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Willy Tarreauc2186022009-10-26 19:48:54 +01001/*
2 * Elastic Binary Trees - macros to manipulate Indirect String data nodes.
Willy Tarreau3a932442010-05-09 19:29:23 +02003 * Version 6.0
4 * (C) 2002-2010 - Willy Tarreau <w@1wt.eu>
Willy Tarreauc2186022009-10-26 19:48:54 +01005 *
6 * This program is free software; you can redistribute it and/or modify
7 * it under the terms of the GNU General Public License as published by
8 * the Free Software Foundation; either version 2 of the License, or
9 * (at your option) any later version.
10 *
11 * This program is distributed in the hope that it will be useful,
12 * but WITHOUT ANY WARRANTY; without even the implied warranty of
13 * MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
14 * GNU General Public License for more details.
15 *
16 * You should have received a copy of the GNU General Public License
17 * along with this program; if not, write to the Free Software
18 * Foundation, Inc., 51 Franklin St, Fifth Floor, Boston, MA 02110-1301 USA
19 */
20
21/* These functions and macros rely on Multi-Byte nodes */
22
23#include <string.h>
24#include "ebtree.h"
25#include "ebpttree.h"
26
27/* These functions and macros rely on Pointer nodes and use the <key> entry as
28 * a pointer to an indirect key. Most operations are performed using ebpt_*.
29 */
30
31/* The following functions are not inlined by default. They are declared
32 * in ebistree.c, which simply relies on their inline version.
33 */
34REGPRM2 struct ebpt_node *ebis_lookup(struct eb_root *root, const char *x);
Willy Tarreauc9a31da2009-12-14 12:40:27 +010035REGPRM3 struct ebpt_node *ebis_lookup_len(struct eb_root *root, const char *x, unsigned int len);
Willy Tarreauc2186022009-10-26 19:48:54 +010036REGPRM2 struct ebpt_node *ebis_insert(struct eb_root *root, struct ebpt_node *new);
37
38/* Find the first occurence of a zero-terminated string <x> in the tree <root>.
39 * It's the caller's reponsibility to use this function only on trees which
40 * only contain zero-terminated strings. If none can be found, return NULL.
41 */
42static forceinline struct ebpt_node *__ebis_lookup(struct eb_root *root, const void *x)
43{
44 struct ebpt_node *node;
45 eb_troot_t *troot;
Willy Tarreau3a932442010-05-09 19:29:23 +020046 int bit;
47 int node_bit;
Willy Tarreauc2186022009-10-26 19:48:54 +010048
49 troot = root->b[EB_LEFT];
50 if (unlikely(troot == NULL))
51 return NULL;
52
53 bit = 0;
54 while (1) {
55 if ((eb_gettag(troot) == EB_LEAF)) {
56 node = container_of(eb_untag(troot, EB_LEAF),
57 struct ebpt_node, node.branches);
58 if (strcmp(node->key, x) == 0)
59 return node;
60 else
61 return NULL;
62 }
63 node = container_of(eb_untag(troot, EB_NODE),
64 struct ebpt_node, node.branches);
Willy Tarreau3a932442010-05-09 19:29:23 +020065 node_bit = node->node.bit;
Willy Tarreauc2186022009-10-26 19:48:54 +010066
Willy Tarreau3a932442010-05-09 19:29:23 +020067 if (node_bit < 0) {
Willy Tarreauc2186022009-10-26 19:48:54 +010068 /* We have a dup tree now. Either it's for the same
69 * value, and we walk down left, or it's a different
70 * one and we don't have our key.
