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afbf86d209
[Coverity CID 75]. * sysdeps/generic/unwind-dw2.c (execute_cfa_program): Print error message for invalid DWARF data instead of crashing.
659 lines
18 KiB
C
659 lines
18 KiB
C
/* Copyright (C) 1995, 1996, 1997, 2000, 2006 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, write to the Free
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Software Foundation, Inc., 59 Temple Place, Suite 330, Boston, MA
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02111-1307 USA. */
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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 <stdlib.h>
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#include <string.h>
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#include <search.h>
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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;
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struct node_t *right;
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unsigned int red:1;
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} *node;
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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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#include <assert.h>
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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 (p->left, d_sofar + (p->left && !p->left->red), d_total);
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check_tree_recurse (p->right, d_sofar + (p->right && !p->right->red), d_total);
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if (p->left)
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assert (!(p->left->red && p->red));
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if (p->right)
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assert (!(p->right->red && p->red));
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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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root->red = 0;
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for(p = root->left; p; p = p->left)
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cnt += !p->red;
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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 = *rootp;
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node *rp, *lp;
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rp = &(*rootp)->right;
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lp = &(*rootp)->left;
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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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|| ((*rp) != NULL && (*lp) != NULL && (*rp)->red && (*lp)->red))
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{
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/* This node becomes red, its successors black. */
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root->red = 1;
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if (*rp)
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(*rp)->red = 0;
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if (*lp)
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(*lp)->red = 0;
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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 && (*parentp)->red)
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{
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node gp = *gparentp;
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node p = *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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p->red = 1;
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gp->red = 1;
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root->red = 0;
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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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p->left = *rp;
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*rp = p;
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gp->right = *lp;
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*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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p->right = *lp;
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*lp = p;
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gp->left = *rp;
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*rp = gp;
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}
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*gparentp = root;
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}
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else
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{
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*gparentp = *parentp;
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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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p->red = 0;
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gp->red = 1;
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if (p_r < 0)
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{
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/* Left edges. */
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gp->left = p->right;
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p->right = gp;
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}
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else
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{
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/* Right edges. */
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gp->right = p->left;
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p->left = 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;
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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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if (rootp == NULL)
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return NULL;
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/* This saves some additional tests below. */
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if (*rootp != NULL)
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(*rootp)->red = 0;
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CHECK_TREE (*rootp);
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nextp = rootp;
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while (*nextp != NULL)
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{
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node root = *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 ? &root->left : &root->right;
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if (*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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*nextp = q; /* link new node to old */
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q->key = key; /* initialize new node */
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q->red = 1;
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q->left = q->right = 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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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 (key, vrootp, compar)
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const void *key;
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void *const *vrootp;
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__compar_fn_t compar;
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{
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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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CHECK_TREE (*rootp);
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while (*rootp != NULL)
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{
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node root = *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 ? &root->left : &root->right;
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}
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return NULL;
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}
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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 = *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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while ((cmp = (*compar) (key, (*rootp)->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 = *rootp;
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rootp = ((cmp < 0)
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? &(*rootp)->left
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: &(*rootp)->right);
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if (*rootp == 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 = *rootp;
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r = root->right;
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q = root->left;
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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 *parent = rootp, *up = &root->right;
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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 *));
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}
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nodestack[sp++] = parent;
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parent = up;
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if ((*up)->left == NULL)
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break;
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up = &(*up)->left;
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}
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unchained = *up;
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}
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/* We know that either the left or right successor of UNCHAINED is NULL.
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R becomes the other one, it is chained into the parent of UNCHAINED. */
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r = unchained->left;
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if (r == NULL)
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r = unchained->right;
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if (sp == 0)
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*rootp = r;
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else
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{
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q = *nodestack[sp-1];
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if (unchained == q->right)
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q->right = r;
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else
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q->left = r;
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}
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if (unchained != root)
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root->key = unchained->key;
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if (!unchained->red)
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{
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/* Now we lost a black edge, which means that the number of black
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edges on every path is no longer constant. We must balance the
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tree. */
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/* NODESTACK now contains all parents of R. R is likely to be NULL
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in the first iteration. */
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/* NULL nodes are considered black throughout - this is necessary for
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correctness. */
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while (sp > 0 && (r == NULL || !r->red))
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{
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node *pp = nodestack[sp - 1];
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p = *pp;
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/* Two symmetric cases. */
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if (r == p->left)
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{
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/* Q is R's brother, P is R's parent. The subtree with root
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R has one black edge less than the subtree with root Q. */
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q = p->right;
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if (q->red)
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{
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/* If Q is red, we know that P is black. We rotate P left
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so that Q becomes the top node in the tree, with P below
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it. P is colored red, Q is colored black.
