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putty-source/tree234.c
Simon Tatham d36a4c3685 Introduced wrapper macros snew(), snewn() and sresize() for the 
malloc functions, which automatically cast to the same type they're
allocating the size of. Should prevent any future errors involving
mallocing the size of the wrong structure type, and will also make
life easier if we ever need to turn the PuTTY core code from real C
into C++-friendly C. I haven't touched the Mac frontend in this
checkin because I couldn't compile or test it.

[originally from svn r3014]
2003-03-29 16:14:26 +00:00

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/*
* tree234.c: reasonably generic counted 2-3-4 tree routines.
*
* This file is copyright 1999-2001 Simon Tatham.
*
* Permission is hereby granted, free of charge, to any person
* obtaining a copy of this software and associated documentation
* files (the "Software"), to deal in the Software without
* restriction, including without limitation the rights to use,
* copy, modify, merge, publish, distribute, sublicense, and/or
* sell copies of the Software, and to permit persons to whom the
* Software is furnished to do so, subject to the following
* conditions:
*
* The above copyright notice and this permission notice shall be
* included in all copies or substantial portions of the Software.
*
* THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND,
* EXPRESS OR IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES
* OF MERCHANTABILITY, FITNESS FOR A PARTICULAR PURPOSE AND
* NONINFRINGEMENT. IN NO EVENT SHALL SIMON TATHAM BE LIABLE FOR
* ANY CLAIM, DAMAGES OR OTHER LIABILITY, WHETHER IN AN ACTION OF
* CONTRACT, TORT OR OTHERWISE, ARISING FROM, OUT OF OR IN
* CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE
* SOFTWARE.
*/
#include <stdio.h>
#include <stdlib.h>
#include <assert.h>
#include "puttymem.h"
#include "tree234.h"
#ifdef TEST
#define LOG(x) (printf x)
#else
#define LOG(x)
#endif
typedef struct node234_Tag node234;
struct tree234_Tag {
node234 *root;
cmpfn234 cmp;
};
struct node234_Tag {
node234 *parent;
node234 *kids[4];
int counts[4];
void *elems[3];
};
/*
* Create a 2-3-4 tree.
*/
tree234 *newtree234(cmpfn234 cmp)
{
tree234 *ret = snew(tree234);
LOG(("created tree %p\n", ret));
ret->root = NULL;
ret->cmp = cmp;
return ret;
}
/*
* Free a 2-3-4 tree (not including freeing the elements).
*/
static void freenode234(node234 * n)
{
if (!n)
return;
freenode234(n->kids[0]);
freenode234(n->kids[1]);
freenode234(n->kids[2]);
freenode234(n->kids[3]);
sfree(n);
}
void freetree234(tree234 * t)
{
freenode234(t->root);
sfree(t);
}
/*
* Internal function to count a node.
*/
static int countnode234(node234 * n)
{
int count = 0;
int i;
if (!n)
return 0;
for (i = 0; i < 4; i++)
count += n->counts[i];
for (i = 0; i < 3; i++)
if (n->elems[i])
count++;
return count;
}
/*
* Count the elements in a tree.
*/
int count234(tree234 * t)
{
if (t->root)
return countnode234(t->root);
else
return 0;
}
/*
* Add an element e to a 2-3-4 tree t. Returns e on success, or if
* an existing element compares equal, returns that.
*/
static void *add234_internal(tree234 * t, void *e, int index)
{
node234 *n, **np, *left, *right;
void *orig_e = e;
int c, lcount, rcount;
LOG(("adding node %p to tree %p\n", e, t));
if (t->root == NULL) {
t->root = snew(node234);
t->root->elems[1] = t->root->elems[2] = NULL;
t->root->kids[0] = t->root->kids[1] = NULL;
t->root->kids[2] = t->root->kids[3] = NULL;
t->root->counts[0] = t->root->counts[1] = 0;
t->root->counts[2] = t->root->counts[3] = 0;
t->root->parent = NULL;
t->root->elems[0] = e;
LOG((" created root %p\n", t->root));
return orig_e;
}
np = &t->root;
while (*np) {
int childnum;
n = *np;
LOG((" node %p: %p/%d [%p] %p/%d [%p] %p/%d [%p] %p/%d\n",
n,
n->kids[0], n->counts[0], n->elems[0],
n->kids[1], n->counts[1], n->elems[1],
n->kids[2], n->counts[2], n->elems[2],
n->kids[3], n->counts[3]));
if (index >= 0) {
if (!n->kids[0]) {
/*
* Leaf node. We want to insert at kid position
* equal to the index:
*
* 0 A 1 B 2 C 3
*/
childnum = index;
} else {
/*
* Internal node. We always descend through it (add
* always starts at the bottom, never in the
* middle).
*/
do { /* this is a do ... while (0) to allow `break' */
if (index <= n->counts[0]) {
childnum = 0;
break;
}
index -= n->counts[0] + 1;
if (index <= n->counts[1]) {
childnum = 1;
break;
}
index -= n->counts[1] + 1;
if (index <= n->counts[2]) {
childnum = 2;
break;
}
index -= n->counts[2] + 1;
if (index <= n->counts[3]) {
childnum = 3;
break;
}
return NULL; /* error: index out of range */
} while (0);
}
} else {
if ((c = t->cmp(e, n->elems[0])) < 0)
childnum = 0;
else if (c == 0)
return n->elems[0]; /* already exists */
else if (n->elems[1] == NULL
|| (c = t->cmp(e, n->elems[1])) < 0) childnum = 1;
else if (c == 0)
return n->elems[1]; /* already exists */
else if (n->elems[2] == NULL
|| (c = t->cmp(e, n->elems[2])) < 0) childnum = 2;
else if (c == 0)
return n->elems[2]; /* already exists */
else
childnum = 3;
}
np = &n->kids[childnum];
LOG((" moving to child %d (%p)\n", childnum, *np));
}
/*
* We need to insert the new element in n at position np.
