Adding upstream version 2.9.5.upstream/2.9.5

Signed-off-by: Daniel Baumann <daniel.baumann@progress-linux.org>

1 files changed, 329 insertions, 0 deletions

diff --git a/include/import/ebistree.h b/include/import/ebistree.h
new file mode 100644
index 0000000..a438fa1
--- /dev/null
+++ b/include/import/ebistree.h
@@ -0,0 +1,329 @@
+/*
+ * Elastic Binary Trees - macros to manipulate Indirect String data nodes.
+ * Version 6.0.6
+ * (C) 2002-2011 - Willy Tarreau <w@1wt.eu>
+ *
+ * This library is free software; you can redistribute it and/or
+ * modify it under the terms of the GNU Lesser General Public
+ * License as published by the Free Software Foundation, version 2.1
+ * exclusively.
+ *
+ * This library is distributed in the hope that it will be useful,
+ * but WITHOUT ANY WARRANTY; without even the implied warranty of
+ * MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the GNU
+ * Lesser General Public License for more details.
+ *
+ * You should have received a copy of the GNU Lesser General Public
+ * License along with this library; if not, write to the Free Software
+ * Foundation, Inc., 51 Franklin Street, Fifth Floor, Boston, MA 02110-1301 USA
+ */
+
+/* These functions and macros rely on Multi-Byte nodes */
+
+#ifndef _EBISTREE_H
+#define _EBISTREE_H
+
+#include <string.h>
+#include "ebtree.h"
+#include "ebpttree.h"
+#include "ebimtree.h"
+
+/* These functions and macros rely on Pointer nodes and use the <key> entry as
+ * a pointer to an indirect key. Most operations are performed using ebpt_*.
+ */
+
+/* The following functions are not inlined by default. They are declared
+ * in ebistree.c, which simply relies on their inline version.
+ */
+struct ebpt_node *ebis_lookup(struct eb_root *root, const char *x);
+struct ebpt_node *ebis_insert(struct eb_root *root, struct ebpt_node *new);
+
+/* Find the first occurrence of a length <len> string <x> in the tree <root>.
+ * It's the caller's responsibility to use this function only on trees which
+ * only contain zero-terminated strings, and that no null character is present
+ * in string <x> in the first <len> chars. If none can be found, return NULL.
+ */
+static forceinline struct ebpt_node *
+ebis_lookup_len(struct eb_root *root, const char *x, unsigned int len)
+{
+	struct ebpt_node *node;
+
+	node = ebim_lookup(root, x, len);
+	if (!node || ((const char *)node->key)[len] != 0)
+		return NULL;
+	return node;
+}
+
+/* Find the first occurrence of a zero-terminated string <x> in the tree <root>.
+ * It's the caller's responsibility to use this function only on trees which
+ * only contain zero-terminated strings. If none can be found, return NULL.
+ */
+static forceinline struct ebpt_node *__ebis_lookup(struct eb_root *root, const void *x)
+{
+	struct ebpt_node *node;
+	eb_troot_t *troot;
+	int bit;
+	int node_bit;
+
+	troot = root->b[EB_LEFT];
+	if (unlikely(troot == NULL))
+		return NULL;
+
+	bit = 0;
+	while (1) {
+		if ((eb_gettag(troot) == EB_LEAF)) {
+			node = container_of(eb_untag(troot, EB_LEAF),
+					    struct ebpt_node, node.branches);
+			if (strcmp(node->key, x) == 0)
+				return node;
+			else
+				return NULL;
+		}
+		node = container_of(eb_untag(troot, EB_NODE),
+				    struct ebpt_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 (strcmp(node->key, x) != 0)
+				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 ebpt_node, node.branches);
+			return node;
+		}
+
+		/* OK, normal data node, let's walk down but don't compare data
+		 * if we already reached the end of the key.
+		 */
+		if (likely(bit >= 0)) {
+			bit = string_equal_bits(x, node->key, bit);
+			if (likely(bit < node_bit)) {
+				if (bit >= 0)
+					return NULL; /* no more common bits */
+
+				/* bit < 0 : we reached the end of the key. If we
+				 * are in a tree with unique keys, we can return
+				 * this node. Otherwise we have to walk it down
+				 * and stop comparing bits.
+				 */
+				if (eb_gettag(root->b[EB_RGHT]))
+					return node;
+			}
+			/* if the bit is larger than the node's, we must bound it
+			 * because we might have compared too many bytes with an
+			 * inappropriate leaf. For a test, build a tree from "0",
+			 * "WW", "W", "S" inserted in this exact sequence and lookup
+			 * "W" => "S" is returned without this assignment.
+			 */
+			else
+				bit = node_bit;
+		}
+
+		troot = node->node.branches.b[(((unsigned char*)x)[node_bit >> 3] >>
+					       (~node_bit & 7)) & 1];
+	}
+}
+
+/* Insert ebpt_node <new> into subtree starting at node root <root>. Only
+ * new->key needs be set with the zero-terminated string key. The ebpt_node is
+ * returned. If root->b[EB_RGHT]==1, the tree may only contain unique keys. The
+ * caller is responsible for properly terminating the key with a zero.
+ */
+static forceinline struct ebpt_node *
+__ebis_insert(struct eb_root *root, struct ebpt_node *new)
+{
+	struct ebpt_node *old;
+	unsigned int side;
+	eb_troot_t *troot;
+	eb_troot_t *root_right;
+	int diff;
+	int bit;
+	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;
+	}
+
+	/* 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)) {
