810 lines
26 KiB
C
810 lines
26 KiB
C
/*
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* Elastic Binary Trees - macros and structures for Multi-Byte 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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#ifndef _EBMBTREE_H
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#define _EBMBTREE_H
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#include <string.h>
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#include "ebtree.h"
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/* Return the structure of type <type> whose member <member> points to <ptr> */
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#define ebmb_entry(ptr, type, member) container_of(ptr, type, member)
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#define EBMB_ROOT EB_ROOT
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#define EBMB_TREE_HEAD EB_TREE_HEAD
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/* This structure carries a node, a leaf, and a key. It must start with the
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* eb_node so that it can be cast into an eb_node. We could also have put some
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* sort of transparent union here to reduce the indirection level, but the fact
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* is, the end user is not meant to manipulate internals, so this is pointless.
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* The 'node.bit' value here works differently from scalar types, as it contains
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* the number of identical bits between the two branches.
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*/
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struct ebmb_node {
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struct eb_node node; /* the tree node, must be at the beginning */
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unsigned char key[0]; /* the key, its size depends on the application */
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};
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/*
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* Exported functions and macros.
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* Many of them are always inlined because they are extremely small, and
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* are generally called at most once or twice in a program.
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*/
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/* Return leftmost node in the tree, or NULL if none */
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static forceinline struct ebmb_node *ebmb_first(struct eb_root *root)
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{
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return ebmb_entry(eb_first(root), struct ebmb_node, node);
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}
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/* Return rightmost node in the tree, or NULL if none */
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static forceinline struct ebmb_node *ebmb_last(struct eb_root *root)
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{
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return ebmb_entry(eb_last(root), struct ebmb_node, node);
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}
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/* Return next node in the tree, or NULL if none */
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static forceinline struct ebmb_node *ebmb_next(struct ebmb_node *ebmb)
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{
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return ebmb_entry(eb_next(&ebmb->node), struct ebmb_node, node);
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}
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/* Return previous node in the tree, or NULL if none */
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static forceinline struct ebmb_node *ebmb_prev(struct ebmb_node *ebmb)
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{
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return ebmb_entry(eb_prev(&ebmb->node), struct ebmb_node, node);
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}
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/* Return next leaf node within a duplicate sub-tree, or NULL if none. */
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static inline struct ebmb_node *ebmb_next_dup(struct ebmb_node *ebmb)
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{
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return ebmb_entry(eb_next_dup(&ebmb->node), struct ebmb_node, node);
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}
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/* Return previous leaf node within a duplicate sub-tree, or NULL if none. */
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static inline struct ebmb_node *ebmb_prev_dup(struct ebmb_node *ebmb)
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{
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return ebmb_entry(eb_prev_dup(&ebmb->node), struct ebmb_node, node);
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}
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/* Return next node in the tree, skipping duplicates, or NULL if none */
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static forceinline struct ebmb_node *ebmb_next_unique(struct ebmb_node *ebmb)
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{
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return ebmb_entry(eb_next_unique(&ebmb->node), struct ebmb_node, node);
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}
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/* Return previous node in the tree, skipping duplicates, or NULL if none */
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static forceinline struct ebmb_node *ebmb_prev_unique(struct ebmb_node *ebmb)
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{
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return ebmb_entry(eb_prev_unique(&ebmb->node), struct ebmb_node, node);
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}
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/* Delete node from the tree if it was linked in. Mark the node unused. Note
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* that this function relies on a non-inlined generic function: eb_delete.
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*/
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static forceinline void ebmb_delete(struct ebmb_node *ebmb)
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{
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eb_delete(&ebmb->node);
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}
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/* The following functions are not inlined by default. They are declared
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* in ebmbtree.c, which simply relies on their inline version.
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*/
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REGPRM3 struct ebmb_node *ebmb_lookup(struct eb_root *root, const void *x, unsigned int len);
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REGPRM3 struct ebmb_node *ebmb_insert(struct eb_root *root, struct ebmb_node *new, unsigned int len);
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REGPRM2 struct ebmb_node *ebmb_lookup_longest(struct eb_root *root, const void *x);
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REGPRM3 struct ebmb_node *ebmb_lookup_prefix(struct eb_root *root, const void *x, unsigned int pfx);
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REGPRM3 struct ebmb_node *ebmb_insert_prefix(struct eb_root *root, struct ebmb_node *new, unsigned int len);
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/* The following functions are less likely to be used directly, because their
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* code is larger. The non-inlined version is preferred.