*/
left = NULL;
lcount = 0;
right = NULL;
rcount = 0;
while (n) {
LOG((" at %p: %p/%d [%p] %p/%d [%p] %p/%d [%p] %p/%d\n",
n,
n->kids[0], n->counts[0], n->elems[0],
n->kids[1], n->counts[1], n->elems[1],
n->kids[2], n->counts[2], n->elems[2],
n->kids[3], n->counts[3]));
LOG((" need to insert %p/%d [%p] %p/%d at position %d\n",
left, lcount, e, right, rcount, np - n->kids));
if (n->elems[1] == NULL) {
/*
* Insert in a 2-node; simple.
*/
if (np == &n->kids[0]) {
LOG((" inserting on left of 2-node\n"));
n->kids[2] = n->kids[1];
n->counts[2] = n->counts[1];
n->elems[1] = n->elems[0];
n->kids[1] = right;
n->counts[1] = rcount;
n->elems[0] = e;
n->kids[0] = left;
n->counts[0] = lcount;
} else { /* np == &n->kids[1] */
LOG((" inserting on right of 2-node\n"));
n->kids[2] = right;
n->counts[2] = rcount;
n->elems[1] = e;
n->kids[1] = left;
n->counts[1] = lcount;
}
if (n->kids[0])
n->kids[0]->parent = n;
if (n->kids[1])
n->kids[1]->parent = n;
if (n->kids[2])
n->kids[2]->parent = n;
LOG((" done\n"));
break;
} else if (n->elems[2] == NULL) {
/*
* Insert in a 3-node; simple.
*/
if (np == &n->kids[0]) {
LOG((" inserting on left of 3-node\n"));
n->kids[3] = n->kids[2];
n->counts[3] = n->counts[2];
n->elems[2] = n->elems[1];
n->kids[2] = n->kids[1];
n->counts[2] = n->counts[1];
n->elems[1] = n->elems[0];
n->kids[1] = right;
n->counts[1] = rcount;
n->elems[0] = e;
n->kids[0] = left;
n->counts[0] = lcount;
} else if (np == &n->kids[1]) {
LOG((" inserting in middle of 3-node\n"));
n->kids[3] = n->kids[2];
n->counts[3] = n->counts[2];
n->elems[2] = n->elems[1];
n->kids[2] = right;
n->counts[2] = rcount;
n->elems[1] = e;
n->kids[1] = left;
n->counts[1] = lcount;
} else { /* np == &n->kids[2] */
LOG((" inserting on right of 3-node\n"));
n->kids[3] = right;
n->counts[3] = rcount;
n->elems[2] = e;
n->kids[2] = left;
n->counts[2] = lcount;
}
if (n->kids[0])
n->kids[0]->parent = n;
if (n->kids[1])
n->kids[1]->parent = n;
if (n->kids[2])
n->kids[2]->parent = n;
if (n->kids[3])
n->kids[3]->parent = n;
LOG((" done\n"));
break;
} else {
node234 *m = snew(node234);
m->parent = n->parent;
LOG((" splitting a 4-node; created new node %p\n", m));
/*
* Insert in a 4-node; split into a 2-node and a
* 3-node, and move focus up a level.
*
* I don't think it matters which way round we put the
* 2 and the 3. For simplicity, we'll put the 3 first
* always.
*/
if (np == &n->kids[0]) {
m->kids[0] = left;
m->counts[0] = lcount;
m->elems[0] = e;
m->kids[1] = right;
m->counts[1] = rcount;
m->elems[1] = n->elems[0];
m->kids[2] = n->kids[1];
m->counts[2] = n->counts[1];
e = n->elems[1];
n->kids[0] = n->kids[2];
n->counts[0] = n->counts[2];
n->elems[0] = n->elems[2];
n->kids[1] = n->kids[3];
n->counts[1] = n->counts[3];
} else if (np == &n->kids[1]) {
m->kids[0] = n->kids[0];
m->counts[0] = n->counts[0];
m->elems[0] = n->elems[0];
m->kids[1] = left;
m->counts[1] = lcount;
m->elems[1] = e;
m->kids[2] = right;
m->counts[2] = rcount;
e = n->elems[1];
n->kids[0] = n->kids[2];
n->counts[0] = n->counts[2];
n->elems[0] = n->elems[2];
n->kids[1] = n->kids[3];
n->counts[1] = n->counts[3];
} else if (np == &n->kids[2]) {
m->kids[0] = n->kids[0];
m->counts[0] = n->counts[0];
m->elems[0] = n->elems[0];
m->kids[1] = n->kids[1];
m->counts[1] = n->counts[1];
m->elems[1] = n->elems[1];
m->kids[2] = left;
m->counts[2] = lcount;
/* e = e; */
n->kids[0] = right;
n->counts[0] = rcount;
n->elems[0] = n->elems[2];
n->kids[1] = n->kids[3];
n->counts[1] = n->counts[3];
} else { /* np == &n->kids[3] */
m->kids[0] = n->kids[0];
m->counts[0] = n->counts[0];
m->elems[0] = n->elems[0];
m->kids[1] = n->kids[1];
m->counts[1] = n->counts[1];
m->elems[1] = n->elems[1];
m->kids[2] = n->kids[2];
m->counts[2] = n->counts[2];
n->kids[0] = left;
n->counts[0] = lcount;
n->elems[0] = e;
n->kids[1] = right;
n->counts[1] = rcount;