+			eb_troot_t *new_left, *new_rght;
+			eb_troot_t *new_leaf, *old_leaf;
+
+			old = container_of(eb_untag(troot, EB_LEAF),
+					    struct ebpt_node, node.branches);
+
+			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);
+			old_leaf = eb_dotag(&old->node.branches, EB_LEAF);
+
+			new->node.node_p = old->node.leaf_p;
+
+			/* Right here, we have 3 possibilities :
+			 * - the tree does not contain the key, and we have
+			 *   new->key < old->key. We insert new above old, on
+			 *   the left ;
+			 *
+			 * - the tree does not contain the key, and we have
+			 *   new->key > old->key. We insert new above old, on
+			 *   the right ;
+			 *
+			 * - the tree does contain the key, which implies it
+			 *   is alone. We add the new key next to it as a
+			 *   first duplicate.
+			 *
+			 * The last two cases can easily be partially merged.
+			 */
+			if (bit >= 0)
+				bit = string_equal_bits(new->key, old->key, bit);
+
+			if (bit < 0) {
+				/* key was already there */
+
+				/* we may refuse to duplicate this key if the tree is
+				 * tagged as containing only unique keys.
+				 */
+				if (eb_gettag(root_right))
+					return old;
+
+				/* new arbitrarily goes to the right and tops the dup tree */
+				old->node.leaf_p = new_left;
+				new->node.leaf_p = new_rght;
+				new->node.branches.b[EB_LEFT] = old_leaf;
+				new->node.branches.b[EB_RGHT] = new_leaf;
+				new->node.bit = -1;
+				root->b[side] = eb_dotag(&new->node.branches, EB_NODE);
+				return new;
+			}
+
+			diff = cmp_bits(new->key, old->key, bit);
+			if (diff < 0) {
+				/* new->key < old->key, new takes the left */
+				new->node.leaf_p = new_left;
+				old->node.leaf_p = new_rght;
+				new->node.branches.b[EB_LEFT] = new_leaf;
+				new->node.branches.b[EB_RGHT] = old_leaf;
+			} else {
+				/* new->key > old->key, new takes the right */
+				old->node.leaf_p = new_left;
+				new->node.leaf_p = new_rght;
+				new->node.branches.b[EB_LEFT] = old_leaf;
+				new->node.branches.b[EB_RGHT] = new_leaf;
+			}
+			break;
+		}
+
+		/* OK we're walking down this link */
+		old = container_of(eb_untag(troot, EB_NODE),
+				   struct ebpt_node, node.branches);
+		old_node_bit = old->node.bit;
+
+		/* Stop going down when we don't have common bits anymore. We
+		 * also stop in front of a duplicates tree because it means we
+		 * have to insert above. Note: we can compare more bits than
+		 * the current node's because as long as they are identical, we
+		 * know we descend along the correct side.
+		 */
+		if (bit >= 0 && (bit < old_node_bit || old_node_bit < 0))
+			bit = string_equal_bits(new->key, old->key, bit);
+
+		if (unlikely(bit < 0)) {
+			/* Perfect match, we must only stop on head of dup tree
+			 * or walk down to a leaf.
+			 */
+			if (old_node_bit < 0) {
+				/* We know here that string_equal_bits matched all
+				 * bits and that we're on top of a dup tree, then
+				 * we can perform the dup insertion and return.
+				 */
+				struct eb_node *ret;
+				ret = eb_insert_dup(&old->node, &new->node);
+				return container_of(ret, struct ebpt_node, node);
+			}
+			/* OK so let's walk down */
+		}
+		else if (bit < old_node_bit || old_node_bit < 0) {
+			/* The tree did not contain the key, or we stopped on top of a dup
+			 * tree, possibly containing the key. In the former case, we insert
+			 * <new> before the node <old>, and set ->bit to designate the lowest
+			 * bit position in <new> which applies to ->branches.b[]. In the later
+			 * case, we add the key to the existing dup tree. Note that we cannot
+			 * enter here if we match an intermediate node's key that is not the
+			 * head of a dup tree.
+			 */
+			eb_troot_t *new_left, *new_rght;
+			eb_troot_t *new_leaf, *old_node;
+
+			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);
+			old_node = eb_dotag(&old->node.branches, EB_NODE);
+
+			new->node.node_p = old->node.node_p;
+
+			/* we can never match all bits here */
+			diff = cmp_bits(new->key, old->key, bit);
+			if (diff < 0) {
+				new->node.leaf_p = new_left;
+				old->node.node_p = new_rght;
+				new->node.branches.b[EB_LEFT] = new_leaf;
+				new->node.branches.b[EB_RGHT] = old_node;
+			}
+			else {
+				old->node.node_p = new_left;
+				new->node.leaf_p = new_rght;
+				new->node.branches.b[EB_LEFT] = old_node;
+				new->node.branches.b[EB_RGHT] = new_leaf;
+			}
+			break;
+		}
+
+		/* walk down */
+		root = &old->node.branches;
+		side = (((unsigned char *)new->key)[old_node_bit >> 3] >> (~old_node_bit & 7)) & 1;
+		troot = root->b[side];
+	}
+
+	/* Ok, now we are inserting <new> between <root> and <old>. <old>'s
+	 * parent is already set to <new>, and the <root>'s branch is still in
+	 * <side>. Update the root's leaf till we have it. Note that we can also
+	 * find the side by checking the side of new->node.node_p.
+	 */
+
+	/* We need the common higher bits between new->key and old->key.
+	 * This number of bits is already in <bit>.
+	 * NOTE: we can't get here whit bit < 0 since we found a dup !
+	 */
+	new->node.bit = bit;
+	root->b[side] = eb_dotag(&new->node.branches, EB_NODE);
+	return new;
+}
+
+#endif /* _EBISTREE_H */