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A tsearch red-black tree node contains 3 pointers (key, left, right) and 1 bit to hold the red-black flag. When allocating new nodes this 1 bit is expanded to a full word. Causing the overhead per node to be 3 times the key size. We can reduce this overhead to just 2 times the key size. malloc returns naturally aligned memory. All nodes are internally allocated with malloc and the left/right node pointers are used as implementation details. So we can use the low bits of the left/right node pointers to store extra information. Replace all direct accesses of the struct node_t node pointers and red-black value with defines that take care of the red-black flag in the low bit of the (left) node pointer. This reduces the size of the nodes on 32-bit systems from 16 to 12 bytes and on 64-bit systems from 32 to 24 bytes. Also fix a call to CHECK_TREE so the code can be build (and tested) with DEBUGGING defined again. V2 changes: - Add assert after malloc to catch any odd pointers from bad interposed mallocs. - Rename implementation flag to USE_MALLOC_LOW_BIT. ChangeLog: * misc/tsearch.c (struct node_t): Reduce to 3 pointers if USE_MALLOC_LOW_BIT. Define pointer/value accessors. (check_tree_recurse): Use newly defined accessors. (check_tree): Likewise. (maybe_split_for_insert): Likewise. (__tfind): Likewise. (__tdelete): Likewise. (trecurse): Likewise. (tdestroy_recurse): Likewise. (__tsearch): Likewise. And add asserts for malloc alignment. (__twalk): Cast root to node in case CHECK_TREE is defined.
751 lines
21 KiB
C
751 lines
21 KiB
C
/* Copyright (C) 1995-2016 Free Software Foundation, Inc.
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This file is part of the GNU C Library.
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Contributed by Bernd Schmidt <crux@Pool.Informatik.RWTH-Aachen.DE>, 1997.
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The GNU C 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; either
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version 2.1 of the License, or (at your option) any later version.
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The GNU C 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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You should have received a copy of the GNU Lesser General Public
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License along with the GNU C Library; if not, see
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<http://www.gnu.org/licenses/>. */
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/* Tree search for red/black trees.
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The algorithm for adding nodes is taken from one of the many "Algorithms"
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books by Robert Sedgewick, although the implementation differs.
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The algorithm for deleting nodes can probably be found in a book named
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"Introduction to Algorithms" by Cormen/Leiserson/Rivest. At least that's
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the book that my professor took most algorithms from during the "Data
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Structures" course...
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Totally public domain. */
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/* Red/black trees are binary trees in which the edges are colored either red
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or black. They have the following properties:
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1. The number of black edges on every path from the root to a leaf is
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constant.
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2. No two red edges are adjacent.
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Therefore there is an upper bound on the length of every path, it's
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O(log n) where n is the number of nodes in the tree. No path can be longer
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than 1+2*P where P is the length of the shortest path in the tree.
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Useful for the implementation:
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3. If one of the children of a node is NULL, then the other one is red
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(if it exists).
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In the implementation, not the edges are colored, but the nodes. The color
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interpreted as the color of the edge leading to this node. The color is
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meaningless for the root node, but we color the root node black for
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convenience. All added nodes are red initially.
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Adding to a red/black tree is rather easy. The right place is searched
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with a usual binary tree search. Additionally, whenever a node N is
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reached that has two red successors, the successors are colored black and
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the node itself colored red. This moves red edges up the tree where they
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pose less of a problem once we get to really insert the new node. Changing
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N's color to red may violate rule 2, however, so rotations may become
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necessary to restore the invariants. Adding a new red leaf may violate
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the same rule, so afterwards an additional check is run and the tree
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possibly rotated.
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Deleting is hairy. There are mainly two nodes involved: the node to be
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deleted (n1), and another node that is to be unchained from the tree (n2).
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If n1 has a successor (the node with a smallest key that is larger than
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n1), then the successor becomes n2 and its contents are copied into n1,
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otherwise n1 becomes n2.