e = n->elems[2];
}
m->kids[3] = n->kids[3] = n->kids[2] = NULL;
m->counts[3] = n->counts[3] = n->counts[2] = 0;
m->elems[2] = n->elems[2] = n->elems[1] = NULL;
if (m->kids[0])
m->kids[0]->parent = m;
if (m->kids[1])
m->kids[1]->parent = m;
if (m->kids[2])
m->kids[2]->parent = m;
if (n->kids[0])
n->kids[0]->parent = n;
if (n->kids[1])
n->kids[1]->parent = n;
LOG((" left (%p): %p/%d [%p] %p/%d [%p] %p/%d\n", m,
m->kids[0], m->counts[0], m->elems[0],
m->kids[1], m->counts[1], m->elems[1],
m->kids[2], m->counts[2]));
LOG((" right (%p): %p/%d [%p] %p/%d\n", n,
n->kids[0], n->counts[0], n->elems[0],
n->kids[1], n->counts[1]));
left = m;
lcount = countnode234(left);
right = n;
rcount = countnode234(right);
}
if (n->parent)
np = (n->parent->kids[0] == n ? &n->parent->kids[0] :
n->parent->kids[1] == n ? &n->parent->kids[1] :
n->parent->kids[2] == n ? &n->parent->kids[2] :
&n->parent->kids[3]);
n = n->parent;
}
/*
* If we've come out of here by `break', n will still be
* non-NULL and all we need to do is go back up the tree
* updating counts. If we've come here because n is NULL, we
* need to create a new root for the tree because the old one
* has just split into two. */
if (n) {
while (n->parent) {
int count = countnode234(n);
int childnum;
childnum = (n->parent->kids[0] == n ? 0 :
n->parent->kids[1] == n ? 1 :
n->parent->kids[2] == n ? 2 : 3);
n->parent->counts[childnum] = count;
n = n->parent;
}
} else {
LOG((" root is overloaded, split into two\n"));
t->root = snew(node234);
t->root->kids[0] = left;
t->root->counts[0] = lcount;
t->root->elems[0] = e;
t->root->kids[1] = right;
t->root->counts[1] = rcount;
t->root->elems[1] = NULL;
t->root->kids[2] = NULL;
t->root->counts[2] = 0;
t->root->elems[2] = NULL;
t->root->kids[3] = NULL;
t->root->counts[3] = 0;
t->root->parent = NULL;
if (t->root->kids[0])
t->root->kids[0]->parent = t->root;
if (t->root->kids[1])
t->root->kids[1]->parent = t->root;
LOG((" new root is %p/%d [%p] %p/%d\n",
t->root->kids[0], t->root->counts[0],
t->root->elems[0], t->root->kids[1], t->root->counts[1]));
}
return orig_e;
}
void *add234(tree234 * t, void *e)
{
if (!t->cmp) /* tree is unsorted */
return NULL;
return add234_internal(t, e, -1);
}
void *addpos234(tree234 * t, void *e, int index)
{
if (index < 0 || /* index out of range */
t->cmp) /* tree is sorted */
return NULL; /* return failure */
return add234_internal(t, e, index); /* this checks the upper bound */
}
/*
* Look up the element at a given numeric index in a 2-3-4 tree.
* Returns NULL if the index is out of range.
*/
void *index234(tree234 * t, int index)
{
node234 *n;
if (!t->root)
return NULL; /* tree is empty */
if (index < 0 || index >= countnode234(t->root))
return NULL; /* out of range */
n = t->root;
while (n) {
if (index < n->counts[0])
n = n->kids[0];
else if (index -= n->counts[0] + 1, index < 0)
return n->elems[0];
else if (index < n->counts[1])
n = n->kids[1];
else if (index -= n->counts[1] + 1, index < 0)
return n->elems[1];
else if (index < n->counts[2])
n = n->kids[2];
else if (index -= n->counts[2] + 1, index < 0)
return n->elems[2];
else
n = n->kids[3];
}
/* We shouldn't ever get here. I wonder how we did. */
return NULL;
}
/*
* Find an element e in a sorted 2-3-4 tree t. Returns NULL if not
* found. e is always passed as the first argument to cmp, so cmp
* can be an asymmetric function if desired. cmp can also be passed
* as NULL, in which case the compare function from the tree proper
* will be used.
*/
void *findrelpos234(tree234 * t, void *e, cmpfn234 cmp,
int relation, int *index)
{
node234 *n;
void *ret;
int c;
int idx, ecount, kcount, cmpret;
if (t->root == NULL)
return NULL;
if (cmp == NULL)
cmp = t->cmp;
n = t->root;
/*
* Attempt to find the element itself.
*/
idx = 0;
ecount = -1;
/*
* Prepare a fake `cmp' result if e is NULL.
