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			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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| 
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| /* These functions and macros rely on Multi-Byte nodes */
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| 
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| #ifndef _EBSTTREE_H
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| #define _EBSTTREE_H
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| 
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| #include "ebtree.h"
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| #include "ebmbtree.h"
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| 
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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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| 
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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 responsibility 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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| 
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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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| 
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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 responsibility 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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| 
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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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| 
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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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| 
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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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| 
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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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| 
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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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| 
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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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| 
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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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| 
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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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| 
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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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| 
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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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| 
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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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| 
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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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| 
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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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| 
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| 			new->node.node_p = old->node.leaf_p;
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| 
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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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| 
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| 			if (bit < 0) {
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| 				/* key was already there */
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| 
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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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| 
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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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| 
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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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| 
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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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| 
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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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| 
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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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| 
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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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| 
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| 			new->node.node_p = old->node.node_p;
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| 
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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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| 
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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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| 
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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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| 
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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 with 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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| 
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| #endif /* _EBSTTREE_H */
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| 
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