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Diffstat (limited to '')
| -rw-r--r-- | src/ebpttree.c | 208 |
1 files changed, 208 insertions, 0 deletions
diff --git a/src/ebpttree.c b/src/ebpttree.c new file mode 100644 index 0000000..558d334 --- /dev/null +++ b/src/ebpttree.c @@ -0,0 +1,208 @@ +/* + * 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 <import/ebpttree.h> + +struct ebpt_node *ebpt_insert(struct eb_root *root, struct ebpt_node *new) +{ + return __ebpt_insert(root, new); +} + +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. + */ +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. + */ +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; +} |