*/
cmpret = 0;
if (e == NULL) {
assert(relation == REL234_LT || relation == REL234_GT);
if (relation == REL234_LT)
cmpret = +1; /* e is a max: always greater */
else if (relation == REL234_GT)
cmpret = -1; /* e is a min: always smaller */
}
while (1) {
for (kcount = 0; kcount < 4; kcount++) {
if (kcount >= 3 || n->elems[kcount] == NULL ||
(c = cmpret ? cmpret : cmp(e, n->elems[kcount])) < 0) {
break;
}
if (n->kids[kcount])
idx += n->counts[kcount];
if (c == 0) {
ecount = kcount;
break;
}
idx++;
}
if (ecount >= 0)
break;
if (n->kids[kcount])
n = n->kids[kcount];
else
break;
}
if (ecount >= 0) {
/*
* We have found the element we're looking for. It's
* n->elems[ecount], at tree index idx. If our search
* relation is EQ, LE or GE we can now go home.
*/
if (relation != REL234_LT && relation != REL234_GT) {
if (index)
*index = idx;
return n->elems[ecount];
}
/*
* Otherwise, we'll do an indexed lookup for the previous
* or next element. (It would be perfectly possible to
* implement these search types in a non-counted tree by
* going back up from where we are, but far more fiddly.)
*/
if (relation == REL234_LT)
idx--;
else
idx++;
} else {
/*
* We've found our way to the bottom of the tree and we
* know where we would insert this node if we wanted to:
* we'd put it in in place of the (empty) subtree
* n->kids[kcount], and it would have index idx
*
* But the actual element isn't there. So if our search
* relation is EQ, we're doomed.
*/
if (relation == REL234_EQ)
return NULL;
/*
* Otherwise, we must do an index lookup for index idx-1
* (if we're going left - LE or LT) or index idx (if we're
* going right - GE or GT).
*/
if (relation == REL234_LT || relation == REL234_LE) {
idx--;
}
}
/*
* We know the index of the element we want; just call index234
* to do the rest. This will return NULL if the index is out of
* bounds, which is exactly what we want.
*/
ret = index234(t, idx);
if (ret && index)
*index = idx;
return ret;
}
void *find234(tree234 * t, void *e, cmpfn234 cmp)
{
return findrelpos234(t, e, cmp, REL234_EQ, NULL);
}
void *findrel234(tree234 * t, void *e, cmpfn234 cmp, int relation)
{
return findrelpos234(t, e, cmp, relation, NULL);
}
void *findpos234(tree234 * t, void *e, cmpfn234 cmp, int *index)
{
return findrelpos234(t, e, cmp, REL234_EQ, index);
}
/*
* Delete an element e in a 2-3-4 tree. Does not free the element,
* merely removes all links to it from the tree nodes.
*/
static void *delpos234_internal(tree234 * t, int index)
{
node234 *n;
void *retval;
int ei = -1;
retval = 0;
n = t->root;
LOG(("deleting item %d from tree %p\n", index, t));
while (1) {
while (n) {
int ki;
node234 *sub;
LOG(
(" node %p: %p/%d [%p] %p/%d [%p] %p/%d [%p] %p/%d index=%d\n",
n, n->kids[0], n->counts[0], n->elems[0], n->kids[1],
n->counts[1], n->elems[1], n->kids[2], n->counts[2],
n->elems[2], n->kids[3], n->counts[3], index));
if (index < n->counts[0]) {
ki = 0;
} else if (index -= n->counts[0] + 1, index < 0) {
ei = 0;
break;
} else if (index < n->counts[1]) {
ki = 1;
} else if (index -= n->counts[1] + 1, index < 0) {
ei = 1;
break;
} else if (index < n->counts[2]) {
ki = 2;
} else if (index -= n->counts[2] + 1, index < 0) {
ei = 2;
break;
} else {
ki = 3;
}
/*
* Recurse down to subtree ki. If it has only one element,
* we have to do some transformation to start with.
*/
LOG((" moving to subtree %d\n", ki));
sub = n->kids[ki];
if (!sub->elems[1]) {
LOG((" subtree has only one element!\n", ki));
if (ki > 0 && n->kids[ki - 1]->elems[1]) {
/*
* Case 3a, left-handed variant. Child ki has
* only one element, but child ki-1 has two or
* more. So we need to move a subtree from ki-1
* to ki.
*
* . C . . B .
* / \ -> / \
* [more] a A b B c d D e [more] a A b c C d D e
*/
node234 *sib = n->kids[ki - 1];
int lastelem = (sib->elems[2] ? 2 :
sib->elems[1] ? 1 : 0);
sub->kids[2] = sub->kids[1];
sub->counts[2] = sub->counts[1];
sub->elems[1] = sub->elems[0];
sub->kids[1] = sub->kids[0];
sub->counts[1] = sub->counts[0];
sub->elems[0] = n->elems[ki - 1];
sub->kids[0] = sib->kids[lastelem + 1];
sub->counts[0] = sib->counts[lastelem + 1];
if (sub->kids[0])
sub->kids[0]->parent = sub;
n->elems[ki - 1] = sib->elems[lastelem];
sib->kids[lastelem + 1] = NULL;
sib->counts[lastelem + 1] = 0;
sib->elems[lastelem] = NULL;
n->counts[ki] = countnode234(sub);
LOG((" case 3a left\n"));
LOG(
(" index and left subtree count before adjustment: %d, %d\n",
index, n->counts[ki - 1]));
index += n->counts[ki - 1];
n->counts[ki - 1] = countnode234(sib);
index -= n->counts[ki - 1];
LOG(
(" index and left subtree count after adjustment: %d, %d\n",
index, n->counts[ki - 1]));
} else if (ki < 3 && n->kids[ki + 1]
&& n->kids[ki + 1]->elems[1]) {
/*
* Case 3a, right-handed variant. ki has only
* one element but ki+1 has two or more. Move a
* subtree from ki+1 to ki.
