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ce3d44a06a
(from ebtree 6.0.5) Last bugfix has introduced a de-optimization in the lookup function because it artificially extended the scope of some local variables, which resulted in higher stack usage and more numerous moves between stack and registers. We can reduce that by moving the return code out of the loop, because gcc notices that it never needs both "troot" and "node" at the same time and can use the same register for both. Doing so has reduced the code size by 39 bytes for the lookup function alone, and has sensibly reduced the instruction dependencies caused by data moves. (cherry picked from commit 59be3cdb96296b65a57aff30cc203269f9a94ebe) It should be backported to 1.4 if previous ebtree fix is backported.
306 lines
9.9 KiB
C
306 lines
9.9 KiB
C
/*
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* Elastic Binary Trees - macros for Indirect Multi-Byte data nodes.
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* Version 6.0.5
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* (C) 2002-2011 - Willy Tarreau <w@1wt.eu>
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*
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* This program is free software; you can redistribute it and/or modify
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* it under the terms of the GNU General Public License as published by
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* the Free Software Foundation; either version 2 of the License, or
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* (at your option) any later version.
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*
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* This program 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
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* GNU General Public License for more details.
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*
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* You should have received a copy of the GNU General Public License
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* along with this program; if not, write to the Free Software
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* Foundation, Inc., 51 Franklin St, Fifth Floor, Boston, MA 02110-1301 USA
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*/
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#include <string.h>
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#include "ebtree.h"
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#include "ebpttree.h"
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/* These functions and macros rely on Pointer nodes and use the <key> entry as
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* a pointer to an indirect key. Most operations are performed using ebpt_*.
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*/
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/* The following functions are not inlined by default. They are declared
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* in ebimtree.c, which simply relies on their inline version.
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*/
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REGPRM3 struct ebpt_node *ebim_lookup(struct eb_root *root, const void *x, unsigned int len);
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REGPRM3 struct ebpt_node *ebim_insert(struct eb_root *root, struct ebpt_node *new, unsigned int len);
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/* Find the first occurence 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 ebpt_node *
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__ebim_lookup(struct eb_root *root, const void *x, unsigned int len)
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{
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struct ebpt_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 ebpt_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 ebpt_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 surprizing 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 (*(unsigned char*)(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 (((*(unsigned char*)(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 ebpt_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 ebpt_node <new> into subtree starting at node root <root>.
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* Only new->key needs be set with the key. The ebpt_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.
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*/
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static forceinline struct ebpt_node *
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__ebim_insert(struct eb_root *root, struct ebpt_node *new, unsigned int len)
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{
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struct ebpt_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 = root;
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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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len <<= 3;
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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 ebpt_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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bit = equal_bits(new->key, old->key, bit, 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.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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/* 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 (diff == 0 && eb_gettag(root_right))
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return old;
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/* new->key >= old->key, new goes 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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if (diff == 0) {
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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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}
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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 ebpt_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 (old_node_bit < 0) {
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/* we're above a duplicate tree, we must compare till the end */
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bit = equal_bits(new->key, old->key, bit, len);
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goto dup_tree;
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}
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else if (bit < old_node_bit) {
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bit = equal_bits(new->key, old->key, bit, old_node_bit);
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}
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if (bit < old_node_bit) { /* we don't have all bits in common */
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/* The tree did not contain the key, so we insert <new> before the node
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* <old>, and set ->bit to designate the lowest bit position in <new>
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* which applies to ->branches.b[].
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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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dup_tree:
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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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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 if (diff > 0) {
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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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else {
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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 ebpt_node, node);
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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 = (((unsigned char *)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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*/
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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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