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*/
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/* Delete node from the tree if it was linked in. Mark the node unused. */
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static forceinline void __ebmb_delete(struct ebmb_node *ebmb)
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{
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__eb_delete(&ebmb->node);
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}
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/* Find the first occurrence of a key of a least <len> bytes matching <x> in the
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* tree <root>. The caller is responsible for ensuring that <len> will not exceed
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* the common parts between the tree's keys and <x>. In case of multiple matches,
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* the leftmost node is returned. This means that this function can be used to
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* lookup string keys by prefix if all keys in the tree are zero-terminated. If
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* no match is found, NULL is returned. Returns first node if <len> is zero.
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*/
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static forceinline struct ebmb_node *__ebmb_lookup(struct eb_root *root, const void *x, unsigned int len)
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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 pos, side;
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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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goto ret_null;
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if (unlikely(len == 0))
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goto walk_down;
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pos = 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 (memcmp(node->key + pos, x, len) != 0)
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goto ret_null;
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else
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goto ret_node;
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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 (memcmp(node->key + pos, x, len) != 0)
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goto ret_null;
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else
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goto walk_left;
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}
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/* OK, normal data node, let's walk down. We check if all full
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* bytes are equal, and we start from the last one we did not
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* completely check. We stop as soon as we reach the last byte,
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* because we must decide to go left/right or abort.
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*/
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node_bit = ~node_bit + (pos << 3) + 8; // = (pos<<3) + (7 - node_bit)
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if (node_bit < 0) {
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/* This surprising construction gives better performance
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* because gcc does not try to reorder the loop. Tested to
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* be fine with 2.95 to 4.2.
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*/
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while (1) {
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if (node->key[pos++] ^ *(unsigned char*)(x++))
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goto ret_null; /* more than one full byte is different */
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if (--len == 0)
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goto walk_left; /* return first node if all bytes matched */
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node_bit += 8;
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if (node_bit >= 0)
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break;
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}
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}
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/* here we know that only the last byte differs, so node_bit < 8.
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* We have 2 possibilities :
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* - more than the last bit differs => return NULL
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* - walk down on side = (x[pos] >> node_bit) & 1
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*/
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side = *(unsigned char *)x >> node_bit;
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if (((node->key[pos] >> node_bit) ^ side) > 1)
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goto ret_null;
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side &= 1;
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troot = node->node.branches.b[side];
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}
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walk_left:
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troot = node->node.branches.b[EB_LEFT];
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walk_down:
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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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ret_node:
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return node;
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ret_null:
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return NULL;
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}
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/* Insert ebmb_node <new> into subtree starting at node root <root>.
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* Only new->key needs be set with the key. The ebmb_node is returned.
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* If root->b[EB_RGHT]==1, the tree may only contain unique keys. The
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* len is specified in bytes. It is absolutely mandatory that this length
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* is the same for all keys in the tree. This function cannot be used to
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* insert strings.
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*/
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static forceinline struct ebmb_node *
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__ebmb_insert(struct eb_root *root, struct ebmb_node *new, unsigned int len)
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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, **up_ptr;
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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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eb_troot_t *new_left, *new_rght;
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eb_troot_t *new_leaf;
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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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/* insert above a 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->node.node_p = old->node.leaf_p;
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up_ptr = &old->node.leaf_p;
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goto check_bit_and_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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if (unlikely(old->node.bit < 0)) {
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/* We're above a duplicate tree, so we must compare the whole value */
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new->node.node_p = old->node.node_p;
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up_ptr = &old->node.node_p;
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check_bit_and_break:
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bit = equal_bits(new->key, old->key, bit, len << 3);
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break;
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}
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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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bit = equal_bits(new->key, old->key, bit, old_node_bit);
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if (unlikely(bit < old_node_bit)) {
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/* The tree did not contain the key, so we insert <new> before the
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* node <old>, and set ->bit to designate the lowest bit position in
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* <new> which applies to ->branches.b[].