*
* . B . . C .
* / \ -> / \
* a A b c C d D e [more] a A b B c d D e [more]
*/
node234 *sib = n->kids[ki + 1];
int j;
sub->elems[1] = n->elems[ki];
sub->kids[2] = sib->kids[0];
sub->counts[2] = sib->counts[0];
if (sub->kids[2])
sub->kids[2]->parent = sub;
n->elems[ki] = sib->elems[0];
sib->kids[0] = sib->kids[1];
sib->counts[0] = sib->counts[1];
for (j = 0; j < 2 && sib->elems[j + 1]; j++) {
sib->kids[j + 1] = sib->kids[j + 2];
sib->counts[j + 1] = sib->counts[j + 2];
sib->elems[j] = sib->elems[j + 1];
}
sib->kids[j + 1] = NULL;
sib->counts[j + 1] = 0;
sib->elems[j] = NULL;
n->counts[ki] = countnode234(sub);
n->counts[ki + 1] = countnode234(sib);
LOG((" case 3a right\n"));
} else {
/*
* Case 3b. ki has only one element, and has no
* neighbour with more than one. So pick a
* neighbour and merge it with ki, taking an
* element down from n to go in the middle.
*
* . B . .
* / \ -> |
* a A b c C d a A b B c C d
*
* (Since at all points we have avoided
* descending to a node with only one element,
* we can be sure that n is not reduced to
* nothingness by this move, _unless_ it was
* the very first node, ie the root of the
* tree. In that case we remove the now-empty
* root and replace it with its single large
* child as shown.)
*/
node234 *sib;
int j;
if (ki > 0) {
ki--;
index += n->counts[ki] + 1;
}
sib = n->kids[ki];
sub = n->kids[ki + 1];
sub->kids[3] = sub->kids[1];
sub->counts[3] = sub->counts[1];
sub->elems[2] = sub->elems[0];
sub->kids[2] = sub->kids[0];
sub->counts[2] = sub->counts[0];
sub->elems[1] = n->elems[ki];
sub->kids[1] = sib->kids[1];
sub->counts[1] = sib->counts[1];
if (sub->kids[1])
sub->kids[1]->parent = sub;
sub->elems[0] = sib->elems[0];
sub->kids[0] = sib->kids[0];
sub->counts[0] = sib->counts[0];
if (sub->kids[0])
sub->kids[0]->parent = sub;
n->counts[ki + 1] = countnode234(sub);
sfree(sib);
/*
* That's built the big node in sub. Now we
* need to remove the reference to sib in n.
*/
for (j = ki; j < 3 && n->kids[j + 1]; j++) {
n->kids[j] = n->kids[j + 1];
n->counts[j] = n->counts[j + 1];
n->elems[j] = j < 2 ? n->elems[j + 1] : NULL;
}
n->kids[j] = NULL;
n->counts[j] = 0;
if (j < 3)
n->elems[j] = NULL;
LOG((" case 3b ki=%d\n", ki));
if (!n->elems[0]) {
/*
* The root is empty and needs to be
* removed.
*/
LOG((" shifting root!\n"));
t->root = sub;
sub->parent = NULL;
sfree(n);
}
}
}
n = sub;
}
if (!retval)
retval = n->elems[ei];
if (ei == -1)
return NULL; /* although this shouldn't happen */
/*
* Treat special case: this is the one remaining item in
* the tree. n is the tree root (no parent), has one
* element (no elems[1]), and has no kids (no kids[0]).
*/
if (!n->parent && !n->elems[1] && !n->kids[0]) {
LOG((" removed last element in tree\n"));
sfree(n);
t->root = NULL;
return retval;
}
/*
* Now we have the element we want, as n->elems[ei], and we
* have also arranged for that element not to be the only
* one in its node. So...
*/
if (!n->kids[0] && n->elems[1]) {
/*
* Case 1. n is a leaf node with more than one element,
* so it's _really easy_. Just delete the thing and
* we're done.
*/
int i;
LOG((" case 1\n"));
for (i = ei; i < 2 && n->elems[i + 1]; i++)
n->elems[i] = n->elems[i + 1];
n->elems[i] = NULL;
/*
* Having done that to the leaf node, we now go back up
* the tree fixing the counts.
*/
while (n->parent) {
int childnum;
childnum = (n->parent->kids[0] == n ? 0 :
n->parent->kids[1] == n ? 1 :
n->parent->kids[2] == n ? 2 : 3);
n->parent->counts[childnum]--;
n = n->parent;
}
return retval; /* finished! */
} else if (n->kids[ei]->elems[1]) {
/*
* Case 2a. n is an internal node, and the root of the
* subtree to the left of e has more than one element.
* So find the predecessor p to e (ie the largest node
* in that subtree), place it where e currently is, and
* then start the deletion process over again on the
* subtree with p as target.
*/
node234 *m = n->kids[ei];
void *target;
LOG((" case 2a\n"));
while (m->kids[0]) {
m = (m->kids[3] ? m->kids[3] :
m->kids[2] ? m->kids[2] :
m->kids[1] ? m->kids[1] : m->kids[0]);
}
target = (m->elems[2] ? m->elems[2] :
m->elems[1] ? m->elems[1] : m->elems[0]);
n->elems[ei] = target;
index = n->counts[ei] - 1;
n = n->kids[ei];
} else if (n->kids[ei + 1]->elems[1]) {
/*
* Case 2b, symmetric to 2a but s/left/right/ and
* s/predecessor/successor/. (And s/largest/smallest/).
