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244 lines
6.5 KiB
C
244 lines
6.5 KiB
C
/* Copyright (C) 1994 DJ Delorie, see COPYING.DJ for details */
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#include <stdlib.h>
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#include <search.h>
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/*-
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* Copyright (c) 1980, 1983 The Regents of the University of California.
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* All rights reserved.
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*
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* Redistribution and use in source and binary forms are permitted
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* provided that: (1) source distributions retain this entire copyright
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* notice and comment, and (2) distributions including binaries display
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* the following acknowledgement: ``This product includes software
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* developed by the University of California, Berkeley and its contributors''
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* in the documentation or other materials provided with the distribution
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* and in all advertising materials mentioning features or use of this
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* software. Neither the name of the University nor the names of its
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* contributors may be used to endorse or promote products derived
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* from this software without specific prior written permission.
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* THIS SOFTWARE IS PROVIDED ``AS IS'' AND WITHOUT ANY EXPRESS OR
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* IMPLIED WARRANTIES, INCLUDING, WITHOUT LIMITATION, THE IMPLIED
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* WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE.
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*/
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/*
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* qsort.c:
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* Our own version of the system qsort routine which is faster by an average
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* of 25%, with lows and highs of 10% and 50%.
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* The THRESHold below is the insertion sort threshold, and has been adjusted
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* for records of size 48 bytes.
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* The MTHREShold is where we stop finding a better median.
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*/
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#define THRESH 4 /* threshold for insertion */
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#define MTHRESH 6 /* threshold for median */
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/*
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* qst:
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* Do a quicksort
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* First, find the median element, and put that one in the first place as the
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* discriminator. (This "median" is just the median of the first, last and
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* middle elements). (Using this median instead of the first element is a big
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* win). Then, the usual partitioning/swapping, followed by moving the
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* discriminator into the right place. Then, figure out the sizes of the two
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* partions, do the smaller one recursively and the larger one via a repeat of
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* this code. Stopping when there are less than THRESH elements in a partition
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* and cleaning up with an insertion sort (in our caller) is a huge win.
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* All data swaps are done in-line, which is space-losing but time-saving.
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* (And there are only three places where this is done).
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*/
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static void __cdecl
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qst(size_t size, int (__cdecl *compar)(const void*, const void*), char *base, char *max)
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{
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char c, *i, *j, *jj;
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size_t ii;
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char *mid, *tmp;
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size_t lo, hi;
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size_t thresh;
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size_t mthresh;
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thresh = size * THRESH;
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mthresh = size * MTHRESH;
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/*
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* At the top here, lo is the number of characters of elements in the
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* current partition. (Which should be max - base).
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* Find the median of the first, last, and middle element and make
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* that the middle element. Set j to largest of first and middle.
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* If max is larger than that guy, then it's that guy, else compare
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* max with loser of first and take larger. Things are set up to
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* prefer the middle, then the first in case of ties.
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*/
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lo = max - base; /* number of elements as chars */
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do {
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mid = i = base + size * ((lo / size) >> 1);
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if (lo >= mthresh)
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{
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j = (compar((jj = base), i) > 0 ? jj : i);
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if (compar(j, (tmp = max - size)) > 0)
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{
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/* switch to first loser */
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j = (j == jj ? i : jj);
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if (compar(j, tmp) < 0)
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j = tmp;
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}
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if (j != i)
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{
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ii = size;
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do {
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c = *i;
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*i++ = *j;
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*j++ = c;
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} while (--ii);
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}
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}
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/*
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* Semi-standard quicksort partitioning/swapping
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*/
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for (i = base, j = max - size; ; )
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{
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while (i < mid && compar(i, mid) <= 0)
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i += size;
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while (j > mid)
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{
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if (compar(mid, j) <= 0)
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{
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j -= size;
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continue;
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}
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tmp = i + size; /* value of i after swap */
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if (i == mid)
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{
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/* j <-> mid, new mid is j */
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mid = jj = j;
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}
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else
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{
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/* i <-> j */
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jj = j;
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j -= size;
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}
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goto swap;
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}
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if (i == mid)
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{
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break;
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}
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else
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{
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/* i <-> mid, new mid is i */
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jj = mid;
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tmp = mid = i; /* value of i after swap */
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j -= size;
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}
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swap:
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ii = size;
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do {
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c = *i;
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*i++ = *jj;
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*jj++ = c;
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} while (--ii);
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i = tmp;
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}
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/*
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* Look at sizes of the two partitions, do the smaller
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* one first by recursion, then do the larger one by
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* making sure lo is its size, base and max are update
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* correctly, and branching back. But only repeat
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* (recursively or by branching) if the partition is
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* of at least size THRESH.
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*/
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i = (j = mid) + size;
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if ((lo = j - base) <= (hi = max - i))
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{
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if (lo >= thresh)
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qst(size, compar, base, j);
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base = i;
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lo = hi;
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}
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else
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{
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if (hi >= thresh)
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qst(size, compar, i, max);
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max = j;
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}
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} while (lo >= thresh);
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}
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/*
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* qsort:
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* First, set up some global parameters for qst to share. Then, quicksort
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* with qst(), and then a cleanup insertion sort ourselves. Sound simple?
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* It's not...
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*
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* @implemented
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*/
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void
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__cdecl
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qsort(void *base0, size_t n, size_t size, int (__cdecl *compar)(const void*, const void*))
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{
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char *base = (char *)base0;
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char c, *i, *j, *lo, *hi;
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char *min, *max;
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size_t thresh;
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if (n <= 1)
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return;
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if (size == 0)
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return;
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thresh = size * THRESH;
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max = base + n * size;
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if (n >= THRESH)
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{
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qst(size, compar, base, max);
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hi = base + thresh;
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}
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else
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{
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hi = max;
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}
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/*
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* First put smallest element, which must be in the first THRESH, in
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* the first position as a sentinel. This is done just by searching
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* the first THRESH elements (or the first n if n < THRESH), finding
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* the min, and swapping it into the first position.
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*/
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for (j = lo = base; (lo += size) < hi; )
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if (compar(j, lo) > 0)
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j = lo;
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if (j != base)
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{
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/* swap j into place */
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for (i = base, hi = base + size; i < hi; )
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{
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c = *j;
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*j++ = *i;
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*i++ = c;
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}
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}
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/*
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* With our sentinel in place, we now run the following hyper-fast
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* insertion sort. For each remaining element, min, from [1] to [n-1],
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* set hi to the index of the element AFTER which this one goes.
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* Then, do the standard insertion sort shift on a character at a time
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* basis for each element in the frob.
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*/
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for (min = base; (hi = min += size) < max; )
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{
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while (compar(hi -= size, min) > 0)
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/* void */;
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if ((hi += size) != min) {
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for (lo = min + size; --lo >= min; )
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{
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c = *lo;
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for (i = j = lo; (j -= size) >= hi; i = j)
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*i = *j;
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*i = c;
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}
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}
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}
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}
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