chiark / gitweb /
Fiddle with CSS+HTML in effort to get more consistent buttons
[disorder] / lib / bits.c
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1/*
2 * This file is part of DisOrder
3 * Copyright (C) 2008 Richard Kettlewell
4 *
5 * This program is free software; you can redistribute it and/or modify
6 * it under the terms of the GNU General Public License as published by
7 * the Free Software Foundation; either version 2 of the License, or
8 * (at your option) any later version.
9 *
10 * This program is distributed in the hope that it will be useful, but
11 * WITHOUT ANY WARRANTY; without even the implied warranty of
12 * MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU
13 * General Public License for more details.
14 *
15 * You should have received a copy of the GNU General Public License
16 * along with this program; if not, write to the Free Software
17 * Foundation, Inc., 59 Temple Place, Suite 330, Boston, MA 02111-1307
18 * USA
19 */
20
21/** @file lib/bits.c
22 * @brief Bit operations
23 */
24
05b75f8d 25#include "common.h"
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26
27#include <math.h>
28
29#include "bits.h"
30
7c8c2fc9 31#if !HAVE_FLS
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32/** @brief Compute index of leftmost 1 bit
33 * @param n Integer
34 * @return Index of leftmost 1 bit or -1
35 *
36 * For positive @p n we return the index of the leftmost bit of @p n. For
37 * instance @c leftmost_bit(1) returns 0, @c leftmost_bit(15) returns 3, etc.
38 *
39 * If @p n is zero then -1 is returned.
40 */
41int leftmost_bit(uint32_t n) {
42 /* See e.g. Hacker's Delight s5-3 (p81) for where the idea comes from.
43 * Warren is computing the number of leading zeroes, but that's not quite
44 * what I wanted. Also this version should be more portable than his, which
45 * inspects the bytes of the floating point number directly.
46 */
47 int x;
48 frexp((double)n, &x);
49 /* This gives: n = m * 2^x, where 0.5 <= m < 1 and x is an integer.
50 *
51 * If we take log2 of either side then we have:
52 * log2(n) = x + log2 m
53 *
54 * We know that 0.5 <= m < 1 => -1 <= log2 m < 0. So we floor either side:
55 *
56 * floor(log2(n)) = x - 1
57 *
58 * What is floor(log2(n))? Well, consider that:
59 *
60 * 2^k <= z < 2^(k+1) => floor(log2(z)) = k.
61 *
62 * But 2^k <= z < 2^(k+1) is the same as saying that the leftmost bit of z is
63 * bit k.
64 *
65 *
66 * Warren adds 0.5 first, to deal with the case when n=0. However frexp()
67 * guarantees to return x=0 when n=0, so we get the right answer without that
68 * step.
69 */
70 return x - 1;
71}
7c8c2fc9 72#endif
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73
74/*
75Local Variables:
76c-basic-offset:2
77comment-column:40
78fill-column:79
79indent-tabs-mode:nil
80End:
81*/