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*/
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new->node.node_p = old->node.node_p;
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up_ptr = &old->node.node_p;
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break;
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}
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/* we don't want to skip bits for further comparisons, so we must limit <bit>.
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* However, since we're going down around <old_node_bit>, we know it will be
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* properly matched, so we can skip this bit.
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*/
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bit = old_node_bit + 1;
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/* walk down */
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root = &old->node.branches;
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side = old_node_bit & 7;
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side ^= 7;
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side = (new->key[old_node_bit >> 3] >> side) & 1;
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troot = root->b[side];
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}
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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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new->node.bit = bit;
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/* Note: we can compare more bits than the current node's because as
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* long as they are identical, we know we descend along the correct
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* side. However we don't want to start to compare past the end.
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*/
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diff = 0;
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if (((unsigned)bit >> 3) < len)
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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.bit = -1; /* mark as new dup tree, just in case */
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if (likely(eb_gettag(root_right))) {
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/* we 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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return old;
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}
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if (eb_gettag(troot) != EB_LEAF) {
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/* there was already a dup tree below */
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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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/* otherwise fall through */
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}
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if (diff >= 0) {
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new->node.branches.b[EB_LEFT] = troot;
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new->node.branches.b[EB_RGHT] = new_leaf;
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new->node.leaf_p = new_rght;
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*up_ptr = new_left;
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}
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else if (diff < 0) {
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new->node.branches.b[EB_LEFT] = new_leaf;
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new->node.branches.b[EB_RGHT] = troot;
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new->node.leaf_p = new_left;
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*up_ptr = new_rght;
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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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root->b[side] = eb_dotag(&new->node.branches, EB_NODE);
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return new;
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}
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/* Find the first occurrence of the longest prefix matching a key <x> in the
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* tree <root>. It's the caller's responsibility to ensure that key <x> is at
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* least as long as the keys in the tree. Note that this can be ensured by
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* having a byte at the end of <x> which cannot be part of any prefix, typically
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* the trailing zero for a string. If none can be found, return NULL.
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*/
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static forceinline struct ebmb_node *__ebmb_lookup_longest(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, *cover;
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int pos, side;
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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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cover = NULL;
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pos = 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 (check_bits(x - pos, node->key, pos, node->node.pfx))
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goto not_found;
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return node;
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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 (check_bits(x - pos, node->key, pos, node->node.pfx))
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goto not_found;
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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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node_bit >>= 1; /* strip cover bit */
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node_bit = ~node_bit + (pos << 3) + 8; // = (pos<<3) + (7 - node_bit)
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if (node_bit < 0) {
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/* This uncommon construction gives better performance
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* because gcc does not try to reorder the loop. Tested to
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* be fine with 2.95 to 4.2.
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*/
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while (1) {
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x++; pos++;
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if (node->key[pos-1] ^ *(unsigned char*)(x-1))
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goto not_found; /* more than one full byte is different */
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node_bit += 8;
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if (node_bit >= 0)
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break;
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}
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}
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/* here we know that only the last byte differs, so 0 <= node_bit <= 7.
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* We have 2 possibilities :
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* - more than the last bit differs => data does not match
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* - walk down on side = (x[pos] >> node_bit) & 1
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*/
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side = *(unsigned char *)x >> node_bit;
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if (((node->key[pos] >> node_bit) ^ side) > 1)
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goto not_found;
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if (!(node->node.bit & 1)) {
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/* This is a cover node, let's keep a reference to it
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* for later. The covering subtree is on the left, and
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* the covered subtree is on the right, so we have to
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* walk down right.