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This action does not change the black edge count for any
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leaf in the tree, but we will be able to recognize one
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of the following situations, which all require that Q
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is black. */
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q->red = 0;
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p->red = 1;
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/* Left rotate p. */
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p->right = q->left;
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q->left = p;
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*pp = q;
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/* Make sure pp is right if the case below tries to use
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it. */
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nodestack[sp++] = pp = &q->left;
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q = p->right;
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}
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/* We know that Q can't be NULL here. We also know that Q is
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black. */
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if ((q->left == NULL || !q->left->red)
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&& (q->right == NULL || !q->right->red))
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{
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/* Q has two black successors. We can simply color Q red.
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The whole subtree with root P is now missing one black
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edge. Note that this action can temporarily make the
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tree invalid (if P is red). But we will exit the loop
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in that case and set P black, which both makes the tree
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valid and also makes the black edge count come out
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right. If P is black, we are at least one step closer
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to the root and we'll try again the next iteration. */
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q->red = 1;
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r = p;
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}
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else
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{
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/* Q is black, one of Q's successors is red. We can
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repair the tree with one operation and will exit the
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loop afterwards. */
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if (q->right == NULL || !q->right->red)
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{
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/* The left one is red. We perform the same action as
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in maybe_split_for_insert where two red edges are
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adjacent but point in different directions:
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Q's left successor (let's call it Q2) becomes the
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top of the subtree we are looking at, its parent (Q)
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and grandparent (P) become its successors. The former
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successors of Q2 are placed below P and Q.
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P becomes black, and Q2 gets the color that P had.
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This changes the black edge count only for node R and
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its successors. */
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node q2 = q->left;
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q2->red = p->red;
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p->right = q2->left;
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q->left = q2->right;
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q2->right = q;
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q2->left = p;
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*pp = q2;
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p->red = 0;
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}
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else
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{
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/* It's the right one. Rotate P left. P becomes black,
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and Q gets the color that P had. Q's right successor
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also becomes black. This changes the black edge
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count only for node R and its successors. */
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q->red = p->red;
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p->red = 0;
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q->right->red = 0;
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/* left rotate p */
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p->right = q->left;
|
|
q->left = p;
|
|
*pp = q;
|
|
}
|
|
|
|
/* We're done. */
|
|
sp = 1;
|
|
r = NULL;
|
|
}
|
|
}
|
|
else
|
|
{
|
|
/* Comments: see above. */
|
|
q = p->left;
|
|
if (q->red)
|
|
{
|
|
q->red = 0;
|
|
p->red = 1;
|
|
p->left = q->right;
|
|
q->right = p;
|
|
*pp = q;
|
|
nodestack[sp++] = pp = &q->right;
|
|
q = p->left;
|
|
}
|
|
if ((q->right == NULL || !q->right->red)
|
|
&& (q->left == NULL || !q->left->red))
|
|
{
|
|
q->red = 1;
|
|
r = p;
|
|
}
|
|
else
|
|
{
|
|
if (q->left == NULL || !q->left->red)
|
|
{
|
|
node q2 = q->right;
|
|
q2->red = p->red;
|
|
p->left = q2->right;
|
|
q->right = q2->left;
|
|
q2->left = q;
|
|
q2->right = p;
|
|
*pp = q2;
|
|
p->red = 0;
|
|
}
|
|
else
|
|
{
|
|
q->red = p->red;
|
|
p->red = 0;
|
|
q->left->red = 0;
|
|
p->left = q->right;
|
|
q->right = p;
|
|
*pp = q;
|
|
}
|
|
sp = 1;
|
|
r = NULL;
|
|
}
|
|
}
|
|
--sp;
|
|
}
|
|
if (r != NULL)
|
|
r->red = 0;
|
|
}
|
|
|
|
free (unchained);
|
|
return retval;
|
|
}
|
|
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 (root->left == NULL && root->right == NULL)
|
|
(*action) (root, leaf, level);
|
|
else
|
|
{
|
|
(*action) (root, preorder, level);
|
|
if (root->left != NULL)
|
|
trecurse (root->left, action, level + 1);
|
|
(*action) (root, postorder, level);
|
|
if (root->right != NULL)
|
|
trecurse (root->right, 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 (root);
|
|
|
|
if (root != NULL && action != NULL)
|
|
trecurse (root, action, 0);
|
|
}
|
|
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 (root->left != NULL)
|
|
tdestroy_recurse (root->left, freefct);
|
|
if (root->right != NULL)
|
|
tdestroy_recurse (root->right, 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)
|