Use round-to-nearest internally in jn, test with ALL_RM_TEST (bug 18602).
[glibc.git] / misc / tsearch.c
blob869dc9bec77e39ca3a6c908e7c8e1aa06006d4fd
1 /* Copyright (C) 1995-2015 Free Software Foundation, Inc.
2 This file is part of the GNU C Library.
3 Contributed by Bernd Schmidt <crux@Pool.Informatik.RWTH-Aachen.DE>, 1997.
5 The GNU C Library is free software; you can redistribute it and/or
6 modify it under the terms of the GNU Lesser General Public
7 License as published by the Free Software Foundation; either
8 version 2.1 of the License, or (at your option) any later version.
10 The GNU C Library is distributed in the hope that it will be useful,
11 but WITHOUT ANY WARRANTY; without even the implied warranty of
12 MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU
13 Lesser General Public License for more details.
15 You should have received a copy of the GNU Lesser General Public
16 License along with the GNU C Library; if not, see
17 <http://www.gnu.org/licenses/>. */
19 /* Tree search for red/black trees.
20 The algorithm for adding nodes is taken from one of the many "Algorithms"
21 books by Robert Sedgewick, although the implementation differs.
22 The algorithm for deleting nodes can probably be found in a book named
23 "Introduction to Algorithms" by Cormen/Leiserson/Rivest. At least that's
24 the book that my professor took most algorithms from during the "Data
25 Structures" course...
27 Totally public domain. */
29 /* Red/black trees are binary trees in which the edges are colored either red
30 or black. They have the following properties:
31 1. The number of black edges on every path from the root to a leaf is
32 constant.
33 2. No two red edges are adjacent.
34 Therefore there is an upper bound on the length of every path, it's
35 O(log n) where n is the number of nodes in the tree. No path can be longer
36 than 1+2*P where P is the length of the shortest path in the tree.
37 Useful for the implementation:
38 3. If one of the children of a node is NULL, then the other one is red
39 (if it exists).
41 In the implementation, not the edges are colored, but the nodes. The color
42 interpreted as the color of the edge leading to this node. The color is
43 meaningless for the root node, but we color the root node black for
44 convenience. All added nodes are red initially.
46 Adding to a red/black tree is rather easy. The right place is searched
47 with a usual binary tree search. Additionally, whenever a node N is
48 reached that has two red successors, the successors are colored black and
49 the node itself colored red. This moves red edges up the tree where they
50 pose less of a problem once we get to really insert the new node. Changing
51 N's color to red may violate rule 2, however, so rotations may become
52 necessary to restore the invariants. Adding a new red leaf may violate
53 the same rule, so afterwards an additional check is run and the tree
54 possibly rotated.
56 Deleting is hairy. There are mainly two nodes involved: the node to be
57 deleted (n1), and another node that is to be unchained from the tree (n2).
58 If n1 has a successor (the node with a smallest key that is larger than
59 n1), then the successor becomes n2 and its contents are copied into n1,
60 otherwise n1 becomes n2.
61 Unchaining a node may violate rule 1: if n2 is black, one subtree is
62 missing one black edge afterwards. The algorithm must try to move this
63 error upwards towards the root, so that the subtree that does not have
64 enough black edges becomes the whole tree. Once that happens, the error
65 has disappeared. It may not be necessary to go all the way up, since it
66 is possible that rotations and recoloring can fix the error before that.
68 Although the deletion algorithm must walk upwards through the tree, we
69 do not store parent pointers in the nodes. Instead, delete allocates a
70 small array of parent pointers and fills it while descending the tree.