*/
node234 *m = n->kids[ei + 1];
void *target;
LOG((" case 2b\n"));
while (m->kids[0]) {
m = m->kids[0];
}
target = m->elems[0];
n->elems[ei] = target;
n = n->kids[ei + 1];
index = 0;
} else {
/*
* Case 2c. n is an internal node, and the subtrees to
* the left and right of e both have only one element.
* So combine the two subnodes into a single big node
* with their own elements on the left and right and e
* in the middle, then restart the deletion process on
* that subtree, with e still as target.
*/
node234 *a = n->kids[ei], *b = n->kids[ei + 1];
int j;
LOG((" case 2c\n"));
a->elems[1] = n->elems[ei];
a->kids[2] = b->kids[0];
a->counts[2] = b->counts[0];
if (a->kids[2])
a->kids[2]->parent = a;
a->elems[2] = b->elems[0];
a->kids[3] = b->kids[1];
a->counts[3] = b->counts[1];
if (a->kids[3])
a->kids[3]->parent = a;
sfree(b);
n->counts[ei] = countnode234(a);
/*
* That's built the big node in a, and destroyed b. Now
* remove the reference to b (and e) in n.
*/
for (j = ei; j < 2 && n->elems[j + 1]; j++) {
n->elems[j] = n->elems[j + 1];
n->kids[j + 1] = n->kids[j + 2];
n->counts[j + 1] = n->counts[j + 2];
}
n->elems[j] = NULL;
n->kids[j + 1] = NULL;
n->counts[j + 1] = 0;
/*
* It's possible, in this case, that we've just removed
* the only element in the root of the tree. If so,
* shift the root.
*/
if (n->elems[0] == NULL) {
LOG((" shifting root!\n"));
t->root = a;
a->parent = NULL;
sfree(n);
}
/*
* Now go round the deletion process again, with n
* pointing at the new big node and e still the same.
*/
n = a;
index = a->counts[0] + a->counts[1] + 1;
}
}
}
void *delpos234(tree234 * t, int index)
{
if (index < 0 || index >= countnode234(t->root))
return NULL;
return delpos234_internal(t, index);
}
void *del234(tree234 * t, void *e)
{
int index;
if (!findrelpos234(t, e, NULL, REL234_EQ, &index))
return NULL; /* it wasn't in there anyway */
return delpos234_internal(t, index); /* it's there; delete it. */
}
#ifdef TEST
/*
* Test code for the 2-3-4 tree. This code maintains an alternative
* representation of the data in the tree, in an array (using the
* obvious and slow insert and delete functions). After each tree
* operation, the verify() function is called, which ensures all
* the tree properties are preserved:
* - node->child->parent always equals node
* - tree->root->parent always equals NULL
* - number of kids == 0 or number of elements + 1;
* - tree has the same depth everywhere
* - every node has at least one element
* - subtree element counts are accurate
* - any NULL kid pointer is accompanied by a zero count
* - in a sorted tree: ordering property between elements of a
* node and elements of its children is preserved
* and also ensures the list represented by the tree is the same
* list it should be. (This last check also doubly verifies the
* ordering properties, because the `same list it should be' is by
* definition correctly ordered. It also ensures all nodes are
* distinct, because the enum functions would get caught in a loop
* if not.)
*/
#include <stdarg.h>
/*
* Error reporting function.
*/
void error(char *fmt, ...)
{
va_list ap;
printf("ERROR: ");
va_start(ap, fmt);
vfprintf(stdout, fmt, ap);
va_end(ap);
printf("\n");
}
/* The array representation of the data. */
void **array;
int arraylen, arraysize;
cmpfn234 cmp;
/* The tree representation of the same data. */
tree234 *tree;
typedef struct {
int treedepth;
int elemcount;
} chkctx;
int chknode(chkctx * ctx, int level, node234 * node,
void *lowbound, void *highbound)
{
int nkids, nelems;
int i;
int count;
/* Count the non-NULL kids. */
for (nkids = 0; nkids < 4 && node->kids[nkids]; nkids++);
/* Ensure no kids beyond the first NULL are non-NULL. */
for (i = nkids; i < 4; i++)
if (node->kids[i]) {
error("node %p: nkids=%d but kids[%d] non-NULL",
node, nkids, i);
} else if (node->counts[i]) {
error("node %p: kids[%d] NULL but count[%d]=%d nonzero",
node, i, i, node->counts[i]);
}
/* Count the non-NULL elements. */
for (nelems = 0; nelems < 3 && node->elems[nelems]; nelems++);
/* Ensure no elements beyond the first NULL are non-NULL. */
for (i = nelems; i < 3; i++)
if (node->elems[i]) {
error("node %p: nelems=%d but elems[%d] non-NULL",
node, nelems, i);
}
if (nkids == 0) {
/*
* If nkids==0, this is a leaf node; verify that the tree
* depth is the same everywhere.
*/
if (ctx->treedepth < 0)
ctx->treedepth = level; /* we didn't know the depth yet */
else if (ctx->treedepth != level)
error("node %p: leaf at depth %d, previously seen depth %d",
node, level, ctx->treedepth);
} else {
/*
* If nkids != 0, then it should be nelems+1, unless nelems
* is 0 in which case nkids should also be 0 (and so we
* shouldn't be in this condition at all).
