2 * This file is part of DisOrder
3 * Copyright (C) 2008 Richard Kettlewell
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
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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.
15 * You should have received a copy of the GNU General Public License
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17 * Foundation, Inc., 59 Temple Place, Suite 330, Boston, MA 02111-1307
22 * @brief Bit operations
32 /** @brief Compute index of leftmost 1 bit
34 * @return Index of leftmost 1 bit or -1
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.
39 * If @p n is zero then -1 is returned.
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.
49 /* This gives: n = m * 2^x, where 0.5 <= m < 1 and x is an integer.
51 * If we take log2 of either side then we have:
52 * log2(n) = x + log2 m
54 * We know that 0.5 <= m < 1 => -1 <= log2 m < 0. So we floor either side:
56 * floor(log2(n)) = x - 1
58 * What is floor(log2(n))? Well, consider that:
60 * 2^k <= z < 2^(k+1) => floor(log2(z)) = k.
62 * But 2^k <= z < 2^(k+1) is the same as saying that the leftmost bit of z is
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