71 Since we know that the length of a path is O(log n), where n is the number
72 of nodes, this is likely to use less memory. */
74 /* Tree rotations look like this:
75 A C
76 / \ / \
77 B C A G
78 / \ / \ --> / \
79 D E F G B F
80 / \
81 D E
83 In this case, A has been rotated left. This preserves the ordering of the
84 binary tree. */
86 #include <stdlib.h>
87 #include <string.h>
88 #include <search.h>
90 typedef struct node_t
92 /* Callers expect this to be the first element in the structure - do not
93 move! */
94 const void *key;
95 struct node_t *left;
96 struct node_t *right;
97 unsigned int red:1;
98 } *node;
99 typedef const struct node_t *const_node;
101 #undef DEBUGGING
103 #ifdef DEBUGGING
105 /* Routines to check tree invariants. */
107 #include <assert.h>
109 #define CHECK_TREE(a) check_tree(a)
111 static void
112 check_tree_recurse (node p, int d_sofar, int d_total)
114 if (p == NULL)
116 assert (d_sofar == d_total);
117 return;
120 check_tree_recurse (p->left, d_sofar + (p->left && !p->left->red), d_total);
121 check_tree_recurse (p->right, d_sofar + (p->right && !p->right->red), d_total);
122 if (p->left)
123 assert (!(p->left->red && p->red));
124 if (p->right)
125 assert (!(p->right->red && p->red));
128 static void
129 check_tree (node root)
131 int cnt = 0;
132 node p;
133 if (root == NULL)
134 return;
135 root->red = 0;
136 for(p = root->left; p; p = p->left)
137 cnt += !p->red;
138 check_tree_recurse (root, 0, cnt);
142 #else
144 #define CHECK_TREE(a)
146 #endif
148 /* Possibly "split" a node with two red successors, and/or fix up two red
149 edges in a row. ROOTP is a pointer to the lowest node we visited, PARENTP
150 and GPARENTP pointers to its parent/grandparent. P_R and GP_R contain the
151 comparison values that determined which way was taken in the tree to reach
152 ROOTP. MODE is 1 if we need not do the split, but must check for two red
153 edges between GPARENTP and ROOTP. */
154 static void
155 maybe_split_for_insert (node *rootp, node *parentp, node *gparentp,
156 int p_r, int gp_r, int mode)
158 node root = *rootp;
159 node *rp, *lp;
160 rp = &(*rootp)->right;
161 lp = &(*rootp)->left;
163 /* See if we have to split this node (both successors red). */
164 if (mode == 1
165 || ((*rp) != NULL && (*lp) != NULL && (*rp)->red && (*lp)->red))
167 /* This node becomes red, its successors black. */
168 root->red = 1;
169 if (*rp)
170 (*rp)->red = 0;
171 if (*lp)
172 (*lp)->red = 0;
174 /* If the parent of this node is also red, we have to do
175 rotations. */
176 if (parentp != NULL && (*parentp)->red)
178 node gp = *gparentp;
179 node p = *parentp;
180 /* There are two main cases:
181 1. The edge types (left or right) of the two red edges differ.
182 2. Both red edges are of the same type.
183 There exist two symmetries of each case, so there is a total of
184 4 cases. */
185 if ((p_r > 0) != (gp_r > 0))
187 /* Put the child at the top of the tree, with its parent
188 and grandparent as successors. */
189 p->red = 1;
190 gp->red = 1;
191 root->red = 0;
192 if (p_r < 0)
194 /* Child is left of parent. */
195 p->left = *rp;
196 *rp = p;
197 gp->right = *lp;
198 *lp = gp;
200 else
202 /* Child is right of parent. */
203 p->right = *lp;
204 *lp = p;
205 gp->left = *rp;
206 *rp = gp;
208 *gparentp = root;
210 else
212 *gparentp = *parentp;
213 /* Parent becomes the top of the tree, grandparent and
214 child are its successors. */
215 p->red = 0;
216 gp->red = 1;
217 if (p_r < 0)
219 /* Left edges. */
220 gp->left = p->right;
221 p->right = gp;
223 else
225 /* Right edges. */
226 gp->right = p->left;
227 p->left = gp;
234 /* Find or insert datum into search tree.
