Add a new function to compute the leftmost bit of a uint32_t

[disorder] / lib / bits.c

diff --git a/lib/bits.c b/lib/bits.c
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+/*
+ * This file is part of DisOrder
+ * Copyright (C) 2008 Richard Kettlewell
+ *
+ * This program is free software; you can redistribute it and/or modify
+ * it under the terms of the GNU General Public License as published by
+ * the Free Software Foundation; either version 2 of the License, or
+ * (at your option) any later version.
+ *
+ * This program is distributed in the hope that it will be useful, but
+ * WITHOUT ANY WARRANTY; without even the implied warranty of
+ * MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the GNU
+ * General Public License for more details.
+ *
+ * You should have received a copy of the GNU General Public License
+ * along with this program; if not, write to the Free Software
+ * Foundation, Inc., 59 Temple Place, Suite 330, Boston, MA 02111-1307
+ * USA
+ */
+
+/** @file lib/bits.c
+ * @brief Bit operations
+ */
+
+#include <config.h>
+#include "types.h"
+
+#include <math.h>
+
+#include "bits.h"
+
+/** @brief Compute index of leftmost 1 bit
+ * @param n Integer
+ * @return Index of leftmost 1 bit or -1
+ *
+ * For positive @p n we return the index of the leftmost bit of @p n.  For
+ * instance @c leftmost_bit(1) returns 0, @c leftmost_bit(15) returns 3, etc.
+ *
+ * If @p n is zero then -1 is returned.
+ */
+int leftmost_bit(uint32_t n) {
+  /* See e.g. Hacker's Delight s5-3 (p81) for where the idea comes from.
+   * Warren is computing the number of leading zeroes, but that's not quite
+   * what I wanted.  Also this version should be more portable than his, which
+   * inspects the bytes of the floating point number directly.
+   */
+  int x;
+  frexp((double)n, &x);
+  /* This gives: n = m * 2^x, where 0.5 <= m < 1 and x is an integer.
+   *
+   * If we take log2 of either side then we have:
+   *    log2(n) = x + log2 m
+   *
+   * We know that 0.5 <= m < 1 => -1 <= log2 m < 0.  So we floor either side:
+   *
+   *    floor(log2(n)) = x - 1
+   *
+   * What is floor(log2(n))?  Well, consider that:
+   *
+   *    2^k <= z < 2^(k+1)  =>  floor(log2(z)) = k.
+   *
+   * But 2^k <= z < 2^(k+1) is the same as saying that the leftmost bit of z is
+   * bit k.
+   *
+   *
+   * Warren adds 0.5 first, to deal with the case when n=0.  However frexp()
+   * guarantees to return x=0 when n=0, so we get the right answer without that
+   * step.
+   */
+  return x - 1;
+}
+
+
+/*
+Local Variables:
+c-basic-offset:2
+comment-column:40
+fill-column:79
+indent-tabs-mode:nil
+End:
+*/