if (eb_gettag(root->b[EB_RGHT]))
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return node;
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}
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/* if the bit is larger than the node's, we must bound it
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* because we might have compared too many bytes with an
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* inappropriate leaf. For a test, build a tree from "0",
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* "WW", "W", "S" inserted in this exact sequence and lookup
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* "W" => "S" is returned without this assignment.
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*/
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else
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bit = node_bit;
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}
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troot = node->node.branches.b[(((unsigned char*)x)[node_bit >> 3] >>
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(~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;
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int diff;
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int bit;
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int old_node_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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if (bit >= 0)
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bit = string_equal_bits(new->key, old->key, bit);
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if (bit < 0) {
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/* key was already there */
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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 (eb_gettag(root_right))
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return old;
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/* new arbitrarily goes to the right and tops the dup tree */
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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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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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diff = cmp_bits(new->key, old->key, bit);
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if (diff < 0) {
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/* new->key < old->key, new takes the left */
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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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/* new->key > old->key, new takes 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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}
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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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old_node_bit = old->node.bit;
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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 (bit >= 0 && (bit < old_node_bit || old_node_bit < 0))
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bit = string_equal_bits(new->key, old->key, bit);
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if (unlikely(bit < 0)) {
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/* Perfect match, we must only stop on head of dup tree
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* or walk down to a leaf.
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*/
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if (old_node_bit < 0) {
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/* We know here that string_equal_bits matched all
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* bits and that we're on top of a dup tree, then
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* we can perform the dup insertion and return.
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*/
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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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/* OK so let's walk down */
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}
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else if (bit < old_node_bit || old_node_bit < 0) {
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/* The tree did not contain the key, or we stopped on top of a dup
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* tree, possibly containing the key. In the former case, we insert
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* <new> before the node <old>, and set ->bit to designate the lowest
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* bit position in <new> which applies to ->branches.b[]. In the later
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* case, we add the key to the existing dup tree. Note that we cannot
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* enter here if we match an intermediate node's key that is not the
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* head of a dup tree.
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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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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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/* we can never match all bits here */
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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 {
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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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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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* NOTE: we can't get here with bit < 0 since we found a dup !
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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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