/* * Elastic Binary Trees - macros to manipulate String data nodes. * Version 6.0.6 * (C) 2002-2011 - Willy Tarreau * * 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 */ /* These functions and macros rely on Multi-Byte nodes */ #ifndef _EBSTTREE_H #define _EBSTTREE_H #include "ebtree.h" #include "ebmbtree.h" /* The following functions are not inlined by default. They are declared * in ebsttree.c, which simply relies on their inline version. */ REGPRM2 struct ebmb_node *ebst_lookup(struct eb_root *root, const char *x); REGPRM2 struct ebmb_node *ebst_insert(struct eb_root *root, struct ebmb_node *new); /* Find the first occurrence of a length string in the tree . * It's the caller's reponsibility to use this function only on trees which * only contain zero-terminated strings, and that no null character is present * in string in the first chars. If none can be found, return NULL. */ static forceinline struct ebmb_node * ebst_lookup_len(struct eb_root *root, const char *x, unsigned int len) { struct ebmb_node *node; node = ebmb_lookup(root, x, len); if (!node || node->key[len] != 0) return NULL; return node; } /* Find the first occurrence of a zero-terminated string in the tree . * 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 ebmb_node *__ebst_lookup(struct eb_root *root, const void *x) { struct ebmb_node *node; eb_troot_t *troot; int bit; int node_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 ebmb_node, node.branches); if (strcmp((char *)node->key, x) == 0) return node; else return NULL; } node = container_of(eb_untag(troot, EB_NODE), struct ebmb_node, node.branches); node_bit = node->node.bit; if (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((char *)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 ebmb_node, node.branches); return node; } /* OK, normal data node, let's walk down but don't compare data * if we already reached the end of the key. */ if (likely(bit >= 0)) { bit = string_equal_bits(x, node->key, bit); if (likely(bit < node_bit)) { if (bit >= 0) return NULL; /* no more common bits */ /* bit < 0 : we reached the end of the key. If we * are in a tree with unique keys, we can return * this node. Otherwise we have to walk it down * and stop comparing bits. */ if (eb_gettag(root->b[EB_RGHT])) return node; } /* if the bit is larger than the node's, we must bound it * because we might have compared too many bytes with an * inappropriate leaf. For a test, build a tree from "0", * "WW", "W", "S" inserted in this exact sequence and lookup * "W" => "S" is returned without this assignment. */ else bit = node_bit; } troot = node->node.branches.b[(((unsigned char*)x)[node_bit >> 3] >> (~node_bit & 7)) & 1]; } } /* Insert ebmb_node into subtree starting at node root . Only * new->key needs be set with the zero-terminated string key. The ebmb_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 ebmb_node * __ebst_insert(struct eb_root *root, struct ebmb_node *new) { struct ebmb_node *old; unsigned int side; eb_troot_t *troot; eb_troot_t *root_right; int diff; int bit; int old_node_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 for the node node we are inserting, * for the node we attach it to, and for the node we are * displacing below . will always point to the future node * (tagged with its type). carries the side the node 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 ebmb_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. */ if (bit >= 0) bit = string_equal_bits(new->key, old->key, bit); if (bit < 0) { /* key was already there */ /* we may refuse to duplicate this key if the tree is * tagged as containing only unique keys. */ if (eb_gettag(root_right)) return old; /* new arbitrarily goes to the right and tops the dup tree */ 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; new->node.bit = -1; root->b[side] = eb_dotag(&new->node.branches, EB_NODE); return new; } diff = cmp_bits(new->key, old->key, bit); if (diff < 0) { /* new->key < old->key, new takes the left */ 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 { /* new->key > old->key, new takes 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; } break; } /* OK we're walking down this link */ old = container_of(eb_untag(troot, EB_NODE), struct ebmb_node, node.branches); old_node_bit = old->node.bit; /* 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 (bit >= 0 && (bit < old_node_bit || old_node_bit < 0)) bit = string_equal_bits(new->key, old->key, bit); if (unlikely(bit < 0)) { /* Perfect match, we must only stop on head of dup tree * or walk down to a leaf. */ if (old_node_bit < 0) { /* We know here that string_equal_bits matched all * bits and that we're on top of a dup tree, then * we can perform the dup insertion and return. */ struct eb_node *ret; ret = eb_insert_dup(&old->node, &new->node); return container_of(ret, struct ebmb_node, node); } /* OK so let's walk down */ } else if (bit < old_node_bit || old_node_bit < 0) { /* The tree did not contain the key, or we stopped on top of a dup * tree, possibly containing the key. In the former case, we insert * before the node , and set ->bit to designate the lowest * bit position in which applies to ->branches.b[]. In the later * case, we add the key to the existing dup tree. Note that we cannot * enter here if we match an intermediate node's key that is not the * head of a dup tree. */ eb_troot_t *new_left, *new_rght; eb_troot_t *new_leaf, *old_node; 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; /* we can never match all bits here */ 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 { 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; } break; } /* walk down */ root = &old->node.branches; side = (new->key[old_node_bit >> 3] >> (~old_node_bit & 7)) & 1; troot = root->b[side]; } /* Ok, now we are inserting between and . 's * parent is already set to , and the 's branch is still in * . 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 . * NOTE: we can't get here whit bit < 0 since we found a dup ! */ new->node.bit = bit; root->b[side] = eb_dotag(&new->node.branches, EB_NODE); return new; } #endif /* _EBSTTREE_H */