*/
int shouldkids = (nelems ? nelems + 1 : 0);
if (nkids != shouldkids) {
error("node %p: %d elems should mean %d kids but has %d",
node, nelems, shouldkids, nkids);
}
}
/*
* nelems should be at least 1.
*/
if (nelems == 0) {
error("node %p: no elems", node, nkids);
}
/*
* Add nelems to the running element count of the whole tree.
*/
ctx->elemcount += nelems;
/*
* Check ordering property: all elements should be strictly >
* lowbound, strictly < highbound, and strictly < each other in
* sequence. (lowbound and highbound are NULL at edges of tree
* - both NULL at root node - and NULL is considered to be <
* everything and > everything. IYSWIM.)
*/
if (cmp) {
for (i = -1; i < nelems; i++) {
void *lower = (i == -1 ? lowbound : node->elems[i]);
void *higher =
(i + 1 == nelems ? highbound : node->elems[i + 1]);
if (lower && higher && cmp(lower, higher) >= 0) {
error("node %p: kid comparison [%d=%s,%d=%s] failed",
node, i, lower, i + 1, higher);
}
}
}
/*
* Check parent pointers: all non-NULL kids should have a
* parent pointer coming back to this node.
*/
for (i = 0; i < nkids; i++)
if (node->kids[i]->parent != node) {
error("node %p kid %d: parent ptr is %p not %p",
node, i, node->kids[i]->parent, node);
}
/*
* Now (finally!) recurse into subtrees.
*/
count = nelems;
for (i = 0; i < nkids; i++) {
void *lower = (i == 0 ? lowbound : node->elems[i - 1]);
void *higher = (i >= nelems ? highbound : node->elems[i]);
int subcount =
chknode(ctx, level + 1, node->kids[i], lower, higher);
if (node->counts[i] != subcount) {
error("node %p kid %d: count says %d, subtree really has %d",
node, i, node->counts[i], subcount);
}
count += subcount;
}
return count;
}
void verify(void)
{
chkctx ctx;
int i;
void *p;
ctx.treedepth = -1; /* depth unknown yet */
ctx.elemcount = 0; /* no elements seen yet */
/*
* Verify validity of tree properties.
*/
if (tree->root) {
if (tree->root->parent != NULL)
error("root->parent is %p should be null", tree->root->parent);
chknode(&ctx, 0, tree->root, NULL, NULL);
}
printf("tree depth: %d\n", ctx.treedepth);
/*
* Enumerate the tree and ensure it matches up to the array.
*/
for (i = 0; NULL != (p = index234(tree, i)); i++) {
if (i >= arraylen)
error("tree contains more than %d elements", arraylen);
if (array[i] != p)
error("enum at position %d: array says %s, tree says %s",
i, array[i], p);
}
if (ctx.elemcount != i) {
error("tree really contains %d elements, enum gave %d",
ctx.elemcount, i);
}
if (i < arraylen) {
error("enum gave only %d elements, array has %d", i, arraylen);
}
i = count234(tree);
if (ctx.elemcount != i) {
error("tree really contains %d elements, count234 gave %d",
ctx.elemcount, i);
}
}
void internal_addtest(void *elem, int index, void *realret)
{
int i, j;
void *retval;
if (arraysize < arraylen + 1) {
arraysize = arraylen + 1 + 256;
array = sresize(array, arraysize, void *);
}
i = index;
/* now i points to the first element >= elem */
retval = elem; /* expect elem returned (success) */
for (j = arraylen; j > i; j--)
array[j] = array[j - 1];
array[i] = elem; /* add elem to array */
arraylen++;
if (realret != retval) {
error("add: retval was %p expected %p", realret, retval);
}
verify();
}
void addtest(void *elem)
{
int i;
void *realret;
realret = add234(tree, elem);
i = 0;
while (i < arraylen && cmp(elem, array[i]) > 0)
i++;
if (i < arraylen && !cmp(elem, array[i])) {
void *retval = array[i]; /* expect that returned not elem */
if (realret != retval) {
error("add: retval was %p expected %p", realret, retval);
}
} else
internal_addtest(elem, i, realret);
}
void addpostest(void *elem, int i)
{
void *realret;
realret = addpos234(tree, elem, i);
internal_addtest(elem, i, realret);
}
void delpostest(int i)
{
int index = i;
void *elem = array[i], *ret;
/* i points to the right element */
while (i < arraylen - 1) {
array[i] = array[i + 1];
i++;
}
arraylen--; /* delete elem from array */
if (tree->cmp)
ret = del234(tree, elem);
else
ret = delpos234(tree, index);
if (ret != elem) {
error("del returned %p, expected %p", ret, elem);
}
verify();
}
void deltest(void *elem)
{
int i;
i = 0;
while (i < arraylen && cmp(elem, array[i]) > 0)
i++;
if (i >= arraylen || cmp(elem, array[i]) != 0)
return; /* don't do it! */
delpostest(i);
}
/* A sample data set and test utility. Designed for pseudo-randomness,
* and yet repeatability. */
/*
* This random number generator uses the `portable implementation'
* given in ANSI C99 draft N869. It assumes `unsigned' is 32 bits;
* change it if not.