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Unchaining a node may violate rule 1: if n2 is black, one subtree is
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missing one black edge afterwards. The algorithm must try to move this
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error upwards towards the root, so that the subtree that does not have
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enough black edges becomes the whole tree. Once that happens, the error
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has disappeared. It may not be necessary to go all the way up, since it
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is possible that rotations and recoloring can fix the error before that.
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Although the deletion algorithm must walk upwards through the tree, we
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do not store parent pointers in the nodes. Instead, delete allocates a
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small array of parent pointers and fills it while descending the tree.
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Since we know that the length of a path is O(log n), where n is the number
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of nodes, this is likely to use less memory. */
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/* Tree rotations look like this:
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A C
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/ \ / \
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B C A G
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/ \ / \ --> / \
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D E F G B F
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/ \
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D E
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In this case, A has been rotated left. This preserves the ordering of the
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binary tree. */
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#include <assert.h>
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#include <stdalign.h>
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#include <stddef.h>
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#include <stdlib.h>
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#include <string.h>
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#include <search.h>
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/* Assume malloc returns naturally aligned (alignof (max_align_t))
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pointers so we can use the low bits to store some extra info. This
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works for the left/right node pointers since they are not user
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visible and always allocated by malloc. The user provides the key
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pointer and so that can point anywhere and doesn't have to be
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aligned. */
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#define USE_MALLOC_LOW_BIT 1
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#ifndef USE_MALLOC_LOW_BIT
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typedef struct node_t
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{
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/* Callers expect this to be the first element in the structure - do not
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move! */
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const void *key;
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struct node_t *left_node;
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struct node_t *right_node;
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unsigned int is_red:1;
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} *node;
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#define RED(N) (N)->is_red
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#define SETRED(N) (N)->is_red = 1
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#define SETBLACK(N) (N)->is_red = 0
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#define SETNODEPTR(NP,P) (*NP) = (P)
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#define LEFT(N) (N)->left_node
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#define LEFTPTR(N) (&(N)->left_node)
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#define SETLEFT(N,L) (N)->left_node = (L)
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#define RIGHT(N) (N)->right_node
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#define RIGHTPTR(N) (&(N)->right_node)
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#define SETRIGHT(N,R) (N)->right_node = (R)
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#define DEREFNODEPTR(NP) (*(NP))
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#else /* USE_MALLOC_LOW_BIT */
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typedef struct node_t
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{
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/* Callers expect this to be the first element in the structure - do not
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move! */
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const void *key;
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uintptr_t left_node; /* Includes whether the node is red in low-bit. */
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uintptr_t right_node;
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} *node;
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#define RED(N) (node)((N)->left_node & ((uintptr_t) 0x1))
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#define SETRED(N) (N)->left_node |= ((uintptr_t) 0x1)
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#define SETBLACK(N) (N)->left_node &= ~((uintptr_t) 0x1)
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#define SETNODEPTR(NP,P) (*NP) = (node)((((uintptr_t)(*NP)) \
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& (uintptr_t) 0x1) | (uintptr_t)(P))
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#define LEFT(N) (node)((N)->left_node & ~((uintptr_t) 0x1))
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#define LEFTPTR(N) (node *)(&(N)->left_node)
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#define SETLEFT(N,L) (N)->left_node = (((N)->left_node & (uintptr_t) 0x1) \
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| (uintptr_t)(L))
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#define RIGHT(N) (node)((N)->right_node)
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#define RIGHTPTR(N) (node *)(&(N)->right_node)
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#define SETRIGHT(N,R) (N)->right_node = (uintptr_t)(R)
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#define DEREFNODEPTR(NP) (node)((uintptr_t)(*(NP)) & ~((uintptr_t) 0x1))
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#endif /* USE_MALLOC_LOW_BIT */
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typedef const struct node_t *const_node;
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#undef DEBUGGING
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#ifdef DEBUGGING
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/* Routines to check tree invariants. */
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#define CHECK_TREE(a) check_tree(a)
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static void
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check_tree_recurse (node p, int d_sofar, int d_total)
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{
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if (p == NULL)
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{
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assert (d_sofar == d_total);
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return;
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}
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check_tree_recurse (LEFT(p), d_sofar + (LEFT(p) && !RED(LEFT(p))),
