| 1 | /* |
| 2 | * tree234.c: reasonably generic counted 2-3-4 tree routines. |
| 3 | * |
| 4 | * This file is copyright 1999-2001 Simon Tatham. |
| 5 | * |
| 6 | * Permission is hereby granted, free of charge, to any person |
| 7 | * obtaining a copy of this software and associated documentation |
| 8 | * files (the "Software"), to deal in the Software without |
| 9 | * restriction, including without limitation the rights to use, |
| 10 | * copy, modify, merge, publish, distribute, sublicense, and/or |
| 11 | * sell copies of the Software, and to permit persons to whom the |
| 12 | * Software is furnished to do so, subject to the following |
| 13 | * conditions: |
| 14 | * |
| 15 | * The above copyright notice and this permission notice shall be |
| 16 | * included in all copies or substantial portions of the Software. |
| 17 | * |
| 18 | * THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, |
| 19 | * EXPRESS OR IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES |
| 20 | * OF MERCHANTABILITY, FITNESS FOR A PARTICULAR PURPOSE AND |
| 21 | * NONINFRINGEMENT. IN NO EVENT SHALL SIMON TATHAM BE LIABLE FOR |
| 22 | * ANY CLAIM, DAMAGES OR OTHER LIABILITY, WHETHER IN AN ACTION OF |
| 23 | * CONTRACT, TORT OR OTHERWISE, ARISING FROM, OUT OF OR IN |
| 24 | * CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE |
| 25 | * SOFTWARE. |
| 26 | */ |
| 27 | |
| 28 | #include <stdio.h> |
| 29 | #include <stdlib.h> |
| 30 | #include <assert.h> |
| 31 | |
| 32 | #include "puttymem.h" |
| 33 | #include "tree234.h" |
| 34 | |
| 35 | #ifdef TEST |
| 36 | #define LOG(x) (printf x) |
| 37 | #else |
| 38 | #define LOG(x) |
| 39 | #endif |
| 40 | |
| 41 | typedef struct node234_Tag node234; |
| 42 | |
| 43 | struct tree234_Tag { |
| 44 | node234 *root; |
| 45 | cmpfn234 cmp; |
| 46 | }; |
| 47 | |
| 48 | struct node234_Tag { |
| 49 | node234 *parent; |
| 50 | node234 *kids[4]; |
| 51 | int counts[4]; |
| 52 | void *elems[3]; |
| 53 | }; |
| 54 | |
| 55 | /* |
| 56 | * Create a 2-3-4 tree. |
| 57 | */ |
| 58 | tree234 *newtree234(cmpfn234 cmp) |
| 59 | { |
| 60 | tree234 *ret = snew(tree234); |
| 61 | LOG(("created tree %p\n", ret)); |
| 62 | ret->root = NULL; |
| 63 | ret->cmp = cmp; |
| 64 | return ret; |
| 65 | } |
| 66 | |
| 67 | /* |
| 68 | * Free a 2-3-4 tree (not including freeing the elements). |
| 69 | */ |
| 70 | static void freenode234(node234 * n) |
| 71 | { |
| 72 | if (!n) |
| 73 | return; |
| 74 | freenode234(n->kids[0]); |
| 75 | freenode234(n->kids[1]); |
| 76 | freenode234(n->kids[2]); |
| 77 | freenode234(n->kids[3]); |
| 78 | sfree(n); |
| 79 | } |
| 80 | |
| 81 | void freetree234(tree234 * t) |
| 82 | { |
| 83 | freenode234(t->root); |
| 84 | sfree(t); |
| 85 | } |
| 86 | |
| 87 | /* |
| 88 | * Internal function to count a node. |
| 89 | */ |
| 90 | static int countnode234(node234 * n) |
| 91 | { |
| 92 | int count = 0; |
| 93 | int i; |
| 94 | if (!n) |
| 95 | return 0; |
| 96 | for (i = 0; i < 4; i++) |
| 97 | count += n->counts[i]; |
| 98 | for (i = 0; i < 3; i++) |
| 99 | if (n->elems[i]) |
| 100 | count++; |
| 101 | return count; |
| 102 | } |
| 103 | |
| 104 | /* |
| 105 | * Count the elements in a tree. |
| 106 | */ |
| 107 | int count234(tree234 * t) |
| 108 | { |
| 109 | if (t->root) |
| 110 | return countnode234(t->root); |
| 111 | else |
| 112 | return 0; |
| 113 | } |
| 114 | |
| 115 | /* |
| 116 | * Add an element e to a 2-3-4 tree t. Returns e on success, or if |
| 117 | * an existing element compares equal, returns that. |
| 118 | */ |
| 119 | static void *add234_internal(tree234 * t, void *e, int index) |
| 120 | { |
| 121 | node234 *n, **np, *left, *right; |
| 122 | void *orig_e = e; |
| 123 | int c, lcount, rcount; |
| 124 | |
| 125 | LOG(("adding node %p to tree %p\n", e, t)); |
| 126 | if (t->root == NULL) { |
| 127 | t->root = snew(node234); |
| 128 | t->root->elems[1] = t->root->elems[2] = NULL; |
| 129 | t->root->kids[0] = t->root->kids[1] = NULL; |
| 130 | t->root->kids[2] = t->root->kids[3] = NULL; |
| 131 | t->root->counts[0] = t->root->counts[1] = 0; |
| 132 | t->root->counts[2] = t->root->counts[3] = 0; |
| 133 | t->root->parent = NULL; |
| 134 | t->root->elems[0] = e; |
| 135 | LOG((" created root %p\n", t->root)); |
| 136 | return orig_e; |
| 137 | } |
| 138 | |
| 139 | np = &t->root; |
| 140 | while (*np) { |
| 141 | int childnum; |
| 142 | n = *np; |
| 143 | LOG((" node %p: %p/%d [%p] %p/%d [%p] %p/%d [%p] %p/%d\n", |
| 144 | n, |
| 145 | n->kids[0], n->counts[0], n->elems[0], |
| 146 | n->kids[1], n->counts[1], n->elems[1], |
| 147 | n->kids[2], n->counts[2], n->elems[2], |
| 148 | n->kids[3], n->counts[3])); |
| 149 | if (index >= 0) { |
| 150 | if (!n->kids[0]) { |
| 151 | /* |
| 152 | * Leaf node. We want to insert at kid position |
| 153 | * equal to the index: |
| 154 | * |
| 155 | * 0 A 1 B 2 C 3 |
| 156 | */ |
| 157 | childnum = index; |
| 158 | } else { |
| 159 | /* |
| 160 | * Internal node. We always descend through it (add |
| 161 | * always starts at the bottom, never in the |
| 162 | * middle). |
| 163 | */ |
| 164 | do { /* this is a do ... while (0) to allow `break' */ |
| 165 | if (index <= n->counts[0]) { |
| 166 | childnum = 0; |
| 167 | break; |
| 168 | } |
| 169 | index -= n->counts[0] + 1; |
| 170 | if (index <= n->counts[1]) { |
| 171 | childnum = 1; |
| 172 | break; |
| 173 | } |
| 174 | index -= n->counts[1] + 1; |
| 175 | if (index <= n->counts[2]) { |
| 176 | childnum = 2; |
| 177 | break; |
| 178 | } |
| 179 | index -= n->counts[2] + 1; |
| 180 | if (index <= n->counts[3]) { |
| 181 | childnum = 3; |
| 182 | break; |
| 183 | } |
| 184 | return NULL; /* error: index out of range */ |
| 185 | } while (0); |
| 186 | } |
| 187 | } else { |
| 188 | if ((c = t->cmp(e, n->elems[0])) < 0) |
| 189 | childnum = 0; |
| 190 | else if (c == 0) |
| 191 | return n->elems[0]; /* already exists */ |
| 192 | else if (n->elems[1] == NULL |
| 193 | || (c = t->cmp(e, n->elems[1])) < 0) childnum = 1; |
| 194 | else if (c == 0) |
| 195 | return n->elems[1]; /* already exists */ |
| 196 | else if (n->elems[2] == NULL |
| 197 | || (c = t->cmp(e, n->elems[2])) < 0) childnum = 2; |
| 198 | else if (c == 0) |
| 199 | return n->elems[2]; /* already exists */ |
| 200 | else |
