/* ffs.c -- find the first set bit in a word.
   Copyright (C) 2011-2021 Free Software Foundation, Inc.

   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 3 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, see <https://www.gnu.org/licenses/>.  */

/* Written by Eric Blake.  */

#include <config.h>

/* Specification.  */
#include <strings.h>

#include <limits.h>

#if defined _MSC_VER && !(__clang_major__ >= 4)
# include <intrin.h>
#endif

int
ffs (int i)
{
#if __GNUC__ > 3 || (__GNUC__ == 3 && __GNUC_MINOR__ >= 4) || (__clang_major__ >= 4)
  return __builtin_ffs (i);
#elif defined _MSC_VER
  /* _BitScanForward
     <https://docs.microsoft.com/en-us/cpp/intrinsics/bitscanforward-bitscanforward64> */
  unsigned long bit;
  if (_BitScanForward (&bit, i))
    return bit + 1;
  else
    return 0;
#else
  /* <https://github.com/gibsjose/BitHacks>
     gives this deBruijn constant for a branch-less computation, although
     that table counted trailing zeros rather than bit position.  This
     requires 32-bit int, we fall back to a naive algorithm on the rare
     platforms where that assumption is not true.  */
  if (CHAR_BIT * sizeof i == 32)
    {
      static unsigned int table[] = {
        1, 2, 29, 3, 30, 15, 25, 4, 31, 23, 21, 16, 26, 18, 5, 9,
        32, 28, 14, 24, 22, 20, 17, 8, 27, 13, 19, 7, 12, 6, 11, 10
      };
      unsigned int u = i;
      unsigned int bit = u & -u;
      return table[(bit * 0x077cb531U) >> 27] - !i;
    }
  else
    {
      unsigned int j;
      for (j = 0; j < CHAR_BIT * sizeof i; j++)
        if (i & (1U << j))
          return j + 1;
      return 0;
    }
#endif
}