235 KEY is the key to be located, ROOTP is the address of tree root,
236 COMPAR the ordering function. */
237 void *
238 __tsearch (const void *key, void **vrootp, __compar_fn_t compar)
240 node q;
241 node *parentp = NULL, *gparentp = NULL;
242 node *rootp = (node *) vrootp;
243 node *nextp;
244 int r = 0, p_r = 0, gp_r = 0; /* No they might not, Mr Compiler. */
246 if (rootp == NULL)
247 return NULL;
249 /* This saves some additional tests below. */
250 if (*rootp != NULL)
251 (*rootp)->red = 0;
253 CHECK_TREE (*rootp);
255 nextp = rootp;
256 while (*nextp != NULL)
258 node root = *rootp;
259 r = (*compar) (key, root->key);
260 if (r == 0)
261 return root;
263 maybe_split_for_insert (rootp, parentp, gparentp, p_r, gp_r, 0);
264 /* If that did any rotations, parentp and gparentp are now garbage.
265 That doesn't matter, because the values they contain are never
266 used again in that case. */
268 nextp = r < 0 ? &root->left : &root->right;
269 if (*nextp == NULL)
270 break;
272 gparentp = parentp;
273 parentp = rootp;
274 rootp = nextp;
276 gp_r = p_r;
277 p_r = r;
280 q = (struct node_t *) malloc (sizeof (struct node_t));
281 if (q != NULL)
283 *nextp = q; /* link new node to old */
284 q->key = key; /* initialize new node */
285 q->red = 1;
286 q->left = q->right = NULL;
288 if (nextp != rootp)
289 /* There may be two red edges in a row now, which we must avoid by
290 rotating the tree. */
291 maybe_split_for_insert (nextp, rootp, parentp, r, p_r, 1);
294 return q;
296 libc_hidden_def (__tsearch)
297 weak_alias (__tsearch, tsearch)
300 /* Find datum in search tree.
301 KEY is the key to be located, ROOTP is the address of tree root,
302 COMPAR the ordering function. */
303 void *
304 __tfind (key, vrootp, compar)
305 const void *key;
306 void *const *vrootp;
307 __compar_fn_t compar;
309 node *rootp = (node *) vrootp;
311 if (rootp == NULL)
312 return NULL;
314 CHECK_TREE (*rootp);
316 while (*rootp != NULL)
318 node root = *rootp;
319 int r;
321 r = (*compar) (key, root->key);
322 if (r == 0)
323 return root;
325 rootp = r < 0 ? &root->left : &root->right;
327 return NULL;
329 libc_hidden_def (__tfind)
330 weak_alias (__tfind, tfind)
333 /* Delete node with given key.
334 KEY is the key to be deleted, ROOTP is the address of the root of tree,
335 COMPAR the comparison function. */
336 void *
337 __tdelete (const void *key, void **vrootp, __compar_fn_t compar)
339 node p, q, r, retval;
340 int cmp;
341 node *rootp = (node *) vrootp;
342 node root, unchained;
343 /* Stack of nodes so we remember the parents without recursion. It's
344 _very_ unlikely that there are paths longer than 40 nodes. The tree
345 would need to have around 250.000 nodes. */
346 int stacksize = 40;
347 int sp = 0;
348 node **nodestack = alloca (sizeof (node *) * stacksize);
350 if (rootp == NULL)
351 return NULL;
352 p = *rootp;
353 if (p == NULL)
354 return NULL;
356 CHECK_TREE (p);
358 while ((cmp = (*compar) (key, (*rootp)->key)) != 0)
360 if (sp == stacksize)
362 node **newstack;
363 stacksize += 20;
364 newstack = alloca (sizeof (node *) * stacksize);
365 nodestack = memcpy (newstack, nodestack, sp * sizeof (node *));
368 nodestack[sp++] = rootp;
369 p = *rootp;
370 rootp = ((cmp < 0)
371 ? &(*rootp)->left
372 : &(*rootp)->right);
373 if (*rootp == NULL)
374 return NULL;
377 /* This is bogus if the node to be deleted is the root... this routine
378 really should return an integer with 0 for success, -1 for failure
379 and errno = ESRCH or something. */
380 retval = p;
382 /* We don't unchain the node we want to delete. Instead, we overwrite
383 it with its successor and unchain the successor. If there is no
384 successor, we really unchain the node to be deleted. */
386 root = *rootp;
388 r = root->right;
389 q = root->left;
391 if (q == NULL || r == NULL)
392 unchained = root;
393 else
395 node *parent = rootp, *up = &root->right;
396 for (;;)
398 if (sp == stacksize)
400 node **newstack;
401 stacksize += 20;
402 newstack = alloca (sizeof (node *) * stacksize);
403 nodestack = memcpy (newstack, nodestack, sp * sizeof (node *));
405 nodestack[sp++] = parent;
406 parent = up;
407 if ((*up)->left == NULL)
408 break;
409 up = &(*up)->left;
411 unchained = *up;
414 /* We know that either the left or right successor of UNCHAINED is NULL.