| 201 | childnum = 3; |
| 202 | } |
| 203 | np = &n->kids[childnum]; |
| 204 | LOG((" moving to child %d (%p)\n", childnum, *np)); |
| 205 | } |
| 206 | |
| 207 | /* |
| 208 | * We need to insert the new element in n at position np. |
| 209 | */ |
| 210 | left = NULL; |
| 211 | lcount = 0; |
| 212 | right = NULL; |
| 213 | rcount = 0; |
| 214 | while (n) { |
| 215 | LOG((" at %p: %p/%d [%p] %p/%d [%p] %p/%d [%p] %p/%d\n", |
| 216 | n, |
| 217 | n->kids[0], n->counts[0], n->elems[0], |
| 218 | n->kids[1], n->counts[1], n->elems[1], |
| 219 | n->kids[2], n->counts[2], n->elems[2], |
| 220 | n->kids[3], n->counts[3])); |
| 221 | LOG((" need to insert %p/%d [%p] %p/%d at position %d\n", |
| 222 | left, lcount, e, right, rcount, np - n->kids)); |
| 223 | if (n->elems[1] == NULL) { |
| 224 | /* |
| 225 | * Insert in a 2-node; simple. |
| 226 | */ |
| 227 | if (np == &n->kids[0]) { |
| 228 | LOG((" inserting on left of 2-node\n")); |
| 229 | n->kids[2] = n->kids[1]; |
| 230 | n->counts[2] = n->counts[1]; |
| 231 | n->elems[1] = n->elems[0]; |
| 232 | n->kids[1] = right; |
| 233 | n->counts[1] = rcount; |
| 234 | n->elems[0] = e; |
| 235 | n->kids[0] = left; |
| 236 | n->counts[0] = lcount; |
| 237 | } else { /* np == &n->kids[1] */ |
| 238 | LOG((" inserting on right of 2-node\n")); |
| 239 | n->kids[2] = right; |
| 240 | n->counts[2] = rcount; |
| 241 | n->elems[1] = e; |
| 242 | n->kids[1] = left; |
| 243 | n->counts[1] = lcount; |
| 244 | } |
| 245 | if (n->kids[0]) |
| 246 | n->kids[0]->parent = n; |
| 247 | if (n->kids[1]) |
| 248 | n->kids[1]->parent = n; |
| 249 | if (n->kids[2]) |
| 250 | n->kids[2]->parent = n; |
| 251 | LOG((" done\n")); |
| 252 | break; |
| 253 | } else if (n->elems[2] == NULL) { |
| 254 | /* |
| 255 | * Insert in a 3-node; simple. |
| 256 | */ |
| 257 | if (np == &n->kids[0]) { |
| 258 | LOG((" inserting on left of 3-node\n")); |
| 259 | n->kids[3] = n->kids[2]; |
| 260 | n->counts[3] = n->counts[2]; |
| 261 | n->elems[2] = n->elems[1]; |
| 262 | n->kids[2] = n->kids[1]; |
| 263 | n->counts[2] = n->counts[1]; |
| 264 | n->elems[1] = n->elems[0]; |
| 265 | n->kids[1] = right; |
| 266 | n->counts[1] = rcount; |
| 267 | n->elems[0] = e; |
| 268 | n->kids[0] = left; |
| 269 | n->counts[0] = lcount; |
| 270 | } else if (np == &n->kids[1]) { |
| 271 | LOG((" inserting in middle of 3-node\n")); |
| 272 | n->kids[3] = n->kids[2]; |
| 273 | n->counts[3] = n->counts[2]; |
| 274 | n->elems[2] = n->elems[1]; |
| 275 | n->kids[2] = right; |
| 276 | n->counts[2] = rcount; |
| 277 | n->elems[1] = e; |
| 278 | n->kids[1] = left; |
| 279 | n->counts[1] = lcount; |
| 280 | } else { /* np == &n->kids[2] */ |
| 281 | LOG((" inserting on right of 3-node\n")); |
| 282 | n->kids[3] = right; |
| 283 | n->counts[3] = rcount; |
| 284 | n->elems[2] = e; |
| 285 | n->kids[2] = left; |
| 286 | n->counts[2] = lcount; |
| 287 | } |
| 288 | if (n->kids[0]) |
| 289 | n->kids[0]->parent = n; |
| 290 | if (n->kids[1]) |
| 291 | n->kids[1]->parent = n; |
| 292 | if (n->kids[2]) |
| 293 | n->kids[2]->parent = n; |
| 294 | if (n->kids[3]) |
| 295 | n->kids[3]->parent = n; |
| 296 | LOG((" done\n")); |
| 297 | break; |
| 298 | } else { |
| 299 | node234 *m = snew(node234); |
| 300 | m->parent = n->parent; |
| 301 | LOG((" splitting a 4-node; created new node %p\n", m)); |
| 302 | /* |
| 303 | * Insert in a 4-node; split into a 2-node and a |
| 304 | * 3-node, and move focus up a level. |
| 305 | * |
| 306 | * I don't think it matters which way round we put the |
| 307 | * 2 and the 3. For simplicity, we'll put the 3 first |
| 308 | * always. |
| 309 | */ |
| 310 | if (np == &n->kids[0]) { |
| 311 | m->kids[0] = left; |
| 312 | m->counts[0] = lcount; |
| 313 | m->elems[0] = e; |
| 314 | m->kids[1] = right; |
| 315 | m->counts[1] = rcount; |
| 316 | m->elems[1] = n->elems[0]; |
| 317 | m->kids[2] = n->kids[1]; |
| 318 | m->counts[2] = n->counts[1]; |
| 319 | e = n->elems[1]; |
| 320 | n->kids[0] = n->kids[2]; |
| 321 | n->counts[0] = n->counts[2]; |
| 322 | n->elems[0] = n->elems[2]; |
| 323 | n->kids[1] = n->kids[3]; |
| 324 | n->counts[1] = n->counts[3]; |
| 325 | } else if (np == &n->kids[1]) { |
| 326 | m->kids[0] = n->kids[0]; |
| 327 | m->counts[0] = n->counts[0]; |
| 328 | m->elems[0] = n->elems[0]; |
| 329 | m->kids[1] = left; |
| 330 | m->counts[1] = lcount; |
| 331 | m->elems[1] = e; |
| 332 | m->kids[2] = right; |
| 333 | m->counts[2] = rcount; |
| 334 | e = n->elems[1]; |
| 335 | n->kids[0] = n->kids[2]; |
| 336 | n->counts[0] = n->counts[2]; |
| 337 | n->elems[0] = n->elems[2]; |
| 338 | n->kids[1] = n->kids[3]; |
| 339 | n->counts[1] = n->counts[3]; |
| 340 | } else if (np == &n->kids[2]) { |
| 341 | m->kids[0] = n->kids[0]; |
| 342 | m->counts[0] = n->counts[0]; |
| 343 | m->elems[0] = n->elems[0]; |
| 344 | m->kids[1] = n->kids[1]; |
| 345 | m->counts[1] = n->counts[1]; |
| 346 | m->elems[1] = n->elems[1]; |
| 347 | m->kids[2] = left; |
| 348 | m->counts[2] = lcount; |
| 349 | /* e = e; */ |
| 350 | n->kids[0] = right; |
| 351 | n->counts[0] = rcount; |
| 352 | n->elems[0] = n->elems[2]; |
| 353 | n->kids[1] = n->kids[3]; |
| 354 | n->counts[1] = n->counts[3]; |
| 355 | } else { /* np == &n->kids[3] */ |
| 356 | m->kids[0] = n->kids[0]; |
| 357 | m->counts[0] = n->counts[0]; |
| 358 | m->elems[0] = n->elems[0]; |
| 359 | m->kids[1] = n->kids[1]; |
| 360 | m->counts[1] = n->counts[1]; |
| 361 | m->elems[1] = n->elems[1]; |
| 362 | m->kids[2] = n->kids[2]; |
| 363 | m->counts[2] = n->counts[2]; |
| 364 | n->kids[0] = left; |
| 365 | n->counts[0] = lcount; |
| 366 | n->elems[0] = e; |
| 367 | n->kids[1] = right; |
| 368 | n->counts[1] = rcount; |
| 369 | e = n->elems[2]; |
| 370 | } |
| 371 | m->kids[3] = n->kids[3] = n->kids[2] = NULL; |
| 372 | m->counts[3] = n->counts[3] = n->counts[2] = 0; |
| 373 | m->elems[2] = n->elems[2] = n->elems[1] = NULL; |
| 374 | if (m->kids[0]) |
| 375 | m->kids[0]->parent = m; |
| 376 | if (m->kids[1]) |
| 377 | m->kids[1]->parent = m; |
| 378 | if (m->kids[2]) |
| 379 | m->kids[2]->parent = m; |
| 380 | if (n->kids[0]) |
| 381 | n->kids[0]->parent = n; |
| 382 | if (n->kids[1]) |
| 383 | n->kids[1]->parent = n; |
| 384 | LOG((" left (%p): %p/%d [%p] %p/%d [%p] %p/%d\n", m, |
