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f9d42b20 RK |
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 | ||
25 | #include <config.h> | |
26 | #include "types.h" | |
27 | ||
28 | #include <math.h> | |
29 | ||
30 | #include "bits.h" | |
31 | ||
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 | */ | |
41 | int 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 | } | |
72 | ||
73 | ||
74 | /* | |
75 | Local Variables: | |
76 | c-basic-offset:2 | |
77 | comment-column:40 | |
78 | fill-column:79 | |
79 | indent-tabs-mode:nil | |
80 | End: | |
81 | */ |