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*/
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cover = node->node.branches.b[EB_LEFT];
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troot = node->node.branches.b[EB_RGHT];
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continue;
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}
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side &= 1;
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troot = node->node.branches.b[side];
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}
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not_found:
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/* Walk down last cover tre if it exists. It does not matter if cover is NULL */
|
|
return ebmb_entry(eb_walk_down(cover, EB_LEFT), struct ebmb_node, node);
|
|
}
|
|
|
|
|
|
/* Find the first occurrence of a prefix matching a key <x> of <pfx> BITS in the
|
|
* tree <root>. It's the caller's responsibility to ensure that key <x> is at
|
|
* least as long as the keys in the tree. Note that this can be ensured by
|
|
* having a byte at the end of <x> which cannot be part of any prefix, typically
|
|
* the trailing zero for a string. If none can be found, return NULL.
|
|
*/
|
|
static forceinline struct ebmb_node *__ebmb_lookup_prefix(struct eb_root *root, const void *x, unsigned int pfx)
|
|
{
|
|
struct ebmb_node *node;
|
|
eb_troot_t *troot;
|
|
int pos, side;
|
|
int node_bit;
|
|
|
|
troot = root->b[EB_LEFT];
|
|
if (unlikely(troot == NULL))
|
|
return NULL;
|
|
|
|
pos = 0;
|
|
while (1) {
|
|
if ((eb_gettag(troot) == EB_LEAF)) {
|
|
node = container_of(eb_untag(troot, EB_LEAF),
|
|
struct ebmb_node, node.branches);
|
|
if (node->node.pfx != pfx)
|
|
return NULL;
|
|
if (check_bits(x - pos, node->key, pos, node->node.pfx))
|
|
return NULL;
|
|
return node;
|
|
}
|
|
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 (node->node.pfx != pfx)
|
|
return NULL;
|
|
if (check_bits(x - pos, node->key, pos, node->node.pfx))
|
|
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;
|
|
}
|
|
|
|
node_bit >>= 1; /* strip cover bit */
|
|
node_bit = ~node_bit + (pos << 3) + 8; // = (pos<<3) + (7 - node_bit)
|
|
if (node_bit < 0) {
|
|
/* This uncommon construction gives better performance
|
|
* because gcc does not try to reorder the loop. Tested to
|
|
* be fine with 2.95 to 4.2.
|
|
*/
|
|
while (1) {
|
|
x++; pos++;
|
|
if (node->key[pos-1] ^ *(unsigned char*)(x-1))
|
|
return NULL; /* more than one full byte is different */
|
|
node_bit += 8;
|
|
if (node_bit >= 0)
|
|
break;
|
|
}
|
|
}
|
|
|
|
/* here we know that only the last byte differs, so 0 <= node_bit <= 7.
|
|
* We have 2 possibilities :
|
|
* - more than the last bit differs => data does not match
|
|
* - walk down on side = (x[pos] >> node_bit) & 1
|
|
*/
|
|
side = *(unsigned char *)x >> node_bit;
|
|
if (((node->key[pos] >> node_bit) ^ side) > 1)
|
|
return NULL;
|
|
|
|
if (!(node->node.bit & 1)) {
|
|
/* This is a cover node, it may be the entry we're
|
|
* looking for. We already know that it matches all the
|
|
* bits, let's compare prefixes and descend the cover
|
|
* subtree if they match.
|
|
*/
|
|
if ((unsigned short)node->node.bit >> 1 == pfx)
|
|
troot = node->node.branches.b[EB_LEFT];
|
|
else
|
|
troot = node->node.branches.b[EB_RGHT];
|
|
continue;
|
|
}
|
|
side &= 1;
|
|
troot = node->node.branches.b[side];
|
|
}
|
|
}
|
|
|
|
|
|
/* Insert ebmb_node <new> into a prefix subtree starting at node root <root>.
|
|
* Only new->key and new->pfx need be set with the key and its prefix length.
|
|
* Note that bits between <pfx> and <len> are theorically ignored and should be
|
|
* zero, as it is not certain yet that they will always be ignored everywhere
|
|
* (eg in bit compare functions).
|
|
* The ebmb_node is returned.
|
|
* If root->b[EB_RGHT]==1, the tree may only contain unique keys. The
|
|
* len is specified in bytes.