| 385 | m->kids[0], m->counts[0], m->elems[0], |
| 386 | m->kids[1], m->counts[1], m->elems[1], |
| 387 | m->kids[2], m->counts[2])); |
| 388 | LOG((" right (%p): %p/%d [%p] %p/%d\n", n, |
| 389 | n->kids[0], n->counts[0], n->elems[0], |
| 390 | n->kids[1], n->counts[1])); |
| 391 | left = m; |
| 392 | lcount = countnode234(left); |
| 393 | right = n; |
| 394 | rcount = countnode234(right); |
| 395 | } |
| 396 | if (n->parent) |
| 397 | np = (n->parent->kids[0] == n ? &n->parent->kids[0] : |
| 398 | n->parent->kids[1] == n ? &n->parent->kids[1] : |
| 399 | n->parent->kids[2] == n ? &n->parent->kids[2] : |
| 400 | &n->parent->kids[3]); |
| 401 | n = n->parent; |
| 402 | } |
| 403 | |
| 404 | /* |
| 405 | * If we've come out of here by `break', n will still be |
| 406 | * non-NULL and all we need to do is go back up the tree |
| 407 | * updating counts. If we've come here because n is NULL, we |
| 408 | * need to create a new root for the tree because the old one |
| 409 | * has just split into two. */ |
| 410 | if (n) { |
| 411 | while (n->parent) { |
| 412 | int count = countnode234(n); |
| 413 | int childnum; |
| 414 | childnum = (n->parent->kids[0] == n ? 0 : |
| 415 | n->parent->kids[1] == n ? 1 : |
| 416 | n->parent->kids[2] == n ? 2 : 3); |
| 417 | n->parent->counts[childnum] = count; |
| 418 | n = n->parent; |
| 419 | } |
| 420 | } else { |
| 421 | LOG((" root is overloaded, split into two\n")); |
| 422 | t->root = snew(node234); |
| 423 | t->root->kids[0] = left; |
| 424 | t->root->counts[0] = lcount; |
| 425 | t->root->elems[0] = e; |
| 426 | t->root->kids[1] = right; |
| 427 | t->root->counts[1] = rcount; |
| 428 | t->root->elems[1] = NULL; |
| 429 | t->root->kids[2] = NULL; |
| 430 | t->root->counts[2] = 0; |
| 431 | t->root->elems[2] = NULL; |
| 432 | t->root->kids[3] = NULL; |
| 433 | t->root->counts[3] = 0; |
| 434 | t->root->parent = NULL; |
| 435 | if (t->root->kids[0]) |
| 436 | t->root->kids[0]->parent = t->root; |
| 437 | if (t->root->kids[1]) |
| 438 | t->root->kids[1]->parent = t->root; |
| 439 | LOG((" new root is %p/%d [%p] %p/%d\n", |
| 440 | t->root->kids[0], t->root->counts[0], |
| 441 | t->root->elems[0], t->root->kids[1], t->root->counts[1])); |
| 442 | } |
| 443 | |
| 444 | return orig_e; |
| 445 | } |
| 446 | |
| 447 | void *add234(tree234 * t, void *e) |
| 448 | { |
| 449 | if (!t->cmp) /* tree is unsorted */ |
| 450 | return NULL; |
| 451 | |
| 452 | return add234_internal(t, e, -1); |
| 453 | } |
| 454 | void *addpos234(tree234 * t, void *e, int index) |
| 455 | { |
| 456 | if (index < 0 || /* index out of range */ |
| 457 | t->cmp) /* tree is sorted */ |
| 458 | return NULL; /* return failure */ |
| 459 | |
| 460 | return add234_internal(t, e, index); /* this checks the upper bound */ |
| 461 | } |
| 462 | |
| 463 | /* |
| 464 | * Look up the element at a given numeric index in a 2-3-4 tree. |
| 465 | * Returns NULL if the index is out of range. |
| 466 | */ |
| 467 | void *index234(tree234 * t, int index) |
| 468 | { |
| 469 | node234 *n; |
| 470 | |
| 471 | if (!t->root) |
| 472 | return NULL; /* tree is empty */ |
| 473 | |
| 474 | if (index < 0 || index >= countnode234(t->root)) |
| 475 | return NULL; /* out of range */ |
| 476 | |
| 477 | n = t->root; |
| 478 | |
| 479 | while (n) { |
| 480 | if (index < n->counts[0]) |
| 481 | n = n->kids[0]; |
| 482 | else if (index -= n->counts[0] + 1, index < 0) |
| 483 | return n->elems[0]; |
| 484 | else if (index < n->counts[1]) |
| 485 | n = n->kids[1]; |
| 486 | else if (index -= n->counts[1] + 1, index < 0) |
| 487 | return n->elems[1]; |
| 488 | else if (index < n->counts[2]) |
| 489 | n = n->kids[2]; |
| 490 | else if (index -= n->counts[2] + 1, index < 0) |
| 491 | return n->elems[2]; |
| 492 | else |
| 493 | n = n->kids[3]; |
| 494 | } |
| 495 | |
| 496 | /* We shouldn't ever get here. I wonder how we did. */ |
| 497 | return NULL; |
| 498 | } |
| 499 | |
| 500 | /* |
| 501 | * Find an element e in a sorted 2-3-4 tree t. Returns NULL if not |
| 502 | * found. e is always passed as the first argument to cmp, so cmp |
| 503 | * can be an asymmetric function if desired. cmp can also be passed |
| 504 | * as NULL, in which case the compare function from the tree proper |
| 505 | * will be used. |
| 506 | */ |
| 507 | void *findrelpos234(tree234 * t, void *e, cmpfn234 cmp, |
| 508 | int relation, int *index) |
| 509 | { |
| 510 | node234 *n; |
| 511 | void *ret; |
| 512 | int c; |
| 513 | int idx, ecount, kcount, cmpret; |
| 514 | |
| 515 | if (t->root == NULL) |
| 516 | return NULL; |
| 517 | |
| 518 | if (cmp == NULL) |
| 519 | cmp = t->cmp; |
| 520 | |
| 521 | n = t->root; |
| 522 | /* |
| 523 | * Attempt to find the element itself. |
| 524 | */ |
| 525 | idx = 0; |
| 526 | ecount = -1; |
| 527 | /* |
| 528 | * Prepare a fake `cmp' result if e is NULL. |
| 529 | */ |
| 530 | cmpret = 0; |
| 531 | if (e == NULL) { |
| 532 | assert(relation == REL234_LT || relation == REL234_GT); |
| 533 | if (relation == REL234_LT) |
| 534 | cmpret = +1; /* e is a max: always greater */ |
| 535 | else if (relation == REL234_GT) |
| 536 | cmpret = -1; /* e is a min: always smaller */ |
| 537 | } |
| 538 | while (1) { |
| 539 | for (kcount = 0; kcount < 4; kcount++) { |
| 540 | if (kcount >= 3 || n->elems[kcount] == NULL || |
| 541 | (c = cmpret ? cmpret : cmp(e, n->elems[kcount])) < 0) { |
| 542 | break; |
| 543 | } |
| 544 | if (n->kids[kcount]) |
| 545 | idx += n->counts[kcount]; |
| 546 | if (c == 0) { |
| 547 | ecount = kcount; |
| 548 | break; |
| 549 | } |
| 550 | idx++; |
| 551 | } |
| 552 | if (ecount >= 0) |
| 553 | break; |
| 554 | if (n->kids[kcount]) |
| 555 | n = n->kids[kcount]; |
| 556 | else |
| 557 | break; |
| 558 | } |
| 559 | |
| 560 | if (ecount >= 0) { |
| 561 | /* |
| 562 | * We have found the element we're looking for. It's |
| 563 | * n->elems[ecount], at tree index idx. If our search |
| 564 | * relation is EQ, LE or GE we can now go home. |
| 565 | */ |
| 566 | if (relation != REL234_LT && relation != REL234_GT) { |
| 567 | if (index) |
| 568 | *index = idx; |
| 569 | return n->elems[ecount]; |
| 570 | } |
| 571 | |
| 572 | /* |
| 573 | * Otherwise, we'll do an indexed lookup for the previous |
| 574 | * or next element. (It would be perfectly possible to |