415 R becomes the other one, it is chained into the parent of UNCHAINED. */
416 r = unchained->left;
417 if (r == NULL)
418 r = unchained->right;
419 if (sp == 0)
420 *rootp = r;
421 else
423 q = *nodestack[sp-1];
424 if (unchained == q->right)
425 q->right = r;
426 else
427 q->left = r;
430 if (unchained != root)
431 root->key = unchained->key;
432 if (!unchained->red)
434 /* Now we lost a black edge, which means that the number of black
435 edges on every path is no longer constant. We must balance the
436 tree. */
437 /* NODESTACK now contains all parents of R. R is likely to be NULL
438 in the first iteration. */
439 /* NULL nodes are considered black throughout - this is necessary for
440 correctness. */
441 while (sp > 0 && (r == NULL || !r->red))
443 node *pp = nodestack[sp - 1];
444 p = *pp;
445 /* Two symmetric cases. */
446 if (r == p->left)
448 /* Q is R's brother, P is R's parent. The subtree with root
449 R has one black edge less than the subtree with root Q. */
450 q = p->right;
451 if (q->red)
453 /* If Q is red, we know that P is black. We rotate P left
454 so that Q becomes the top node in the tree, with P below
455 it. P is colored red, Q is colored black.
456 This action does not change the black edge count for any
457 leaf in the tree, but we will be able to recognize one
458 of the following situations, which all require that Q
459 is black. */
460 q->red = 0;
461 p->red = 1;
462 /* Left rotate p. */
463 p->right = q->left;
464 q->left = p;
465 *pp = q;
466 /* Make sure pp is right if the case below tries to use
467 it. */
468 nodestack[sp++] = pp = &q->left;
469 q = p->right;
471 /* We know that Q can't be NULL here. We also know that Q is
472 black. */
473 if ((q->left == NULL || !q->left->red)
474 && (q->right == NULL || !q->right->red))
476 /* Q has two black successors. We can simply color Q red.
477 The whole subtree with root P is now missing one black
478 edge. Note that this action can temporarily make the
479 tree invalid (if P is red). But we will exit the loop
480 in that case and set P black, which both makes the tree
481 valid and also makes the black edge count come out
482 right. If P is black, we are at least one step closer
483 to the root and we'll try again the next iteration. */
484 q->red = 1;
485 r = p;
487 else
489 /* Q is black, one of Q's successors is red. We can
490 repair the tree with one operation and will exit the
491 loop afterwards. */
492 if (q->right == NULL || !q->right->red)
494 /* The left one is red. We perform the same action as
495 in maybe_split_for_insert where two red edges are
496 adjacent but point in different directions:
497 Q's left successor (let's call it Q2) becomes the
498 top of the subtree we are looking at, its parent (Q)
499 and grandparent (P) become its successors. The former
500 successors of Q2 are placed below P and Q.