|
|
*/
|
|
static forceinline struct ebmb_node *
|
|
__ebmb_insert_prefix(struct eb_root *root, struct ebmb_node *new, unsigned int len)
|
|
{
|
|
struct ebmb_node *old;
|
|
unsigned int side;
|
|
eb_troot_t *troot, **up_ptr;
|
|
eb_troot_t *root_right;
|
|
int diff;
|
|
int bit;
|
|
eb_troot_t *new_left, *new_rght;
|
|
eb_troot_t *new_leaf;
|
|
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;
|
|
}
|
|
|
|
len <<= 3;
|
|
if (len > new->node.pfx)
|
|
len = new->node.pfx;
|
|
|
|
/* 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 <new> for the node node we are inserting, <root>
|
|
* for the node we attach it to, and <old> for the node we are
|
|
* displacing below <new>. <troot> will always point to the future node
|
|
* (tagged with its type). <side> carries the side the node <new> 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)) {
|
|
/* Insert above a leaf. Note that this leaf could very
|
|
* well be part of a cover node.
|
|
*/
|
|
old = container_of(eb_untag(troot, EB_LEAF),
|
|
struct ebmb_node, node.branches);
|
|
new->node.node_p = old->node.leaf_p;
|
|
up_ptr = &old->node.leaf_p;
|
|
goto check_bit_and_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;
|
|
/* Note that old_node_bit can be :
|
|
* < 0 : dup tree
|
|
* = 2N : cover node for N bits
|
|
* = 2N+1 : normal node at N bits
|
|
*/
|
|
|
|
if (unlikely(old_node_bit < 0)) {
|
|
/* We're above a duplicate tree, so we must compare the whole value */
|
|
new->node.node_p = old->node.node_p;
|
|
up_ptr = &old->node.node_p;
|
|
check_bit_and_break:
|
|
/* No need to compare everything if the leaves are shorter than the new one. */
|
|
if (len > old->node.pfx)
|
|
len = old->node.pfx;
|
|
bit = equal_bits(new->key, old->key, bit, len);
|
|
break;
|
|
}
|
|
|
|
/* WARNING: for the two blocks below, <bit> is counted in half-bits */
|
|
|
|
bit = equal_bits(new->key, old->key, bit, old_node_bit >> 1);
|
|
bit = (bit << 1) + 1; // assume comparisons with normal nodes
|
|
|
|
/* we must always check that our prefix is larger than the nodes
|
|
* we visit, otherwise we have to stop going down. The following
|
|
* test is able to stop before both normal and cover nodes.
|
|
*/
|
|
if (bit >= (new->node.pfx << 1) && (new->node.pfx << 1) < old_node_bit) {
|
|
/* insert cover node here on the left */
|
|
new->node.node_p = old->node.node_p;
|
|
up_ptr = &old->node.node_p;
|
|
new->node.bit = new->node.pfx << 1;
|
|
diff = -1;
|
|
goto insert_above;
|
|
}
|
|
|
|
if (unlikely(bit < old_node_bit)) {
|
|
/* The tree did not contain the key, so we insert <new> before the
|
|
* node <old>, and set ->bit to designate the lowest bit position in
|
|
* <new> which applies to ->branches.b[]. We know that the bit is not
|
|
* greater than the prefix length thanks to the test above.
|
|
*/
|
|
new->node.node_p = old->node.node_p;
|
|
up_ptr = &old->node.node_p;
|
|
new->node.bit = bit;
|
|
diff = cmp_bits(new->key, old->key, bit >> 1);
|
|
goto insert_above;
|
|
}
|
|
|
|
if (!(old_node_bit & 1)) {
|
|
/* if we encounter a cover node with our exact prefix length, it's
|
|
* necessarily the same value, so we insert there as a duplicate on
|
|
* the left. For that, we go down on the left and the leaf detection
|
|
* code will finish the job.
|
|
*/
|
|
if ((new->node.pfx << 1) == old_node_bit) {
|
|
root = &old->node.branches;
|
|
side = EB_LEFT;
|
|
troot = root->b[side];
|
|
continue;
|
|
}
|
|
|
|
/* cover nodes are always walked through on the right */
|
|
side = EB_RGHT;
|
|
bit = old_node_bit >> 1; /* recheck that bit */
|
|
root = &old->node.branches;
|
|
troot = root->b[side];
|
|
continue;
|
|
}
|
|
|
|
/* we don't want to skip bits for further comparisons, so we must limit <bit>.
|
|
* However, since we're going down around <old_node_bit>, we know it will be
|
|
* properly matched, so we can skip this bit.