| 575 | * implement these search types in a non-counted tree by |
| 576 | * going back up from where we are, but far more fiddly.) |
| 577 | */ |
| 578 | if (relation == REL234_LT) |
| 579 | idx--; |
| 580 | else |
| 581 | idx++; |
| 582 | } else { |
| 583 | /* |
| 584 | * We've found our way to the bottom of the tree and we |
| 585 | * know where we would insert this node if we wanted to: |
| 586 | * we'd put it in in place of the (empty) subtree |
| 587 | * n->kids[kcount], and it would have index idx |
| 588 | * |
| 589 | * But the actual element isn't there. So if our search |
| 590 | * relation is EQ, we're doomed. |
| 591 | */ |
| 592 | if (relation == REL234_EQ) |
| 593 | return NULL; |
| 594 | |
| 595 | /* |
| 596 | * Otherwise, we must do an index lookup for index idx-1 |
| 597 | * (if we're going left - LE or LT) or index idx (if we're |
| 598 | * going right - GE or GT). |
| 599 | */ |
| 600 | if (relation == REL234_LT || relation == REL234_LE) { |
| 601 | idx--; |
| 602 | } |
| 603 | } |
| 604 | |
| 605 | /* |
| 606 | * We know the index of the element we want; just call index234 |
| 607 | * to do the rest. This will return NULL if the index is out of |
| 608 | * bounds, which is exactly what we want. |
| 609 | */ |
| 610 | ret = index234(t, idx); |
| 611 | if (ret && index) |
| 612 | *index = idx; |
| 613 | return ret; |
| 614 | } |
| 615 | void *find234(tree234 * t, void *e, cmpfn234 cmp) |
| 616 | { |
| 617 | return findrelpos234(t, e, cmp, REL234_EQ, NULL); |
| 618 | } |
| 619 | void *findrel234(tree234 * t, void *e, cmpfn234 cmp, int relation) |
| 620 | { |
| 621 | return findrelpos234(t, e, cmp, relation, NULL); |
| 622 | } |
| 623 | void *findpos234(tree234 * t, void *e, cmpfn234 cmp, int *index) |
| 624 | { |
| 625 | return findrelpos234(t, e, cmp, REL234_EQ, index); |
| 626 | } |
| 627 | |
| 628 | /* |
| 629 | * Delete an element e in a 2-3-4 tree. Does not free the element, |
| 630 | * merely removes all links to it from the tree nodes. |
| 631 | */ |
| 632 | static void *delpos234_internal(tree234 * t, int index) |
| 633 | { |
| 634 | node234 *n; |
| 635 | void *retval; |
| 636 | int ei = -1; |
| 637 | |
| 638 | retval = 0; |
| 639 | |
| 640 | n = t->root; |
| 641 | LOG(("deleting item %d from tree %p\n", index, t)); |
| 642 | while (1) { |
| 643 | while (n) { |
| 644 | int ki; |
| 645 | node234 *sub; |
| 646 | |
| 647 | LOG( |
| 648 | (" node %p: %p/%d [%p] %p/%d [%p] %p/%d [%p] %p/%d index=%d\n", |
| 649 | n, n->kids[0], n->counts[0], n->elems[0], n->kids[1], |
| 650 | n->counts[1], n->elems[1], n->kids[2], n->counts[2], |
| 651 | n->elems[2], n->kids[3], n->counts[3], index)); |
| 652 | if (index < n->counts[0]) { |
| 653 | ki = 0; |
| 654 | } else if (index -= n->counts[0] + 1, index < 0) { |
| 655 | ei = 0; |
| 656 | break; |
| 657 | } else if (index < n->counts[1]) { |
| 658 | ki = 1; |
| 659 | } else if (index -= n->counts[1] + 1, index < 0) { |
| 660 | ei = 1; |
| 661 | break; |
| 662 | } else if (index < n->counts[2]) { |
| 663 | ki = 2; |
| 664 | } else if (index -= n->counts[2] + 1, index < 0) { |
| 665 | ei = 2; |
| 666 | break; |
| 667 | } else { |
| 668 | ki = 3; |
| 669 | } |
| 670 | /* |
| 671 | * Recurse down to subtree ki. If it has only one element, |
| 672 | * we have to do some transformation to start with. |
| 673 | */ |
| 674 | LOG((" moving to subtree %d\n", ki)); |
| 675 | sub = n->kids[ki]; |
| 676 | if (!sub->elems[1]) { |
| 677 | LOG((" subtree has only one element!\n", ki)); |
| 678 | if (ki > 0 && n->kids[ki - 1]->elems[1]) { |
| 679 | /* |
| 680 | * Case 3a, left-handed variant. Child ki has |
| 681 | * only one element, but child ki-1 has two or |
| 682 | * more. So we need to move a subtree from ki-1 |
| 683 | * to ki. |
| 684 | * |
| 685 | * . C . . B . |
| 686 | * / \ -> / \ |
| 687 | * [more] a A b B c d D e [more] a A b c C d D e |
| 688 | */ |
| 689 | node234 *sib = n->kids[ki - 1]; |
| 690 | int lastelem = (sib->elems[2] ? 2 : |
| 691 | sib->elems[1] ? 1 : 0); |
| 692 | sub->kids[2] = sub->kids[1]; |
| 693 | sub->counts[2] = sub->counts[1]; |
| 694 | sub->elems[1] = sub->elems[0]; |
| 695 | sub->kids[1] = sub->kids[0]; |
| 696 | sub->counts[1] = sub->counts[0]; |
| 697 | sub->elems[0] = n->elems[ki - 1]; |
| 698 | sub->kids[0] = sib->kids[lastelem + 1]; |
| 699 | sub->counts[0] = sib->counts[lastelem + 1]; |
| 700 | if (sub->kids[0]) |
| 701 | sub->kids[0]->parent = sub; |
| 702 | n->elems[ki - 1] = sib->elems[lastelem]; |
| 703 | sib->kids[lastelem + 1] = NULL; |
| 704 | sib->counts[lastelem + 1] = 0; |
| 705 | sib->elems[lastelem] = NULL; |
| 706 | n->counts[ki] = countnode234(sub); |
| 707 | LOG((" case 3a left\n")); |
| 708 | LOG( |
| 709 | (" index and left subtree count before adjustment: %d, %d\n", |
| 710 | index, n->counts[ki - 1])); |
| 711 | index += n->counts[ki - 1]; |
| 712 | n->counts[ki - 1] = countnode234(sib); |
| 713 | index -= n->counts[ki - 1]; |
| 714 | LOG( |
| 715 | (" index and left subtree count after adjustment: %d, %d\n", |
| 716 | index, n->counts[ki - 1])); |
| 717 | } else if (ki < 3 && n->kids[ki + 1] |
| 718 | && n->kids[ki + 1]->elems[1]) { |
| 719 | /* |
| 720 | * Case 3a, right-handed variant. ki has only |
| 721 | * one element but ki+1 has two or more. Move a |
| 722 | * subtree from ki+1 to ki. |
| 723 | * |
| 724 | * . B . . C . |
| 725 | * / \ -> / \ |
| 726 | * a A b c C d D e [more] a A b B c d D e [more] |
| 727 | */ |
| 728 | node234 *sib = n->kids[ki + 1]; |
| 729 | int j; |
| 730 | sub->elems[1] = n->elems[ki]; |
| 731 | sub->kids[2] = sib->kids[0]; |
| 732 | sub->counts[2] = sib->counts[0]; |
| 733 | if (sub->kids[2]) |
| 734 | sub->kids[2]->parent = sub; |
| 735 | n->elems[ki] = sib->elems[0]; |
| 736 | sib->kids[0] = sib->kids[1]; |
| 737 | sib->counts[0] = sib->counts[1]; |
| 738 | for (j = 0; j < 2 && sib->elems[j + 1]; j++) { |
| 739 | sib->kids[j + 1] = sib->kids[j + 2]; |
| 740 | sib->counts[j + 1] = sib->counts[j + 2]; |
| 741 | sib->elems[j] = sib->elems[j + 1]; |
| 742 | } |
| 743 | sib->kids[j + 1] = NULL; |
| 744 | sib->counts[j + 1] = 0; |
| 745 | sib->elems[j] = NULL; |
| 746 | n->counts[ki] = countnode234(sub); |
| 747 | n->counts[ki + 1] = countnode234(sib); |
| 748 | LOG((" case 3a right\n")); |
