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1 | /* |
2 | * This file is part of DisOrder | |
3 | * Copyright (C) 2008 Richard Kettlewell | |
4 | * | |
e7eb3a27 | 5 | * This program is free software: you can redistribute it and/or modify |
f9d42b20 | 6 | * it under the terms of the GNU General Public License as published by |
e7eb3a27 | 7 | * the Free Software Foundation, either version 3 of the License, or |
f9d42b20 | 8 | * (at your option) any later version. |
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9 | * |
10 | * This program is distributed in the hope that it will be useful, | |
11 | * but WITHOUT ANY WARRANTY; without even the implied warranty of | |
12 | * MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the | |
13 | * GNU General Public License for more details. | |
14 | * | |
f9d42b20 | 15 | * You should have received a copy of the GNU General Public License |
e7eb3a27 | 16 | * along with this program. If not, see <http://www.gnu.org/licenses/>. |
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17 | */ |
18 | ||
19 | /** @file lib/bits.c | |
20 | * @brief Bit operations | |
21 | */ | |
22 | ||
05b75f8d | 23 | #include "common.h" |
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24 | |
25 | #include <math.h> | |
26 | ||
27 | #include "bits.h" | |
28 | ||
7c8c2fc9 | 29 | #if !HAVE_FLS |
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30 | /** @brief Compute index of leftmost 1 bit |
31 | * @param n Integer | |
32 | * @return Index of leftmost 1 bit or -1 | |
33 | * | |
34 | * For positive @p n we return the index of the leftmost bit of @p n. For | |
35 | * instance @c leftmost_bit(1) returns 0, @c leftmost_bit(15) returns 3, etc. | |
36 | * | |
37 | * If @p n is zero then -1 is returned. | |
38 | */ | |
39 | int leftmost_bit(uint32_t n) { | |
40 | /* See e.g. Hacker's Delight s5-3 (p81) for where the idea comes from. | |
41 | * Warren is computing the number of leading zeroes, but that's not quite | |
42 | * what I wanted. Also this version should be more portable than his, which | |
43 | * inspects the bytes of the floating point number directly. | |
44 | */ | |
45 | int x; | |
46 | frexp((double)n, &x); | |
47 | /* This gives: n = m * 2^x, where 0.5 <= m < 1 and x is an integer. | |
48 | * | |
49 | * If we take log2 of either side then we have: | |
50 | * log2(n) = x + log2 m | |
51 | * | |
52 | * We know that 0.5 <= m < 1 => -1 <= log2 m < 0. So we floor either side: | |
53 | * | |
54 | * floor(log2(n)) = x - 1 | |
55 | * | |
56 | * What is floor(log2(n))? Well, consider that: | |
57 | * | |
58 | * 2^k <= z < 2^(k+1) => floor(log2(z)) = k. | |
59 | * | |
60 | * But 2^k <= z < 2^(k+1) is the same as saying that the leftmost bit of z is | |
61 | * bit k. | |
62 | * | |
63 | * | |
64 | * Warren adds 0.5 first, to deal with the case when n=0. However frexp() | |
65 | * guarantees to return x=0 when n=0, so we get the right answer without that | |
66 | * step. | |
67 | */ | |
68 | return x - 1; | |
69 | } | |
7c8c2fc9 | 70 | #endif |
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71 | |
72 | /* | |
73 | Local Variables: | |
74 | c-basic-offset:2 | |
75 | comment-column:40 | |
76 | fill-column:79 | |
77 | indent-tabs-mode:nil | |
78 | End: | |
79 | */ |