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d_total);
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check_tree_recurse (RIGHT(p), d_sofar + (RIGHT(p) && !RED(RIGHT(p))),
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d_total);
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if (LEFT(p))
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assert (!(RED(LEFT(p)) && RED(p)));
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if (RIGHT(p))
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assert (!(RED(RIGHT(p)) && RED(p)));
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}
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static void
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check_tree (node root)
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{
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int cnt = 0;
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node p;
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if (root == NULL)
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return;
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SETBLACK(root);
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for(p = LEFT(root); p; p = LEFT(p))
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cnt += !RED(p);
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check_tree_recurse (root, 0, cnt);
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}
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#else
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#define CHECK_TREE(a)
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#endif
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/* Possibly "split" a node with two red successors, and/or fix up two red
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edges in a row. ROOTP is a pointer to the lowest node we visited, PARENTP
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and GPARENTP pointers to its parent/grandparent. P_R and GP_R contain the
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comparison values that determined which way was taken in the tree to reach
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ROOTP. MODE is 1 if we need not do the split, but must check for two red
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edges between GPARENTP and ROOTP. */
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static void
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maybe_split_for_insert (node *rootp, node *parentp, node *gparentp,
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int p_r, int gp_r, int mode)
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{
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node root = DEREFNODEPTR(rootp);
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node *rp, *lp;
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node rpn, lpn;
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rp = RIGHTPTR(root);
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rpn = RIGHT(root);
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lp = LEFTPTR(root);
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lpn = LEFT(root);
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/* See if we have to split this node (both successors red). */
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if (mode == 1
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|| ((rpn) != NULL && (lpn) != NULL && RED(rpn) && RED(lpn)))
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{
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/* This node becomes red, its successors black. */
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SETRED(root);
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if (rpn)
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SETBLACK(rpn);
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if (lpn)
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SETBLACK(lpn);
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/* If the parent of this node is also red, we have to do
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rotations. */
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if (parentp != NULL && RED(DEREFNODEPTR(parentp)))
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{
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node gp = DEREFNODEPTR(gparentp);
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node p = DEREFNODEPTR(parentp);
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/* There are two main cases:
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1. The edge types (left or right) of the two red edges differ.
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2. Both red edges are of the same type.
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There exist two symmetries of each case, so there is a total of
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4 cases. */
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if ((p_r > 0) != (gp_r > 0))
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{
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/* Put the child at the top of the tree, with its parent
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and grandparent as successors. */
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SETRED(p);
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SETRED(gp);
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SETBLACK(root);
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if (p_r < 0)
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{
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/* Child is left of parent. */
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SETLEFT(p,rpn);
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SETNODEPTR(rp,p);
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SETRIGHT(gp,lpn);
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SETNODEPTR(lp,gp);
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}
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else
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{
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/* Child is right of parent. */
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SETRIGHT(p,lpn);
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SETNODEPTR(lp,p);
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SETLEFT(gp,rpn);
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SETNODEPTR(rp,gp);
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}
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SETNODEPTR(gparentp,root);
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}
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else
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{
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SETNODEPTR(gparentp,p);
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/* Parent becomes the top of the tree, grandparent and
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child are its successors. */
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SETBLACK(p);
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SETRED(gp);
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if (p_r < 0)
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{
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/* Left edges. */
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SETLEFT(gp,RIGHT(p));
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SETRIGHT(p,gp);
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}
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else
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{
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/* Right edges. */
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SETRIGHT(gp,LEFT(p));
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SETLEFT(p,gp);
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}
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}
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}
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}
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}
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/* Find or insert datum into search tree.