| 749 | } else { |
| 750 | /* |
| 751 | * Case 3b. ki has only one element, and has no |
| 752 | * neighbour with more than one. So pick a |
| 753 | * neighbour and merge it with ki, taking an |
| 754 | * element down from n to go in the middle. |
| 755 | * |
| 756 | * . B . . |
| 757 | * / \ -> | |
| 758 | * a A b c C d a A b B c C d |
| 759 | * |
| 760 | * (Since at all points we have avoided |
| 761 | * descending to a node with only one element, |
| 762 | * we can be sure that n is not reduced to |
| 763 | * nothingness by this move, _unless_ it was |
| 764 | * the very first node, ie the root of the |
| 765 | * tree. In that case we remove the now-empty |
| 766 | * root and replace it with its single large |
| 767 | * child as shown.) |
| 768 | */ |
| 769 | node234 *sib; |
| 770 | int j; |
| 771 | |
| 772 | if (ki > 0) { |
| 773 | ki--; |
| 774 | index += n->counts[ki] + 1; |
| 775 | } |
| 776 | sib = n->kids[ki]; |
| 777 | sub = n->kids[ki + 1]; |
| 778 | |
| 779 | sub->kids[3] = sub->kids[1]; |
| 780 | sub->counts[3] = sub->counts[1]; |
| 781 | sub->elems[2] = sub->elems[0]; |
| 782 | sub->kids[2] = sub->kids[0]; |
| 783 | sub->counts[2] = sub->counts[0]; |
| 784 | sub->elems[1] = n->elems[ki]; |
| 785 | sub->kids[1] = sib->kids[1]; |
| 786 | sub->counts[1] = sib->counts[1]; |
| 787 | if (sub->kids[1]) |
| 788 | sub->kids[1]->parent = sub; |
| 789 | sub->elems[0] = sib->elems[0]; |
| 790 | sub->kids[0] = sib->kids[0]; |
| 791 | sub->counts[0] = sib->counts[0]; |
| 792 | if (sub->kids[0]) |
| 793 | sub->kids[0]->parent = sub; |
| 794 | |
| 795 | n->counts[ki + 1] = countnode234(sub); |
| 796 | |
| 797 | sfree(sib); |
| 798 | |
| 799 | /* |
| 800 | * That's built the big node in sub. Now we |
| 801 | * need to remove the reference to sib in n. |
| 802 | */ |
| 803 | for (j = ki; j < 3 && n->kids[j + 1]; j++) { |
| 804 | n->kids[j] = n->kids[j + 1]; |
| 805 | n->counts[j] = n->counts[j + 1]; |
| 806 | n->elems[j] = j < 2 ? n->elems[j + 1] : NULL; |
| 807 | } |
| 808 | n->kids[j] = NULL; |
| 809 | n->counts[j] = 0; |
| 810 | if (j < 3) |
| 811 | n->elems[j] = NULL; |
| 812 | LOG((" case 3b ki=%d\n", ki)); |
| 813 | |
| 814 | if (!n->elems[0]) { |
| 815 | /* |
| 816 | * The root is empty and needs to be |
| 817 | * removed. |
| 818 | */ |
| 819 | LOG((" shifting root!\n")); |
| 820 | t->root = sub; |
| 821 | sub->parent = NULL; |
| 822 | sfree(n); |
| 823 | } |
| 824 | } |
| 825 | } |
| 826 | n = sub; |
| 827 | } |
| 828 | if (!retval) |
| 829 | retval = n->elems[ei]; |
| 830 | |
| 831 | if (ei == -1) |
| 832 | return NULL; /* although this shouldn't happen */ |
| 833 | |
| 834 | /* |
| 835 | * Treat special case: this is the one remaining item in |
| 836 | * the tree. n is the tree root (no parent), has one |
| 837 | * element (no elems[1]), and has no kids (no kids[0]). |
| 838 | */ |
| 839 | if (!n->parent && !n->elems[1] && !n->kids[0]) { |
| 840 | LOG((" removed last element in tree\n")); |
| 841 | sfree(n); |
| 842 | t->root = NULL; |
| 843 | return retval; |
| 844 | } |
| 845 | |
| 846 | /* |
| 847 | * Now we have the element we want, as n->elems[ei], and we |
| 848 | * have also arranged for that element not to be the only |
| 849 | * one in its node. So... |
| 850 | */ |
| 851 | |
| 852 | if (!n->kids[0] && n->elems[1]) { |
| 853 | /* |
| 854 | * Case 1. n is a leaf node with more than one element, |
| 855 | * so it's _really easy_. Just delete the thing and |
| 856 | * we're done. |
| 857 | */ |
| 858 | int i; |
| 859 | LOG((" case 1\n")); |
| 860 | for (i = ei; i < 2 && n->elems[i + 1]; i++) |
| 861 | n->elems[i] = n->elems[i + 1]; |
| 862 | n->elems[i] = NULL; |
| 863 | /* |
| 864 | * Having done that to the leaf node, we now go back up |
| 865 | * the tree fixing the counts. |
| 866 | */ |
| 867 | while (n->parent) { |
| 868 | int childnum; |
| 869 | childnum = (n->parent->kids[0] == n ? 0 : |
| 870 | n->parent->kids[1] == n ? 1 : |
| 871 | n->parent->kids[2] == n ? 2 : 3); |
| 872 | n->parent->counts[childnum]--; |
| 873 | n = n->parent; |
| 874 | } |
| 875 | return retval; /* finished! */ |
| 876 | } else if (n->kids[ei]->elems[1]) { |
| 877 | /* |
| 878 | * Case 2a. n is an internal node, and the root of the |
| 879 | * subtree to the left of e has more than one element. |
| 880 | * So find the predecessor p to e (ie the largest node |
| 881 | * in that subtree), place it where e currently is, and |
| 882 | * then start the deletion process over again on the |
| 883 | * subtree with p as target. |
| 884 | */ |
| 885 | node234 *m = n->kids[ei]; |
| 886 | void *target; |
| 887 | LOG((" case 2a\n")); |
| 888 | while (m->kids[0]) { |
| 889 | m = (m->kids[3] ? m->kids[3] : |
| 890 | m->kids[2] ? m->kids[2] : |
| 891 | m->kids[1] ? m->kids[1] : m->kids[0]); |
| 892 | } |
| 893 | target = (m->elems[2] ? m->elems[2] : |
| 894 | m->elems[1] ? m->elems[1] : m->elems[0]); |
| 895 | n->elems[ei] = target; |
| 896 | index = n->counts[ei] - 1; |
| 897 | n = n->kids[ei]; |
| 898 | } else if (n->kids[ei + 1]->elems[1]) { |
| 899 | /* |
| 900 | * Case 2b, symmetric to 2a but s/left/right/ and |
| 901 | * s/predecessor/successor/. (And s/largest/smallest/). |
| 902 | */ |
| 903 | node234 *m = n->kids[ei + 1]; |
| 904 | void *target; |
| 905 | LOG((" case 2b\n")); |
| 906 | while (m->kids[0]) { |
| 907 | m = m->kids[0]; |
| 908 | } |
| 909 | target = m->elems[0]; |
| 910 | n->elems[ei] = target; |
| 911 | n = n->kids[ei + 1]; |
| 912 | index = 0; |
| 913 | } else { |
| 914 | /* |
| 915 | * Case 2c. n is an internal node, and the subtrees to |
| 916 | * the left and right of e both have only one element. |
| 917 | * So combine the two subnodes into a single big node |
| 918 | * with their own elements on the left and right and e |
| 919 | * in the middle, then restart the deletion process on |
| 920 | * that subtree, with e still as target. |
| 921 | */ |
| 922 | node234 *a = n->kids[ei], *b = n->kids[ei + 1]; |
| 923 | int j; |
| 924 | |
| 925 | LOG((" case 2c\n")); |
| 926 | a->elems[1] = n->elems[ei]; |
| 927 | a->kids[2] = b->kids[0]; |
| 928 | a->counts[2] = b->counts[0]; |
| 929 | if (a->kids[2]) |
| 930 | a->kids[2]->parent = a; |
| 931 | a->elems[2] = b->elems[0]; |
| 932 | a->kids[3] = b->kids[1]; |
