Reintegrate tools from trunk

1 files changed, 1479 insertions, 0 deletions

diff --git a/tools/plink/tree234.c b/tools/plink/tree234.c
new file mode 100644
index 000000000..4e2da9dd6
--- /dev/null
+++ b/tools/plink/tree234.c
@@ -0,0 +1,1479 @@
+/*

+ * 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;

+    }

+

+    n = NULL; /* placate gcc; will always be set below since t->root != NULL */

+    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