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KEY is the key to be located, ROOTP is the address of tree root,
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COMPAR the ordering function. */
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void *
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__tsearch (const void *key, void **vrootp, __compar_fn_t compar)
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{
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node q, root;
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node *parentp = NULL, *gparentp = NULL;
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node *rootp = (node *) vrootp;
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node *nextp;
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int r = 0, p_r = 0, gp_r = 0; /* No they might not, Mr Compiler. */
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#ifdef USE_MALLOC_LOW_BIT
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static_assert (alignof (max_align_t) > 1, "malloc must return aligned ptrs");
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#endif
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if (rootp == NULL)
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return NULL;
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/* This saves some additional tests below. */
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root = DEREFNODEPTR(rootp);
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if (root != NULL)
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SETBLACK(root);
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CHECK_TREE (root);
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nextp = rootp;
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while (DEREFNODEPTR(nextp) != NULL)
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{
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root = DEREFNODEPTR(rootp);
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r = (*compar) (key, root->key);
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if (r == 0)
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return root;
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maybe_split_for_insert (rootp, parentp, gparentp, p_r, gp_r, 0);
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/* If that did any rotations, parentp and gparentp are now garbage.
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That doesn't matter, because the values they contain are never
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used again in that case. */
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nextp = r < 0 ? LEFTPTR(root) : RIGHTPTR(root);
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if (DEREFNODEPTR(nextp) == NULL)
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break;
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gparentp = parentp;
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parentp = rootp;
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rootp = nextp;
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gp_r = p_r;
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p_r = r;
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}
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q = (struct node_t *) malloc (sizeof (struct node_t));
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if (q != NULL)
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{
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/* Make sure the malloc implementation returns naturally aligned
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memory blocks when expected. Or at least even pointers, so we
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can use the low bit as red/black flag. Even though we have a
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static_assert to make sure alignof (max_align_t) > 1 there could
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be an interposed malloc implementation that might cause havoc by
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not obeying the malloc contract. */
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#ifdef USE_MALLOC_LOW_BIT
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assert (((uintptr_t) q & (uintptr_t) 0x1) == 0);
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#endif
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SETNODEPTR(nextp,q); /* link new node to old */
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q->key = key; /* initialize new node */
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SETRED(q);
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SETLEFT(q,NULL);
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SETRIGHT(q,NULL);
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if (nextp != rootp)
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/* There may be two red edges in a row now, which we must avoid by
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rotating the tree. */
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maybe_split_for_insert (nextp, rootp, parentp, r, p_r, 1);
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}
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return q;
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}
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libc_hidden_def (__tsearch)
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weak_alias (__tsearch, tsearch)
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/* Find datum in search tree.
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KEY is the key to be located, ROOTP is the address of tree root,
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COMPAR the ordering function. */
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void *
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__tfind (const void *key, void *const *vrootp, __compar_fn_t compar)
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{
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node root;
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node *rootp = (node *) vrootp;
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if (rootp == NULL)
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return NULL;
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root = DEREFNODEPTR(rootp);
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CHECK_TREE (root);
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while (DEREFNODEPTR(rootp) != NULL)
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{
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root = DEREFNODEPTR(rootp);
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int r;
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r = (*compar) (key, root->key);
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if (r == 0)
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return root;
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rootp = r < 0 ? LEFTPTR(root) : RIGHTPTR(root);
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}
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return NULL;
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}
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libc_hidden_def (__tfind)
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weak_alias (__tfind, tfind)
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/* Delete node with given key.