| 933 | a->counts[3] = b->counts[1]; |
| 934 | if (a->kids[3]) |
| 935 | a->kids[3]->parent = a; |
| 936 | sfree(b); |
| 937 | n->counts[ei] = countnode234(a); |
| 938 | /* |
| 939 | * That's built the big node in a, and destroyed b. Now |
| 940 | * remove the reference to b (and e) in n. |
| 941 | */ |
| 942 | for (j = ei; j < 2 && n->elems[j + 1]; j++) { |
| 943 | n->elems[j] = n->elems[j + 1]; |
| 944 | n->kids[j + 1] = n->kids[j + 2]; |
| 945 | n->counts[j + 1] = n->counts[j + 2]; |
| 946 | } |
| 947 | n->elems[j] = NULL; |
| 948 | n->kids[j + 1] = NULL; |
| 949 | n->counts[j + 1] = 0; |
| 950 | /* |
| 951 | * It's possible, in this case, that we've just removed |
| 952 | * the only element in the root of the tree. If so, |
| 953 | * shift the root. |
| 954 | */ |
| 955 | if (n->elems[0] == NULL) { |
| 956 | LOG((" shifting root!\n")); |
| 957 | t->root = a; |
| 958 | a->parent = NULL; |
| 959 | sfree(n); |
| 960 | } |
| 961 | /* |
| 962 | * Now go round the deletion process again, with n |
| 963 | * pointing at the new big node and e still the same. |
| 964 | */ |
| 965 | n = a; |
| 966 | index = a->counts[0] + a->counts[1] + 1; |
| 967 | } |
| 968 | } |
| 969 | } |
| 970 | void *delpos234(tree234 * t, int index) |
| 971 | { |
| 972 | if (index < 0 || index >= countnode234(t->root)) |
| 973 | return NULL; |
| 974 | return delpos234_internal(t, index); |
| 975 | } |
| 976 | void *del234(tree234 * t, void *e) |
| 977 | { |
| 978 | int index; |
| 979 | if (!findrelpos234(t, e, NULL, REL234_EQ, &index)) |
| 980 | return NULL; /* it wasn't in there anyway */ |
| 981 | return delpos234_internal(t, index); /* it's there; delete it. */ |
| 982 | } |
| 983 | |
| 984 | #ifdef TEST |
| 985 | |
| 986 | /* |
| 987 | * Test code for the 2-3-4 tree. This code maintains an alternative |
| 988 | * representation of the data in the tree, in an array (using the |
| 989 | * obvious and slow insert and delete functions). After each tree |
| 990 | * operation, the verify() function is called, which ensures all |
| 991 | * the tree properties are preserved: |
| 992 | * - node->child->parent always equals node |
| 993 | * - tree->root->parent always equals NULL |
| 994 | * - number of kids == 0 or number of elements + 1; |
| 995 | * - tree has the same depth everywhere |
| 996 | * - every node has at least one element |
| 997 | * - subtree element counts are accurate |
| 998 | * - any NULL kid pointer is accompanied by a zero count |
| 999 | * - in a sorted tree: ordering property between elements of a |
| 1000 | * node and elements of its children is preserved |
| 1001 | * and also ensures the list represented by the tree is the same |
| 1002 | * list it should be. (This last check also doubly verifies the |
| 1003 | * ordering properties, because the `same list it should be' is by |
| 1004 | * definition correctly ordered. It also ensures all nodes are |
| 1005 | * distinct, because the enum functions would get caught in a loop |
| 1006 | * if not.) |
| 1007 | */ |
| 1008 | |
| 1009 | #include <stdarg.h> |
| 1010 | |
| 1011 | /* |
| 1012 | * Error reporting function. |
| 1013 | */ |
| 1014 | void error(char *fmt, ...) |
| 1015 | { |
| 1016 | va_list ap; |
| 1017 | printf("ERROR: "); |
| 1018 | va_start(ap, fmt); |
| 1019 | vfprintf(stdout, fmt, ap); |
| 1020 | va_end(ap); |
| 1021 | printf("\n"); |
| 1022 | } |
| 1023 | |
| 1024 | /* The array representation of the data. */ |
| 1025 | void **array; |
| 1026 | int arraylen, arraysize; |
| 1027 | cmpfn234 cmp; |
| 1028 | |
| 1029 | /* The tree representation of the same data. */ |
| 1030 | tree234 *tree; |
| 1031 | |
| 1032 | typedef struct { |
| 1033 | int treedepth; |
| 1034 | int elemcount; |
| 1035 | } chkctx; |
| 1036 | |
| 1037 | int chknode(chkctx * ctx, int level, node234 * node, |
| 1038 | void *lowbound, void *highbound) |
| 1039 | { |
| 1040 | int nkids, nelems; |
| 1041 | int i; |
| 1042 | int count; |
| 1043 | |
| 1044 | /* Count the non-NULL kids. */ |
| 1045 | for (nkids = 0; nkids < 4 && node->kids[nkids]; nkids++); |
| 1046 | /* Ensure no kids beyond the first NULL are non-NULL. */ |
| 1047 | for (i = nkids; i < 4; i++) |
| 1048 | if (node->kids[i]) { |
| 1049 | error("node %p: nkids=%d but kids[%d] non-NULL", |
| 1050 | node, nkids, i); |
| 1051 | } else if (node->counts[i]) { |
| 1052 | error("node %p: kids[%d] NULL but count[%d]=%d nonzero", |
| 1053 | node, i, i, node->counts[i]); |
| 1054 | } |
| 1055 | |
| 1056 | /* Count the non-NULL elements. */ |
| 1057 | for (nelems = 0; nelems < 3 && node->elems[nelems]; nelems++); |
| 1058 | /* Ensure no elements beyond the first NULL are non-NULL. */ |
| 1059 | for (i = nelems; i < 3; i++) |
| 1060 | if (node->elems[i]) { |
| 1061 | error("node %p: nelems=%d but elems[%d] non-NULL", |
| 1062 | node, nelems, i); |
| 1063 | } |
| 1064 | |
| 1065 | if (nkids == 0) { |
| 1066 | /* |
| 1067 | * If nkids==0, this is a leaf node; verify that the tree |
| 1068 | * depth is the same everywhere. |
| 1069 | */ |
| 1070 | if (ctx->treedepth < 0) |
| 1071 | ctx->treedepth = level; /* we didn't know the depth yet */ |
| 1072 | else if (ctx->treedepth != level) |
| 1073 | error("node %p: leaf at depth %d, previously seen depth %d", |
| 1074 | node, level, ctx->treedepth); |
| 1075 | } else { |
| 1076 | /* |
| 1077 | * If nkids != 0, then it should be nelems+1, unless nelems |
| 1078 | * is 0 in which case nkids should also be 0 (and so we |
| 1079 | * shouldn't be in this condition at all). |
| 1080 | */ |
| 1081 | int shouldkids = (nelems ? nelems + 1 : 0); |
| 1082 | if (nkids != shouldkids) { |
| 1083 | error("node %p: %d elems should mean %d kids but has %d", |
| 1084 | node, nelems, shouldkids, nkids); |
| 1085 | } |
| 1086 | } |
| 1087 | |
| 1088 | /* |
| 1089 | * nelems should be at least 1. |
| 1090 | */ |
| 1091 | if (nelems == 0) { |
| 1092 | error("node %p: no elems", node, nkids); |
| 1093 | } |
| 1094 | |
| 1095 | /* |
| 1096 | * Add nelems to the running element count of the whole tree. |
| 1097 | */ |
| 1098 | ctx->elemcount += nelems; |
| 1099 | |
| 1100 | /* |
| 1101 | * Check ordering property: all elements should be strictly > |
| 1102 | * lowbound, strictly < highbound, and strictly < each other in |