501 P becomes black, and Q2 gets the color that P had.
502 This changes the black edge count only for node R and
503 its successors. */
504 node q2 = q->left;
505 q2->red = p->red;
506 p->right = q2->left;
507 q->left = q2->right;
508 q2->right = q;
509 q2->left = p;
510 *pp = q2;
511 p->red = 0;
513 else
515 /* It's the right one. Rotate P left. P becomes black,
516 and Q gets the color that P had. Q's right successor
517 also becomes black. This changes the black edge
518 count only for node R and its successors. */
519 q->red = p->red;
520 p->red = 0;
522 q->right->red = 0;
524 /* left rotate p */
525 p->right = q->left;
526 q->left = p;
527 *pp = q;
530 /* We're done. */
531 sp = 1;
532 r = NULL;
535 else
537 /* Comments: see above. */
538 q = p->left;
539 if (q->red)
541 q->red = 0;
542 p->red = 1;
543 p->left = q->right;
544 q->right = p;
545 *pp = q;
546 nodestack[sp++] = pp = &q->right;
547 q = p->left;
549 if ((q->right == NULL || !q->right->red)
550 && (q->left == NULL || !q->left->red))
552 q->red = 1;
553 r = p;
555 else
557 if (q->left == NULL || !q->left->red)
559 node q2 = q->right;
560 q2->red = p->red;
561 p->left = q2->right;
562 q->right = q2->left;
563 q2->left = q;
564 q2->right = p;
565 *pp = q2;
566 p->red = 0;
568 else
570 q->red = p->red;
571 p->red = 0;
572 q->left->red = 0;
573 p->left = q->right;
574 q->right = p;
575 *pp = q;
577 sp = 1;
578 r = NULL;
581 --sp;
583 if (r != NULL)
584 r->red = 0;
587 free (unchained);
588 return retval;
590 libc_hidden_def (__tdelete)
591 weak_alias (__tdelete, tdelete)
594 /* Walk the nodes of a tree.
595 ROOT is the root of the tree to be walked, ACTION the function to be
596 called at each node. LEVEL is the level of ROOT in the whole tree. */
597 static void
598 internal_function
599 trecurse (const void *vroot, __action_fn_t action, int level)
601 const_node root = (const_node) vroot;
603 if (root->left == NULL && root->right == NULL)
604 (*action) (root, leaf, level);
605 else
607 (*action) (root, preorder, level);
608 if (root->left != NULL)
609 trecurse (root->left, action, level + 1);
610 (*action) (root, postorder, level);
611 if (root->right != NULL)
612 trecurse (root->right, action, level + 1);
613 (*action) (root, endorder, level);
618 /* Walk the nodes of a tree.
619 ROOT is the root of the tree to be walked, ACTION the function to be
620 called at each node. */
621 void
622 __twalk (const void *vroot, __action_fn_t action)
624 const_node root = (const_node) vroot;
626 CHECK_TREE (root);
628 if (root != NULL && action != NULL)
629 trecurse (root, action, 0);
631 libc_hidden_def (__twalk)
632 weak_alias (__twalk, twalk)
636 /* The standardized functions miss an important functionality: the
637 tree cannot be removed easily. We provide a function to do this. */
638 static void
639 internal_function
640 tdestroy_recurse (node root, __free_fn_t freefct)
642 if (root->left != NULL)
643 tdestroy_recurse (root->left, freefct);
644 if (root->right != NULL)
645 tdestroy_recurse (root->right, freefct);
646 (*freefct) ((void *) root->key);
647 /* Free the node itself. */
648 free (root);
651 void
652 __tdestroy (void *vroot, __free_fn_t freefct)
654 node root = (node) vroot;
656 CHECK_TREE (root);
658 if (root != NULL)
659 tdestroy_recurse (root, freefct);
661 weak_alias (__tdestroy, tdestroy)