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KEY is the key to be deleted, ROOTP is the address of the root of tree,
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COMPAR the comparison function. */
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void *
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__tdelete (const void *key, void **vrootp, __compar_fn_t compar)
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{
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node p, q, r, retval;
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int cmp;
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node *rootp = (node *) vrootp;
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node root, unchained;
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/* Stack of nodes so we remember the parents without recursion. It's
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_very_ unlikely that there are paths longer than 40 nodes. The tree
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would need to have around 250.000 nodes. */
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int stacksize = 40;
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int sp = 0;
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node **nodestack = alloca (sizeof (node *) * stacksize);
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if (rootp == NULL)
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return NULL;
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p = DEREFNODEPTR(rootp);
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if (p == NULL)
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return NULL;
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CHECK_TREE (p);
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root = DEREFNODEPTR(rootp);
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while ((cmp = (*compar) (key, root->key)) != 0)
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{
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if (sp == stacksize)
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{
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node **newstack;
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stacksize += 20;
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newstack = alloca (sizeof (node *) * stacksize);
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nodestack = memcpy (newstack, nodestack, sp * sizeof (node *));
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}
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nodestack[sp++] = rootp;
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p = DEREFNODEPTR(rootp);
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if (cmp < 0)
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{
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rootp = LEFTPTR(p);
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root = LEFT(p);
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}
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else
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{
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rootp = RIGHTPTR(p);
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root = RIGHT(p);
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}
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if (root == NULL)
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return NULL;
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}
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/* This is bogus if the node to be deleted is the root... this routine
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really should return an integer with 0 for success, -1 for failure
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and errno = ESRCH or something. */
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retval = p;
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/* We don't unchain the node we want to delete. Instead, we overwrite
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it with its successor and unchain the successor. If there is no
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successor, we really unchain the node to be deleted. */
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root = DEREFNODEPTR(rootp);
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r = RIGHT(root);
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q = LEFT(root);
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if (q == NULL || r == NULL)
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unchained = root;
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else
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{
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node *parentp = rootp, *up = RIGHTPTR(root);
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node upn;
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for (;;)
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{
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if (sp == stacksize)
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{
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node **newstack;
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stacksize += 20;
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newstack = alloca (sizeof (node *) * stacksize);
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nodestack = memcpy (newstack, nodestack, sp * sizeof (node *));
|
|
}
|
|
nodestack[sp++] = parentp;
|
|
parentp = up;
|
|
upn = DEREFNODEPTR(up);
|
|
if (LEFT(upn) == NULL)
|
|
break;
|
|
up = LEFTPTR(upn);
|
|
}
|
|
unchained = DEREFNODEPTR(up);
|
|
}
|
|
|
|
/* We know that either the left or right successor of UNCHAINED is NULL.
|
|
R becomes the other one, it is chained into the parent of UNCHAINED. */
|
|
r = LEFT(unchained);
|
|
if (r == NULL)
|
|
r = RIGHT(unchained);
|
|
if (sp == 0)
|
|
SETNODEPTR(rootp,r);
|
|
else
|
|
{
|
|
q = DEREFNODEPTR(nodestack[sp-1]);
|
|
if (unchained == RIGHT(q))
|
|
SETRIGHT(q,r);
|
|
else
|
|
SETLEFT(q,r);
|
|
}
|
|
|
|
if (unchained != root)
|
|
root->key = unchained->key;
|
|
if (!RED(unchained))
|
|
{
|
|
/* Now we lost a black edge, which means that the number of black
|
|
edges on every path is no longer constant. We must balance the
|
|
tree. */
|
|
/* NODESTACK now contains all parents of R. R is likely to be NULL
|
|
in the first iteration. */
|
|
/* NULL nodes are considered black throughout - this is necessary for
|
|
correctness. */
|
|
while (sp > 0 && (r == NULL || !RED(r)))
|
|
{
|
|
node *pp = nodestack[sp - 1];
|
|
p = DEREFNODEPTR(pp);
|
|
/* Two symmetric cases. */
|
|
if (r == LEFT(p))
|
|
{
|
|
/* Q is R's brother, P is R's parent. The subtree with root
|
|
R has one black edge less than the subtree with root Q. */
|
|
q = RIGHT(p);
|
|
if (RED(q))
|
|
{
|
|
/* If Q is red, we know that P is black. We rotate P left
|
|
so that Q becomes the top node in the tree, with P below
|
|
it. P is colored red, Q is colored black.