| 1103 | * sequence. (lowbound and highbound are NULL at edges of tree |
| 1104 | * - both NULL at root node - and NULL is considered to be < |
| 1105 | * everything and > everything. IYSWIM.) |
| 1106 | */ |
| 1107 | if (cmp) { |
| 1108 | for (i = -1; i < nelems; i++) { |
| 1109 | void *lower = (i == -1 ? lowbound : node->elems[i]); |
| 1110 | void *higher = |
| 1111 | (i + 1 == nelems ? highbound : node->elems[i + 1]); |
| 1112 | if (lower && higher && cmp(lower, higher) >= 0) { |
| 1113 | error("node %p: kid comparison [%d=%s,%d=%s] failed", |
| 1114 | node, i, lower, i + 1, higher); |
| 1115 | } |
| 1116 | } |
| 1117 | } |
| 1118 | |
| 1119 | /* |
| 1120 | * Check parent pointers: all non-NULL kids should have a |
| 1121 | * parent pointer coming back to this node. |
| 1122 | */ |
| 1123 | for (i = 0; i < nkids; i++) |
| 1124 | if (node->kids[i]->parent != node) { |
| 1125 | error("node %p kid %d: parent ptr is %p not %p", |
| 1126 | node, i, node->kids[i]->parent, node); |
| 1127 | } |
| 1128 | |
| 1129 | |
| 1130 | /* |
| 1131 | * Now (finally!) recurse into subtrees. |
| 1132 | */ |
| 1133 | count = nelems; |
| 1134 | |
| 1135 | for (i = 0; i < nkids; i++) { |
| 1136 | void *lower = (i == 0 ? lowbound : node->elems[i - 1]); |
| 1137 | void *higher = (i >= nelems ? highbound : node->elems[i]); |
| 1138 | int subcount = |
| 1139 | chknode(ctx, level + 1, node->kids[i], lower, higher); |
| 1140 | if (node->counts[i] != subcount) { |
| 1141 | error("node %p kid %d: count says %d, subtree really has %d", |
| 1142 | node, i, node->counts[i], subcount); |
| 1143 | } |
| 1144 | count += subcount; |
| 1145 | } |
| 1146 | |
| 1147 | return count; |
| 1148 | } |
| 1149 | |
| 1150 | void verify(void) |
| 1151 | { |
| 1152 | chkctx ctx; |
| 1153 | int i; |
| 1154 | void *p; |
| 1155 | |
| 1156 | ctx.treedepth = -1; /* depth unknown yet */ |
| 1157 | ctx.elemcount = 0; /* no elements seen yet */ |
| 1158 | /* |
| 1159 | * Verify validity of tree properties. |
| 1160 | */ |
| 1161 | if (tree->root) { |
| 1162 | if (tree->root->parent != NULL) |
| 1163 | error("root->parent is %p should be null", tree->root->parent); |
| 1164 | chknode(&ctx, 0, tree->root, NULL, NULL); |
| 1165 | } |
| 1166 | printf("tree depth: %d\n", ctx.treedepth); |
| 1167 | /* |
| 1168 | * Enumerate the tree and ensure it matches up to the array. |
| 1169 | */ |
| 1170 | for (i = 0; NULL != (p = index234(tree, i)); i++) { |
| 1171 | if (i >= arraylen) |
| 1172 | error("tree contains more than %d elements", arraylen); |
| 1173 | if (array[i] != p) |
| 1174 | error("enum at position %d: array says %s, tree says %s", |
| 1175 | i, array[i], p); |
| 1176 | } |
| 1177 | if (ctx.elemcount != i) { |
| 1178 | error("tree really contains %d elements, enum gave %d", |
| 1179 | ctx.elemcount, i); |
| 1180 | } |
| 1181 | if (i < arraylen) { |
| 1182 | error("enum gave only %d elements, array has %d", i, arraylen); |
| 1183 | } |
| 1184 | i = count234(tree); |
| 1185 | if (ctx.elemcount != i) { |
| 1186 | error("tree really contains %d elements, count234 gave %d", |
| 1187 | ctx.elemcount, i); |
| 1188 | } |
| 1189 | } |
| 1190 | |
| 1191 | void internal_addtest(void *elem, int index, void *realret) |
| 1192 | { |
| 1193 | int i, j; |
| 1194 | void *retval; |
| 1195 | |
| 1196 | if (arraysize < arraylen + 1) { |
| 1197 | arraysize = arraylen + 1 + 256; |
| 1198 | array = sresize(array, arraysize, void *); |
| 1199 | } |
| 1200 | |
| 1201 | i = index; |
| 1202 | /* now i points to the first element >= elem */ |
| 1203 | retval = elem; /* expect elem returned (success) */ |
| 1204 | for (j = arraylen; j > i; j--) |
| 1205 | array[j] = array[j - 1]; |
| 1206 | array[i] = elem; /* add elem to array */ |
| 1207 | arraylen++; |
| 1208 | |
| 1209 | if (realret != retval) { |
| 1210 | error("add: retval was %p expected %p", realret, retval); |
| 1211 | } |
| 1212 | |
| 1213 | verify(); |
| 1214 | } |
| 1215 | |
| 1216 | void addtest(void *elem) |
| 1217 | { |
| 1218 | int i; |
| 1219 | void *realret; |
| 1220 | |
| 1221 | realret = add234(tree, elem); |
| 1222 | |
| 1223 | i = 0; |
| 1224 | while (i < arraylen && cmp(elem, array[i]) > 0) |
| 1225 | i++; |
| 1226 | if (i < arraylen && !cmp(elem, array[i])) { |
| 1227 | void *retval = array[i]; /* expect that returned not elem */ |
| 1228 | if (realret != retval) { |
| 1229 | error("add: retval was %p expected %p", realret, retval); |
| 1230 | } |
| 1231 | } else |
| 1232 | internal_addtest(elem, i, realret); |
| 1233 | } |
| 1234 | |
| 1235 | void addpostest(void *elem, int i) |
| 1236 | { |
| 1237 | void *realret; |
| 1238 | |
| 1239 | realret = addpos234(tree, elem, i); |
| 1240 | |
| 1241 | internal_addtest(elem, i, realret); |
| 1242 | } |
| 1243 | |
| 1244 | void delpostest(int i) |
| 1245 | { |
| 1246 | int index = i; |
| 1247 | void *elem = array[i], *ret; |
| 1248 | |
| 1249 | /* i points to the right element */ |
| 1250 | while (i < arraylen - 1) { |
| 1251 | array[i] = array[i + 1]; |
| 1252 | i++; |
| 1253 | } |
| 1254 | arraylen--; /* delete elem from array */ |
| 1255 | |
| 1256 | if (tree->cmp) |
| 1257 | ret = del234(tree, elem); |
| 1258 | else |
| 1259 | ret = delpos234(tree, index); |
| 1260 | |
| 1261 | if (ret != elem) { |
| 1262 | error("del returned %p, expected %p", ret, elem); |
| 1263 | } |
| 1264 | |
| 1265 | verify(); |
| 1266 | } |
| 1267 | |
| 1268 | void deltest(void *elem) |
| 1269 | { |
| 1270 | int i; |
| 1271 | |
| 1272 | i = 0; |
| 1273 | while (i < arraylen && cmp(elem, array[i]) > 0) |
| 1274 | i++; |
| 1275 | if (i >= arraylen || cmp(elem, array[i]) != 0) |
| 1276 | return; /* don't do it! */ |
| 1277 | delpostest(i); |
| 1278 | } |
| 1279 | |
| 1280 | /* A sample data set and test utility. Designed for pseudo-randomness, |
| 1281 | * and yet repeatability. */ |
| 1282 | |
| 1283 | /* |
| 1284 | * This random number generator uses the `portable implementation' |
| 1285 | * given in ANSI C99 draft N869. It assumes `unsigned' is 32 bits; |
| 1286 | * change it if not. |
| 1287 | */ |
| 1288 | int randomnumber(unsigned *seed) |
| 1289 | { |
| 1290 | *seed *= 1103515245; |
| 1291 | *seed += 12345; |
| 1292 | return ((*seed) / 65536) % 32768; |
| 1293 | } |
| 1294 | |
| 1295 | int mycmp(void *av, void *bv) |
| 1296 | { |