71 */
72 if (strcmp(node->key, x) != 0)
73 return NULL;
74
75 troot = node->node.branches.b[EB_LEFT];
76 while (eb_gettag(troot) != EB_LEAF)
77 troot = (eb_untag(troot, EB_NODE))->b[EB_LEFT];
78 node = container_of(eb_untag(troot, EB_LEAF),
79 struct ebpt_node, node.branches);
80 return node;
81 }
82
83 /* OK, normal data node, let's walk down */
84 bit = string_equal_bits(x, node->key, bit);
Willy Tarreau3a932442010-05-09 19:29:23 +020085 if (bit < node_bit)
Willy Tarreauc2186022009-10-26 19:48:54 +010086 return NULL; /* no more common bits */
87
Willy Tarreau3a932442010-05-09 19:29:23 +020088 troot = node->node.branches.b[(((unsigned char*)x)[node_bit >> 3] >>
89 (~node_bit & 7)) & 1];
Willy Tarreauc2186022009-10-26 19:48:54 +010090 }
91}
92
93/* Insert ebpt_node <new> into subtree starting at node root <root>. Only
94 * new->key needs be set with the zero-terminated string key. The ebpt_node is
95 * returned. If root->b[EB_RGHT]==1, the tree may only contain unique keys. The
96 * caller is responsible for properly terminating the key with a zero.
97 */
98static forceinline struct ebpt_node *
99__ebis_insert(struct eb_root *root, struct ebpt_node *new)
100{
101 struct ebpt_node *old;
102 unsigned int side;
103 eb_troot_t *troot;
104 eb_troot_t *root_right = root;
105 int diff;
106 int bit;
Willy Tarreau3a932442010-05-09 19:29:23 +0200107 int old_node_bit;
Willy Tarreauc2186022009-10-26 19:48:54 +0100108
109 side = EB_LEFT;
110 troot = root->b[EB_LEFT];
111 root_right = root->b[EB_RGHT];
112 if (unlikely(troot == NULL)) {
113 /* Tree is empty, insert the leaf part below the left branch */
114 root->b[EB_LEFT] = eb_dotag(&new->node.branches, EB_LEAF);
115 new->node.leaf_p = eb_dotag(root, EB_LEFT);
116 new->node.node_p = NULL; /* node part unused */
117 return new;
118 }
119
120 /* The tree descent is fairly easy :
121 * - first, check if we have reached a leaf node
122 * - second, check if we have gone too far
123 * - third, reiterate
124 * Everywhere, we use <new> for the node node we are inserting, <root>
125 * for the node we attach it to, and <old> for the node we are
126 * displacing below <new>. <troot> will always point to the future node
127 * (tagged with its type). <side> carries the side the node <new> is
128 * attached to below its parent, which is also where previous node
129 * was attached.
130 */
131
132 bit = 0;
133 while (1) {
134 if (unlikely(eb_gettag(troot) == EB_LEAF)) {
135 eb_troot_t *new_left, *new_rght;
136 eb_troot_t *new_leaf, *old_leaf;
137
138 old = container_of(eb_untag(troot, EB_LEAF),
139 struct ebpt_node, node.branches);
140
141 new_left = eb_dotag(&new->node.branches, EB_LEFT);
142 new_rght = eb_dotag(&new->node.branches, EB_RGHT);
143 new_leaf = eb_dotag(&new->node.branches, EB_LEAF);
144 old_leaf = eb_dotag(&old->node.branches, EB_LEAF);
145
146 new->node.node_p = old->node.leaf_p;
147
148 /* Right here, we have 3 possibilities :
149 * - the tree does not contain the key, and we have
150 * new->key < old->key. We insert new above old, on
151 * the left ;
152 *
153 * - the tree does not contain the key, and we have
154 * new->key > old->key. We insert new above old, on
155 * the right ;
156 *
157 * - the tree does contain the key, which implies it
158 * is alone. We add the new key next to it as a
159 * first duplicate.
160 *
161 * The last two cases can easily be partially merged.
162 */
163 bit = string_equal_bits(new->key, old->key, bit);
164 diff = cmp_bits(new->key, old->key, bit);
165
166 if (diff < 0) {
167 new->node.leaf_p = new_left;
168 old->node.leaf_p = new_rght;
169 new->node.branches.b[EB_LEFT] = new_leaf;
170 new->node.branches.b[EB_RGHT] = old_leaf;
171 } else {
172 /* we may refuse to duplicate this key if the tree is
173 * tagged as containing only unique keys.