|
|
*/
|
|
old_node_bit >>= 1;
|
|
bit = old_node_bit + 1;
|
|
|
|
/* walk down */
|
|
root = &old->node.branches;
|
|
side = old_node_bit & 7;
|
|
side ^= 7;
|
|
side = (new->key[old_node_bit >> 3] >> side) & 1;
|
|
troot = root->b[side];
|
|
}
|
|
|
|
/* Right here, we have 4 possibilities :
|
|
* - the tree does not contain any leaf matching the
|
|
* key, and we have new->key < old->key. We insert
|
|
* new above old, on the left ;
|
|
*
|
|
* - the tree does not contain any leaf matching the
|
|
* key, and we have new->key > old->key. We insert
|
|
* new above old, on the right ;
|
|
*
|
|
* - the tree does contain the key with the same prefix
|
|
* length. We add the new key next to it as a first
|
|
* duplicate (since it was alone).
|
|
*
|
|
* The last two cases can easily be partially merged.
|
|
*
|
|
* - the tree contains a leaf matching the key, we have
|
|
* to insert above it as a cover node. The leaf with
|
|
* the shortest prefix becomes the left subtree and
|
|
* the leaf with the longest prefix becomes the right
|
|
* one. The cover node gets the min of both prefixes
|
|
* as its new bit.
|
|
*/
|
|
|
|
/* first we want to ensure that we compare the correct bit, which means
|
|
* the largest common to both nodes.
|
|
*/
|
|
if (bit > new->node.pfx)
|
|
bit = new->node.pfx;
|
|
if (bit > old->node.pfx)
|
|
bit = old->node.pfx;
|
|
|
|
new->node.bit = (bit << 1) + 1; /* assume normal node by default */
|
|
|
|
/* if one prefix is included in the second one, we don't compare bits
|
|
* because they won't necessarily match, we just proceed with a cover
|
|
* node insertion.
|
|
*/
|
|
diff = 0;
|
|
if (bit < old->node.pfx && bit < new->node.pfx)
|
|
diff = cmp_bits(new->key, old->key, bit);
|
|
|
|
if (diff == 0) {
|
|
/* Both keys match. Either it's a duplicate entry or we have to
|
|
* put the shortest prefix left and the largest one right below
|
|
* a new cover node. By default, diff==0 means we'll be inserted
|
|
* on the right.
|
|
*/
|
|
new->node.bit--; /* anticipate cover node insertion */
|
|
if (new->node.pfx == old->node.pfx) {
|
|
new->node.bit = -1; /* mark as new dup tree, just in case */
|
|
|
|
if (unlikely(eb_gettag(root_right))) {
|
|
/* we refuse to duplicate this key if the tree is
|
|
* tagged as containing only unique keys.
|
|
*/
|
|
return old;
|
|
}
|
|
|
|
if (eb_gettag(troot) != EB_LEAF) {
|
|
/* there was already a dup tree below */
|
|
struct eb_node *ret;
|
|
ret = eb_insert_dup(&old->node, &new->node);
|
|
return container_of(ret, struct ebmb_node, node);
|
|
}
|
|
/* otherwise fall through to insert first duplicate */
|
|
}
|
|
/* otherwise we just rely on the tests below to select the right side */
|
|
else if (new->node.pfx < old->node.pfx)
|
|
diff = -1; /* force insertion to left side */
|
|
}
|
|
|
|
insert_above:
|
|
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);
|
|
|
|
if (diff >= 0) {
|
|
new->node.branches.b[EB_LEFT] = troot;
|
|
new->node.branches.b[EB_RGHT] = new_leaf;
|
|
new->node.leaf_p = new_rght;
|
|
*up_ptr = new_left;
|
|
}
|
|
else {
|
|
new->node.branches.b[EB_LEFT] = new_leaf;
|
|
new->node.branches.b[EB_RGHT] = troot;
|
|
new->node.leaf_p = new_left;
|
|
*up_ptr = new_rght;
|
|
}
|
|
|
|
root->b[side] = eb_dotag(&new->node.branches, EB_NODE);
|
|
return new;
|
|
}
|
|
|
|
|
|
|
|
#endif /* _EBMBTREE_H */
|
|
|