| 1297 | char const *a = (char const *) av; |
| 1298 | char const *b = (char const *) bv; |
| 1299 | return strcmp(a, b); |
| 1300 | } |
| 1301 | |
| 1302 | #define lenof(x) ( sizeof((x)) / sizeof(*(x)) ) |
| 1303 | |
| 1304 | char *strings[] = { |
| 1305 | "a", "ab", "absque", "coram", "de", |
| 1306 | "palam", "clam", "cum", "ex", "e", |
| 1307 | "sine", "tenus", "pro", "prae", |
| 1308 | "banana", "carrot", "cabbage", "broccoli", "onion", "zebra", |
| 1309 | "penguin", "blancmange", "pangolin", "whale", "hedgehog", |
| 1310 | "giraffe", "peanut", "bungee", "foo", "bar", "baz", "quux", |
| 1311 | "murfl", "spoo", "breen", "flarn", "octothorpe", |
| 1312 | "snail", "tiger", "elephant", "octopus", "warthog", "armadillo", |
| 1313 | "aardvark", "wyvern", "dragon", "elf", "dwarf", "orc", "goblin", |
| 1314 | "pixie", "basilisk", "warg", "ape", "lizard", "newt", "shopkeeper", |
| 1315 | "wand", "ring", "amulet" |
| 1316 | }; |
| 1317 | |
| 1318 | #define NSTR lenof(strings) |
| 1319 | |
| 1320 | int findtest(void) |
| 1321 | { |
| 1322 | const static int rels[] = { |
| 1323 | REL234_EQ, REL234_GE, REL234_LE, REL234_LT, REL234_GT |
| 1324 | }; |
| 1325 | const static char *const relnames[] = { |
| 1326 | "EQ", "GE", "LE", "LT", "GT" |
| 1327 | }; |
| 1328 | int i, j, rel, index; |
| 1329 | char *p, *ret, *realret, *realret2; |
| 1330 | int lo, hi, mid, c; |
| 1331 | |
| 1332 | for (i = 0; i < NSTR; i++) { |
| 1333 | p = strings[i]; |
| 1334 | for (j = 0; j < sizeof(rels) / sizeof(*rels); j++) { |
| 1335 | rel = rels[j]; |
| 1336 | |
| 1337 | lo = 0; |
| 1338 | hi = arraylen - 1; |
| 1339 | while (lo <= hi) { |
| 1340 | mid = (lo + hi) / 2; |
| 1341 | c = strcmp(p, array[mid]); |
| 1342 | if (c < 0) |
| 1343 | hi = mid - 1; |
| 1344 | else if (c > 0) |
| 1345 | lo = mid + 1; |
| 1346 | else |
| 1347 | break; |
| 1348 | } |
| 1349 | |
| 1350 | if (c == 0) { |
| 1351 | if (rel == REL234_LT) |
| 1352 | ret = (mid > 0 ? array[--mid] : NULL); |
| 1353 | else if (rel == REL234_GT) |
| 1354 | ret = (mid < arraylen - 1 ? array[++mid] : NULL); |
| 1355 | else |
| 1356 | ret = array[mid]; |
| 1357 | } else { |
| 1358 | assert(lo == hi + 1); |
| 1359 | if (rel == REL234_LT || rel == REL234_LE) { |
| 1360 | mid = hi; |
| 1361 | ret = (hi >= 0 ? array[hi] : NULL); |
| 1362 | } else if (rel == REL234_GT || rel == REL234_GE) { |
| 1363 | mid = lo; |
| 1364 | ret = (lo < arraylen ? array[lo] : NULL); |
| 1365 | } else |
| 1366 | ret = NULL; |
| 1367 | } |
| 1368 | |
| 1369 | realret = findrelpos234(tree, p, NULL, rel, &index); |
| 1370 | if (realret != ret) { |
| 1371 | error("find(\"%s\",%s) gave %s should be %s", |
| 1372 | p, relnames[j], realret, ret); |
| 1373 | } |
| 1374 | if (realret && index != mid) { |
| 1375 | error("find(\"%s\",%s) gave %d should be %d", |
| 1376 | p, relnames[j], index, mid); |
| 1377 | } |
| 1378 | if (realret && rel == REL234_EQ) { |
| 1379 | realret2 = index234(tree, index); |
| 1380 | if (realret2 != realret) { |
| 1381 | error("find(\"%s\",%s) gave %s(%d) but %d -> %s", |
| 1382 | p, relnames[j], realret, index, index, realret2); |
| 1383 | } |
| 1384 | } |
| 1385 | #if 0 |
| 1386 | printf("find(\"%s\",%s) gave %s(%d)\n", p, relnames[j], |
| 1387 | realret, index); |
| 1388 | #endif |
| 1389 | } |
| 1390 | } |
| 1391 | |
| 1392 | realret = findrelpos234(tree, NULL, NULL, REL234_GT, &index); |
| 1393 | if (arraylen && (realret != array[0] || index != 0)) { |
| 1394 | error("find(NULL,GT) gave %s(%d) should be %s(0)", |
| 1395 | realret, index, array[0]); |
| 1396 | } else if (!arraylen && (realret != NULL)) { |
| 1397 | error("find(NULL,GT) gave %s(%d) should be NULL", realret, index); |
| 1398 | } |
| 1399 | |
| 1400 | realret = findrelpos234(tree, NULL, NULL, REL234_LT, &index); |
| 1401 | if (arraylen |
| 1402 | && (realret != array[arraylen - 1] || index != arraylen - 1)) { |
| 1403 | error("find(NULL,LT) gave %s(%d) should be %s(0)", realret, index, |
| 1404 | array[arraylen - 1]); |
| 1405 | } else if (!arraylen && (realret != NULL)) { |
| 1406 | error("find(NULL,LT) gave %s(%d) should be NULL", realret, index); |
| 1407 | } |
| 1408 | } |
| 1409 | |
| 1410 | int main(void) |
| 1411 | { |
| 1412 | int in[NSTR]; |
| 1413 | int i, j, k; |
| 1414 | unsigned seed = 0; |
| 1415 | |
| 1416 | for (i = 0; i < NSTR; i++) |
| 1417 | in[i] = 0; |
| 1418 | array = NULL; |
| 1419 | arraylen = arraysize = 0; |
| 1420 | tree = newtree234(mycmp); |
| 1421 | cmp = mycmp; |
| 1422 | |
| 1423 | verify(); |
| 1424 | for (i = 0; i < 10000; i++) { |
| 1425 | j = randomnumber(&seed); |
| 1426 | j %= NSTR; |
| 1427 | printf("trial: %d\n", i); |
| 1428 | if (in[j]) { |
| 1429 | printf("deleting %s (%d)\n", strings[j], j); |
| 1430 | deltest(strings[j]); |
| 1431 | in[j] = 0; |
| 1432 | } else { |
| 1433 | printf("adding %s (%d)\n", strings[j], j); |
| 1434 | addtest(strings[j]); |
| 1435 | in[j] = 1; |
| 1436 | } |
| 1437 | findtest(); |
| 1438 | } |
| 1439 | |
| 1440 | while (arraylen > 0) { |
| 1441 | j = randomnumber(&seed); |
| 1442 | j %= arraylen; |
| 1443 | deltest(array[j]); |
| 1444 | } |
| 1445 | |
| 1446 | freetree234(tree); |
| 1447 | |
| 1448 | /* |
| 1449 | * Now try an unsorted tree. We don't really need to test |
| 1450 | * delpos234 because we know del234 is based on it, so it's |
| 1451 | * already been tested in the above sorted-tree code; but for |
| 1452 | * completeness we'll use it to tear down our unsorted tree |
| 1453 | * once we've built it. |
| 1454 | */ |
| 1455 | tree = newtree234(NULL); |
| 1456 | cmp = NULL; |
| 1457 | verify(); |
| 1458 | for (i = 0; i < 1000; i++) { |
| 1459 | printf("trial: %d\n", i); |
| 1460 | j = randomnumber(&seed); |
| 1461 | j %= NSTR; |
| 1462 | k = randomnumber(&seed); |
| 1463 | k %= count234(tree) + 1; |
| 1464 | printf("adding string %s at index %d\n", strings[j], k); |
| 1465 | addpostest(strings[j], k); |
| 1466 | } |
| 1467 | while (count234(tree) > 0) { |
| 1468 | printf("cleanup: tree size %d\n", count234(tree)); |
| 1469 | j = randomnumber(&seed); |
| 1470 | j %= count234(tree); |
| 1471 | printf("deleting string %s from index %d\n", array[j], j); |
| 1472 | delpostest(j); |
| 1473 | } |
| 1474 | |
| 1475 | return 0; |
| 1476 | } |
| 1477 | |
| 1478 | #endif |