|
|
This action does not change the black edge count for any
|
|
leaf in the tree, but we will be able to recognize one
|
|
of the following situations, which all require that Q
|
|
is black. */
|
|
SETBLACK(q);
|
|
SETRED(p);
|
|
/* Left rotate p. */
|
|
SETRIGHT(p,LEFT(q));
|
|
SETLEFT(q,p);
|
|
SETNODEPTR(pp,q);
|
|
/* Make sure pp is right if the case below tries to use
|
|
it. */
|
|
nodestack[sp++] = pp = LEFTPTR(q);
|
|
q = RIGHT(p);
|
|
}
|
|
/* We know that Q can't be NULL here. We also know that Q is
|
|
black. */
|
|
if ((LEFT(q) == NULL || !RED(LEFT(q)))
|
|
&& (RIGHT(q) == NULL || !RED(RIGHT(q))))
|
|
{
|
|
/* Q has two black successors. We can simply color Q red.
|
|
The whole subtree with root P is now missing one black
|
|
edge. Note that this action can temporarily make the
|
|
tree invalid (if P is red). But we will exit the loop
|
|
in that case and set P black, which both makes the tree
|
|
valid and also makes the black edge count come out
|
|
right. If P is black, we are at least one step closer
|
|
to the root and we'll try again the next iteration. */
|
|
SETRED(q);
|
|
r = p;
|
|
}
|
|
else
|
|
{
|
|
/* Q is black, one of Q's successors is red. We can
|
|
repair the tree with one operation and will exit the
|
|
loop afterwards. */
|
|
if (RIGHT(q) == NULL || !RED(RIGHT(q)))
|
|
{
|
|
/* The left one is red. We perform the same action as
|
|
in maybe_split_for_insert where two red edges are
|
|
adjacent but point in different directions:
|
|
Q's left successor (let's call it Q2) becomes the
|
|
top of the subtree we are looking at, its parent (Q)
|
|
and grandparent (P) become its successors. The former
|
|
successors of Q2 are placed below P and Q.
|
|
P becomes black, and Q2 gets the color that P had.
|
|
This changes the black edge count only for node R and
|
|
its successors. */
|
|
node q2 = LEFT(q);
|
|
if (RED(p))
|
|
SETRED(q2);
|
|
else
|
|
SETBLACK(q2);
|
|
SETRIGHT(p,LEFT(q2));
|
|
SETLEFT(q,RIGHT(q2));
|
|
SETRIGHT(q2,q);
|
|
SETLEFT(q2,p);
|
|
SETNODEPTR(pp,q2);
|
|
SETBLACK(p);
|
|
}
|
|
else
|
|
{
|
|
/* It's the right one. Rotate P left. P becomes black,
|
|
and Q gets the color that P had. Q's right successor
|
|
also becomes black. This changes the black edge
|
|
count only for node R and its successors. */
|
|
if (RED(p))
|
|
SETRED(q);
|
|
else
|
|
SETBLACK(q);
|
|
SETBLACK(p);
|
|
|
|
SETBLACK(RIGHT(q));
|
|
|
|
/* left rotate p */
|
|
SETRIGHT(p,LEFT(q));
|
|
SETLEFT(q,p);
|
|
SETNODEPTR(pp,q);
|
|
}
|
|
|
|
/* We're done. */
|
|
sp = 1;
|
|
r = NULL;
|
|
}
|
|
}
|
|
else
|
|
{
|
|
/* Comments: see above. */
|
|
q = LEFT(p);
|
|
if (RED(q))
|
|
{
|
|
SETBLACK(q);
|
|
SETRED(p);
|
|
SETLEFT(p,RIGHT(q));
|
|
SETRIGHT(q,p);
|
|
SETNODEPTR(pp,q);
|
|