174 */
175 if (diff == 0 && eb_gettag(root_right))
176 return old;
177
178 /* new->key >= old->key, new goes the right */
179 old->node.leaf_p = new_left;
180 new->node.leaf_p = new_rght;
181 new->node.branches.b[EB_LEFT] = old_leaf;
182 new->node.branches.b[EB_RGHT] = new_leaf;
183
184 if (diff == 0) {
185 new->node.bit = -1;
186 root->b[side] = eb_dotag(&new->node.branches, EB_NODE);
187 return new;
188 }
189 }
190 break;
191 }
192
193 /* OK we're walking down this link */
194 old = container_of(eb_untag(troot, EB_NODE),
195 struct ebpt_node, node.branches);
Willy Tarreau3a932442010-05-09 19:29:23 +0200196 old_node_bit = old->node.bit;
Willy Tarreauc2186022009-10-26 19:48:54 +0100197
198 /* Stop going down when we don't have common bits anymore. We
199 * also stop in front of a duplicates tree because it means we
200 * have to insert above. Note: we can compare more bits than
201 * the current node's because as long as they are identical, we
202 * know we descend along the correct side.
203 */
Willy Tarreau3a932442010-05-09 19:29:23 +0200204 if (old_node_bit < 0) {
Willy Tarreauc2186022009-10-26 19:48:54 +0100205 /* we're above a duplicate tree, we must compare till the end */
206 bit = string_equal_bits(new->key, old->key, bit);
207 goto dup_tree;
208 }
Willy Tarreau3a932442010-05-09 19:29:23 +0200209 else if (bit < old_node_bit) {
Willy Tarreauc2186022009-10-26 19:48:54 +0100210 bit = string_equal_bits(new->key, old->key, bit);
211 }
212
Willy Tarreau3a932442010-05-09 19:29:23 +0200213 if (bit < old_node_bit) { /* we don't have all bits in common */
Willy Tarreauc2186022009-10-26 19:48:54 +0100214 /* The tree did not contain the key, so we insert <new> before the node
215 * <old>, and set ->bit to designate the lowest bit position in <new>
216 * which applies to ->branches.b[].
217 */
218 eb_troot_t *new_left, *new_rght;
219 eb_troot_t *new_leaf, *old_node;
220 dup_tree:
221 new_left = eb_dotag(&new->node.branches, EB_LEFT);
222 new_rght = eb_dotag(&new->node.branches, EB_RGHT);
223 new_leaf = eb_dotag(&new->node.branches, EB_LEAF);
224 old_node = eb_dotag(&old->node.branches, EB_NODE);
225
226 new->node.node_p = old->node.node_p;
227
228 diff = cmp_bits(new->key, old->key, bit);
229 if (diff < 0) {
230 new->node.leaf_p = new_left;
231 old->node.node_p = new_rght;
232 new->node.branches.b[EB_LEFT] = new_leaf;
233 new->node.branches.b[EB_RGHT] = old_node;
234 }
235 else if (diff > 0) {
236 old->node.node_p = new_left;
237 new->node.leaf_p = new_rght;
238 new->node.branches.b[EB_LEFT] = old_node;
239 new->node.branches.b[EB_RGHT] = new_leaf;
240 }
241 else {
242 struct eb_node *ret;
243 ret = eb_insert_dup(&old->node, &new->node);
244 return container_of(ret, struct ebpt_node, node);
245 }
246 break;
247 }
248
249 /* walk down */
250 root = &old->node.branches;
Willy Tarreau3a932442010-05-09 19:29:23 +0200251 side = (((unsigned char *)new->key)[old_node_bit >> 3] >> (~old_node_bit & 7)) & 1;
Willy Tarreauc2186022009-10-26 19:48:54 +0100252 troot = root->b[side];
253 }
254
255 /* Ok, now we are inserting <new> between <root> and <old>. <old>'s
256 * parent is already set to <new>, and the <root>'s branch is still in
257 * <side>. Update the root's leaf till we have it. Note that we can also
258 * find the side by checking the side of new->node.node_p.
259 */
260
261 /* We need the common higher bits between new->key and old->key.
262 * This number of bits is already in <bit>.
263 */
264 new->node.bit = bit;
265 root->b[side] = eb_dotag(&new->node.branches, EB_NODE);
266 return new;
267}
268