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 3 of the License, or
8 * (at your option) any later version.
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.
15 * You should have received a copy of the GNU General Public License
16 * along with this program. If not, see <http://www.gnu.org/licenses/>.
20 * @brief Bit operations
30 /** @brief Compute index of leftmost 1 bit
32 * @return Index of leftmost 1 bit or -1
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.
37 * If @p n is zero then -1 is returned.
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.
47 /* This gives: n = m * 2^x, where 0.5 <= m < 1 and x is an integer.
49 * If we take log2 of either side then we have:
50 * log2(n) = x + log2 m
52 * We know that 0.5 <= m < 1 => -1 <= log2 m < 0. So we floor either side:
54 * floor(log2(n)) = x - 1
56 * What is floor(log2(n))? Well, consider that:
58 * 2^k <= z < 2^(k+1) => floor(log2(z)) = k.
60 * But 2^k <= z < 2^(k+1) is the same as saying that the leftmost bit of z is
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