*/
int randomnumber(unsigned *seed)
{
*seed *= 1103515245;
*seed += 12345;
return ((*seed) / 65536) % 32768;
}
int mycmp(void *av, void *bv)
{
char const *a = (char const *) av;
char const *b = (char const *) bv;
return strcmp(a, b);
}
#define lenof(x) ( sizeof((x)) / sizeof(*(x)) )
char *strings[] = {
"a", "ab", "absque", "coram", "de",
"palam", "clam", "cum", "ex", "e",
"sine", "tenus", "pro", "prae",
"banana", "carrot", "cabbage", "broccoli", "onion", "zebra",
"penguin", "blancmange", "pangolin", "whale", "hedgehog",
"giraffe", "peanut", "bungee", "foo", "bar", "baz", "quux",
"murfl", "spoo", "breen", "flarn", "octothorpe",
"snail", "tiger", "elephant", "octopus", "warthog", "armadillo",
"aardvark", "wyvern", "dragon", "elf", "dwarf", "orc", "goblin",
"pixie", "basilisk", "warg", "ape", "lizard", "newt", "shopkeeper",
"wand", "ring", "amulet"
};
#define NSTR lenof(strings)
int findtest(void)
{
const static int rels[] = {
REL234_EQ, REL234_GE, REL234_LE, REL234_LT, REL234_GT
};
const static char *const relnames[] = {
"EQ", "GE", "LE", "LT", "GT"
};
int i, j, rel, index;
char *p, *ret, *realret, *realret2;
int lo, hi, mid, c;
for (i = 0; i < NSTR; i++) {
p = strings[i];
for (j = 0; j < sizeof(rels) / sizeof(*rels); j++) {
rel = rels[j];
lo = 0;
hi = arraylen - 1;
while (lo <= hi) {
mid = (lo + hi) / 2;
c = strcmp(p, array[mid]);
if (c < 0)
hi = mid - 1;
else if (c > 0)
lo = mid + 1;
else
break;
}
if (c == 0) {
if (rel == REL234_LT)
ret = (mid > 0 ? array[--mid] : NULL);
else if (rel == REL234_GT)
ret = (mid < arraylen - 1 ? array[++mid] : NULL);
else
ret = array[mid];
} else {
assert(lo == hi + 1);
if (rel == REL234_LT || rel == REL234_LE) {
mid = hi;
ret = (hi >= 0 ? array[hi] : NULL);
} else if (rel == REL234_GT || rel == REL234_GE) {
mid = lo;
ret = (lo < arraylen ? array[lo] : NULL);
} else
ret = NULL;
}
realret = findrelpos234(tree, p, NULL, rel, &index);
if (realret != ret) {
error("find(\"%s\",%s) gave %s should be %s",
p, relnames[j], realret, ret);
}
if (realret && index != mid) {
error("find(\"%s\",%s) gave %d should be %d",
p, relnames[j], index, mid);
}
if (realret && rel == REL234_EQ) {
realret2 = index234(tree, index);
if (realret2 != realret) {
error("find(\"%s\",%s) gave %s(%d) but %d -> %s",
p, relnames[j], realret, index, index, realret2);
}
}
#if 0
printf("find(\"%s\",%s) gave %s(%d)\n", p, relnames[j],
realret, index);
#endif
}
}
realret = findrelpos234(tree, NULL, NULL, REL234_GT, &index);
if (arraylen && (realret != array[0] || index != 0)) {
error("find(NULL,GT) gave %s(%d) should be %s(0)",
realret, index, array[0]);
} else if (!arraylen && (realret != NULL)) {
error("find(NULL,GT) gave %s(%d) should be NULL", realret, index);
}
realret = findrelpos234(tree, NULL, NULL, REL234_LT, &index);
if (arraylen
&& (realret != array[arraylen - 1] || index != arraylen - 1)) {
error("find(NULL,LT) gave %s(%d) should be %s(0)", realret, index,
array[arraylen - 1]);
} else if (!arraylen && (realret != NULL)) {
error("find(NULL,LT) gave %s(%d) should be NULL", realret, index);
}
}
int main(void)
{
int in[NSTR];
int i, j, k;
unsigned seed = 0;
for (i = 0; i < NSTR; i++)
in[i] = 0;
array = NULL;
arraylen = arraysize = 0;
tree = newtree234(mycmp);
cmp = mycmp;
verify();
for (i = 0; i < 10000; i++) {
j = randomnumber(&seed);
j %= NSTR;
printf("trial: %d\n", i);
if (in[j]) {
printf("deleting %s (%d)\n", strings[j], j);
deltest(strings[j]);
in[j] = 0;
} else {
printf("adding %s (%d)\n", strings[j], j);
addtest(strings[j]);
in[j] = 1;
}
findtest();
}
while (arraylen > 0) {
j = randomnumber(&seed);
j %= arraylen;
deltest(array[j]);
}
freetree234(tree);
/*
* Now try an unsorted tree. We don't really need to test
* delpos234 because we know del234 is based on it, so it's
* already been tested in the above sorted-tree code; but for
* completeness we'll use it to tear down our unsorted tree
* once we've built it.
*/
tree = newtree234(NULL);
cmp = NULL;
verify();
for (i = 0; i < 1000; i++) {
printf("trial: %d\n", i);
j = randomnumber(&seed);
j %= NSTR;
k = randomnumber(&seed);
k %= count234(tree) + 1;
printf("adding string %s at index %d\n", strings[j], k);
addpostest(strings[j], k);
}
while (count234(tree) > 0) {
printf("cleanup: tree size %d\n", count234(tree));
j = randomnumber(&seed);
j %= count234(tree);
printf("deleting string %s from index %d\n", array[j], j);
delpostest(j);
}
return 0;
}
#endif
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