nodestack[sp++] = pp = RIGHTPTR(q);
|
|
q = LEFT(p);
|
|
}
|
|
if ((RIGHT(q) == NULL || !RED(RIGHT(q)))
|
|
&& (LEFT(q) == NULL || !RED(LEFT(q))))
|
|
{
|
|
SETRED(q);
|
|
r = p;
|
|
}
|
|
else
|
|
{
|
|
if (LEFT(q) == NULL || !RED(LEFT(q)))
|
|
{
|
|
node q2 = RIGHT(q);
|
|
if (RED(p))
|
|
SETRED(q2);
|
|
else
|
|
SETBLACK(q2);
|
|
SETLEFT(p,RIGHT(q2));
|
|
SETRIGHT(q,LEFT(q2));
|
|
SETLEFT(q2,q);
|
|
SETRIGHT(q2,p);
|
|
SETNODEPTR(pp,q2);
|
|
SETBLACK(p);
|
|
}
|
|
else
|
|
{
|
|
if (RED(p))
|
|
SETRED(q);
|
|
else
|
|
SETBLACK(q);
|
|
SETBLACK(p);
|
|
SETBLACK(LEFT(q));
|
|
SETLEFT(p,RIGHT(q));
|
|
SETRIGHT(q,p);
|
|
SETNODEPTR(pp,q);
|
|
}
|
|
sp = 1;
|
|
r = NULL;
|
|
}
|
|
}
|
|
--sp;
|
|
}
|
|
if (r != NULL)
|
|
SETBLACK(r);
|
|
}
|
|
|
|
free (unchained);
|
|
return retval;
|
|
}
|
|
libc_hidden_def (__tdelete)
|
|
weak_alias (__tdelete, tdelete)
|
|
|
|
|
|
/* Walk the nodes of a tree.
|
|
ROOT is the root of the tree to be walked, ACTION the function to be
|
|
called at each node. LEVEL is the level of ROOT in the whole tree. */
|
|
static void
|
|
internal_function
|
|
trecurse (const void *vroot, __action_fn_t action, int level)
|
|
{
|
|
const_node root = (const_node) vroot;
|
|
|
|
if (LEFT(root) == NULL && RIGHT(root) == NULL)
|
|
(*action) (root, leaf, level);
|
|
else
|
|
{
|
|
(*action) (root, preorder, level);
|
|
if (LEFT(root) != NULL)
|
|
trecurse (LEFT(root), action, level + 1);
|
|
(*action) (root, postorder, level);
|
|
if (RIGHT(root) != NULL)
|
|
trecurse (RIGHT(root), action, level + 1);
|
|
(*action) (root, endorder, level);
|
|
}
|
|
}
|
|
|
|
|
|
/* Walk the nodes of a tree.
|
|
ROOT is the root of the tree to be walked, ACTION the function to be
|
|
called at each node. */
|
|
void
|
|
__twalk (const void *vroot, __action_fn_t action)
|
|
{
|
|
const_node root = (const_node) vroot;
|
|
|
|
CHECK_TREE ((node) root);
|
|
|
|
if (root != NULL && action != NULL)
|
|
trecurse (root, action, 0);
|
|
}
|
|
libc_hidden_def (__twalk)
|
|
weak_alias (__twalk, twalk)
|
|
|
|
|
|
|
|
/* The standardized functions miss an important functionality: the
|
|
tree cannot be removed easily. We provide a function to do this. */
|
|
static void
|
|
internal_function
|
|
tdestroy_recurse (node root, __free_fn_t freefct)
|
|
{
|
|
if (LEFT(root) != NULL)
|
|
tdestroy_recurse (LEFT(root), freefct);
|
|
if (RIGHT(root) != NULL)
|
|
tdestroy_recurse (RIGHT(root), freefct);
|
|
(*freefct) ((void *) root->key);
|
|
/* Free the node itself. */
|
|
free (root);
|
|
}
|
|
|
|
void
|
|
__tdestroy (void *vroot, __free_fn_t freefct)
|
|
{
|
|
node root = (node) vroot;
|
|
|
|
CHECK_TREE (root);
|
|
|
|
if (root != NULL)
|
|
tdestroy_recurse (root, freefct);
|
|
}
|
|
weak_alias (__tdestroy, tdestroy)
|