/*
* CDDL HEADER START
*
* The contents of this file are subject to the terms of the
* Common Development and Distribution License, Version 1.0 only
* (the "License"). You may not use this file except in compliance
* with the License.
*
* You can obtain a copy of the license at usr/src/OPENSOLARIS.LICENSE
* or http://www.opensolaris.org/os/licensing.
* See the License for the specific language governing permissions
* and limitations under the License.
*
* When distributing Covered Code, include this CDDL HEADER in each
* file and include the License file at usr/src/OPENSOLARIS.LICENSE.
* If applicable, add the following below this CDDL HEADER, with the
* fields enclosed by brackets "[]" replaced with your own identifying
* information: Portions Copyright [yyyy] [name of copyright owner]
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* CDDL HEADER END
*/
/*
* Copyright 1990 Sun Microsystems, Inc. All rights reserved.
* Use is subject to license terms.
*/
#pragma ident "%Z%%M% %I% %E% SMI"
/*
* qsort.c:
* Our own version of the system qsort routine which is faster by an average
* of 25%, with lows and highs of 10% and 50%.
* The THRESHold below is the insertion sort threshold, and has been adjusted
* for records of size 48 bytes.
* The MTHREShold is where we stop finding a better median.
*/
#define THRESH 4 /* threshold for insertion */
#define MTHRESH 6 /* threshold for median */
static int (*qcmp)(); /* the comparison routine */
static int qsz; /* size of each record */
static int thresh; /* THRESHold in chars */
static int mthresh; /* MTHRESHold in chars */
static void qst(char *, char *);
/*
* qsort:
* First, set up some global parameters for qst to share. Then, quicksort
* with qst(), and then a cleanup insertion sort ourselves. Sound simple?
* It's not...
*/
void
qsort(char *base, int n, int size, int (*compar)())
{
char c, *i, *j, *lo, *hi;
char *min, *max;
if (n <= 1)
return;
qsz = size;
qcmp = compar;
thresh = qsz * THRESH;
mthresh = qsz * MTHRESH;
max = base + n * qsz;
if (n >= THRESH) {
qst(base, max);
hi = base + thresh;
} else {
hi = max;
}
/*
* First put smallest element, which must be in the first THRESH, in
* the first position as a sentinel. This is done just by searching
* the first THRESH elements (or the first n if n < THRESH), finding
* the min, and swapping it into the first position.
*/
for (j = lo = base; (lo += qsz) < hi; )
if (qcmp(j, lo) > 0)
j = lo;
if (j != base) {
/* swap j into place */
for (i = base, hi = base + qsz; i < hi; ) {
c = *j;
*j++ = *i;
*i++ = c;
}
}
/*
* With our sentinel in place, we now run the following hyper-fast
* insertion sort. For each remaining element, min, from [1] to [n-1],
* set hi to the index of the element AFTER which this one goes.
* Then, do the standard insertion sort shift on a character at a time
* basis for each element in the frob.
*/
for (min = base; (hi = min += qsz) < max; ) {
while (qcmp(hi -= qsz, min) > 0)
/* void */;
if ((hi += qsz) != min) {
for (lo = min + qsz; --lo >= min; ) {
c = *lo;
for (i = j = lo; (j -= qsz) >= hi; i = j)
*i = *j;
*i = c;
}
}
}
}
/*
* qst:
* Do a quicksort
* First, find the median element, and put that one in the first place as the
* discriminator. (This "median" is just the median of the first, last and
* middle elements). (Using this median instead of the first element is a big
* win). Then, the usual partitioning/swapping, followed by moving the
* discriminator into the right place. Then, figure out the sizes of the two
* partions, do the smaller one recursively and the larger one via a repeat of
* this code. Stopping when there are less than THRESH elements in a partition
* and cleaning up with an insertion sort (in our caller) is a huge win.
* All data swaps are done in-line, which is space-losing but time-saving.
* (And there are only three places where this is done).
*/
static void
qst(char *base, char *max)
{
char c, *i, *j, *jj;
int ii;
char *mid, *tmp;
int lo, hi;
/*
* At the top here, lo is the number of characters of elements in the
* current partition. (Which should be max - base).
* Find the median of the first, last, and middle element and make
* that the middle element. Set j to largest of first and middle.
* If max is larger than that guy, then it's that guy, else compare
* max with loser of first and take larger. Things are set up to
* prefer the middle, then the first in case of ties.
*/
lo = max - base; /* number of elements as chars */
do {
mid = i = base + qsz * ((lo / qsz) >> 1);
if (lo >= mthresh) {
j = (qcmp((jj = base), i) > 0 ? jj : i);
if (qcmp(j, (tmp = max - qsz)) > 0) {
/* switch to first loser */
j = (j == jj ? i : jj);
if (qcmp(j, tmp) < 0)
j = tmp;
}
if (j != i) {
ii = qsz;
do {
c = *i;
*i++ = *j;
*j++ = c;
} while (--ii);
}
}
/*
* Semi-standard quicksort partitioning/swapping
*/
for (i = base, j = max - qsz; ; ) {
while (i < mid && qcmp(i, mid) <= 0)
i += qsz;
while (j > mid) {
if (qcmp(mid, j) <= 0) {
j -= qsz;
continue;
}
tmp = i + qsz; /* value of i after swap */
if (i == mid) {
/* j <-> mid, new mid is j */
mid = jj = j;
} else {
/* i <-> j */
jj = j;
j -= qsz;
}
goto swap;
}
if (i == mid) {
break;
} else {
/* i <-> mid, new mid is i */
jj = mid;
tmp = mid = i; /* value of i after swap */
j -= qsz;
}
swap:
ii = qsz;
do {
c = *i;
*i++ = *jj;
*jj++ = c;
} while (--ii);
i = tmp;
}
/*
* Look at sizes of the two partitions, do the smaller
* one first by recursion, then do the larger one by
* making sure lo is its size, base and max are update
* correctly, and branching back. But only repeat
* (recursively or by branching) if the partition is
* of at least size THRESH.
*/
i = (j = mid) + qsz;
if ((lo = j - base) <= (hi = max - i)) {
if (lo >= thresh)
qst(base, j);
base = i;
lo = hi;
} else {
if (hi >= thresh)
qst(i, max);
max = j;
}
} while (lo >= thresh);
}