reactos/sdk/lib/crt/stdlib/qsort.c

/* Copyright (C) 1994 DJ Delorie, see COPYING.DJ for details */

#include <stdlib.h>
#include <search.h>

/*-
 * Copyright (c) 1980, 1983 The Regents of the University of California.
 * All rights reserved.
 *
 * Redistribution and use in source and binary forms are permitted
 * provided that: (1) source distributions retain this entire copyright
 * notice and comment, and (2) distributions including binaries display
 * the following acknowledgement:  ``This product includes software
 * developed by the University of California, Berkeley and its contributors''
 * in the documentation or other materials provided with the distribution
 * and in all advertising materials mentioning features or use of this
 * software. Neither the name of the University nor the names of its
 * contributors may be used to endorse or promote products derived
 * from this software without specific prior written permission.
 * THIS SOFTWARE IS PROVIDED ``AS IS'' AND WITHOUT ANY EXPRESS OR
 * IMPLIED WARRANTIES, INCLUDING, WITHOUT LIMITATION, THE IMPLIED
 * WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE.
 */

/*
 * 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 */

/*
 * 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 __cdecl
qst(size_t size, int (__cdecl *compar)(const void*, const void*), char *base, char *max)
{
  char c, *i, *j, *jj;
  size_t ii;
  char *mid, *tmp;
  size_t lo, hi;
  size_t thresh;
  size_t mthresh;

  thresh = size * THRESH;
  mthresh = size * MTHRESH;

  /*
   * 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 + size * ((lo / size) >> 1);
    if (lo >= mthresh)
    {
      j = (compar((jj = base), i) > 0 ? jj : i);
      if (compar(j, (tmp = max - size)) > 0)
      {
	/* switch to first loser */
	j = (j == jj ? i : jj);
	if (compar(j, tmp) < 0)
	  j = tmp;
      }
      if (j != i)
      {
	ii = size;
	do	{
	  c = *i;
	  *i++ = *j;
	  *j++ = c;
	} while (--ii);
      }
    }
    /*
     * Semi-standard quicksort partitioning/swapping
     */
    for (i = base, j = max - size; ; )
    {
      while (i < mid && compar(i, mid) <= 0)
	i += size;
      while (j > mid)
      {
	if (compar(mid, j) <= 0)
	{
	  j -= size;
	  continue;
	}
	tmp = i + size;	/* value of i after swap */
	if (i == mid)
	{
	  /* j <-> mid, new mid is j */
	  mid = jj = j;
	}
	else
	{
	  /* i <-> j */
	  jj = j;
	  j -= size;
	}
	goto swap;
      }
      if (i == mid)
      {
	break;
      }
      else
      {
	/* i <-> mid, new mid is i */
	jj = mid;
	tmp = mid = i;		/* value of i after swap */
	j -= size;
      }
    swap:
      ii = size;
      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) + size;
    if ((lo = j - base) <= (hi = max - i))
    {
      if (lo >= thresh)
	qst(size, compar, base, j);
      base = i;
      lo = hi;
    }
    else
    {
      if (hi >= thresh)
	qst(size, compar, i, max);
      max = j;
    }
  } while (lo >= thresh);
}

/*
 * 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...
 *
 * @implemented
 */
void
__cdecl
qsort(void *base0, size_t n, size_t size, int (__cdecl *compar)(const void*, const void*))
{
  char *base = (char *)base0;
  char c, *i, *j, *lo, *hi;
  char *min, *max;
  size_t thresh;

  if (n <= 1)
    return;

  if (size == 0)
    return;
  thresh = size * THRESH;
  max = base + n * size;
  if (n >= THRESH)
  {
    qst(size, compar, 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 += size) < hi; )
    if (compar(j, lo) > 0)
      j = lo;
  if (j != base)
  {
    /* swap j into place */
    for (i = base, hi = base + size; 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 += size) < max; )
  {
    while (compar(hi -= size, min) > 0)
      /* void */;
    if ((hi += size) != min) {
      for (lo = min + size; --lo >= min; )
      {
	c = *lo;
	for (i = j = lo; (j -= size) >= hi; i = j)
	  *i = *j;
	*i = c;
      }
    }
  }
}