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03e7853581
This used to be a minor optimization on ix86 where registers are scarce and the calling convention not very efficient, but this platform is not relevant enough anymore to warrant all this dirt in the code for the sake of saving 1 or 2% of performance. Modern platforms don't use this at all since their calling convention already defaults to using several registers so better get rid of this once for all.
325 lines
10 KiB
C
325 lines
10 KiB
C
/*
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* Elastic Binary Trees - macros to manipulate String 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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/* These functions and macros rely on Multi-Byte nodes */
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#ifndef _EBSTTREE_H
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#define _EBSTTREE_H
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#include "ebtree.h"
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#include "ebmbtree.h"
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/* The following functions are not inlined by default. They are declared
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* in ebsttree.c, which simply relies on their inline version.
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*/
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struct ebmb_node *ebst_lookup(struct eb_root *root, const char *x);
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struct ebmb_node *ebst_insert(struct eb_root *root, struct ebmb_node *new);
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/* Find the first occurrence of a length <len> string <x> in the tree <root>.
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* It's the caller's reponsibility to use this function only on trees which
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* only contain zero-terminated strings, and that no null character is present
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* in string <x> in the first <len> chars. If none can be found, return NULL.
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*/
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static forceinline struct ebmb_node *
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ebst_lookup_len(struct eb_root *root, const char *x, unsigned int len)
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{
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struct ebmb_node *node;
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node = ebmb_lookup(root, x, len);
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if (!node || node->key[len] != 0)
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return NULL;
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return node;
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}
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/* Find the first occurrence of a zero-terminated string <x> in the tree <root>.
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* It's the caller's reponsibility to use this function only on trees which
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* only contain zero-terminated strings. If none can be found, return NULL.
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*/
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static forceinline struct ebmb_node *__ebst_lookup(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;
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int bit;
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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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bit = 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 (strcmp((char *)node->key, x) == 0)
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return node;
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else
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return NULL;
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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 (strcmp((char *)node->key, x) != 0)
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return NULL;
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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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/* OK, normal data node, let's walk down but don't compare data
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* if we already reached the end of the key.
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*/
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if (likely(bit >= 0)) {
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bit = string_equal_bits(x, node->key, bit);
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if (likely(bit < node_bit)) {
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if (bit >= 0)
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return NULL; /* no more common bits */
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/* bit < 0 : we reached the end of the key. If we
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* are in a tree with unique keys, we can return
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* this node. Otherwise we have to walk it down
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* and stop comparing bits.
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*/
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if (eb_gettag(root->b[EB_RGHT]))
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return node;
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}
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/* if the bit is larger than the node's, we must bound it
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* because we might have compared too many bytes with an
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* inappropriate leaf. For a test, build a tree from "0",
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* "WW", "W", "S" inserted in this exact sequence and lookup
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* "W" => "S" is returned without this assignment.
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*/
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else
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bit = node_bit;
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}
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troot = node->node.branches.b[(((unsigned char*)x)[node_bit >> 3] >>
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(~node_bit & 7)) & 1];
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}
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}
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/* Insert ebmb_node <new> into subtree starting at node root <root>. Only
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* new->key needs be set with the zero-terminated string key. The ebmb_node is
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* returned. If root->b[EB_RGHT]==1, the tree may only contain unique keys. The
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* caller is responsible for properly terminating the key with a zero.
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*/
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static forceinline struct ebmb_node *
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__ebst_insert(struct eb_root *root, struct ebmb_node *new)
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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;
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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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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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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 ebmb_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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if (bit >= 0)
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bit = string_equal_bits(new->key, old->key, bit);
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if (bit < 0) {
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/* key was already there */
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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 (eb_gettag(root_right))
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return old;
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/* new arbitrarily goes to the right and tops the dup tree */
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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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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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diff = cmp_bits(new->key, old->key, bit);
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if (diff < 0) {
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/* new->key < old->key, new takes the left */
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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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/* new->key > old->key, new takes 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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}
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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 ebmb_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 (bit >= 0 && (bit < old_node_bit || old_node_bit < 0))
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bit = string_equal_bits(new->key, old->key, bit);
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if (unlikely(bit < 0)) {
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/* Perfect match, we must only stop on head of dup tree
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* or walk down to a leaf.
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*/
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if (old_node_bit < 0) {
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/* We know here that string_equal_bits matched all
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* bits and that we're on top of a dup tree, then
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* we can perform the dup insertion and return.
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*/
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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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/* OK so let's walk down */
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}
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else if (bit < old_node_bit || old_node_bit < 0) {
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/* The tree did not contain the key, or we stopped on top of a dup
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* tree, possibly containing the key. In the former case, we insert
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* <new> before the node <old>, and set ->bit to designate the lowest
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* bit position in <new> which applies to ->branches.b[]. In the later
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* case, we add the key to the existing dup tree. Note that we cannot
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* enter here if we match an intermediate node's key that is not the
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* head of a dup tree.
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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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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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/* we can never match all bits here */
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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 {
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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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break;
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}
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/* walk down */
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root = &old->node.branches;
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side = (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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* NOTE: we can't get here whit bit < 0 since we found a dup !
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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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#endif /* _EBSTTREE_H */
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