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/* memrchr -- find the last occurrence of a byte in a memory block

   Copyright (C) 1991, 1993, 1996, 1997, 1999, 2000, 2003 Free
   Software Foundation, Inc.

   Based on strlen implementation by Torbjorn Granlund (tege@sics.se),
   with help from Dan Sahlin (dan@sics.se) and
   commentary by Jim Blandy (jimb@ai.mit.edu);
   adaptation to memchr suggested by Dick Karpinski (dick@cca.ucsf.edu),
   and implemented by Roland McGrath (roland@ai.mit.edu).

   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, 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
   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.  */

# include <config.h>

#include <stdlib.h>
#include <string.h>

#if defined (_LIBC)
# include <memcopy.h>
# define reg_char char

#include <limits.h>

#define LONG_MAX_32_BITS 2147483647

#include <sys/types.h>

#undef __memrchr
#undef memrchr

#ifndef weak_alias
# define __memrchr memrchr

/* Search no more than N bytes of S for C.  */
void *
__memrchr (void const *s, int c_in, size_t n)
  const unsigned char *char_ptr;
  const unsigned long int *longword_ptr;
  unsigned long int longword, magic_bits, charmask;
  unsigned reg_char c;

  c = (unsigned char) c_in;

  /* Handle the last few characters by reading one character at a time.
     Do this until CHAR_PTR is aligned on a longword boundary.  */
  for (char_ptr = (const unsigned char *) s + n;
       n > 0 && ((unsigned long int) char_ptr
             & (sizeof (longword) - 1)) != 0;
    if (*--char_ptr == c)
      return (void *) char_ptr;

  /* All these elucidatory comments refer to 4-byte longwords,
     but the theory applies equally well to 8-byte longwords.  */

  longword_ptr = (unsigned long int *) char_ptr;

  /* Bits 31, 24, 16, and 8 of this number are zero.  Call these bits
     the "holes."  Note that there is a hole just to the left of
     each byte, with an extra at the end:

     bits:  01111110 11111110 11111110 11111111

     The 1-bits make sure that carries propagate to the next 0-bit.
     The 0-bits provide holes for carries to fall into.  */

  if (sizeof (longword) != 4 && sizeof (longword) != 8)
    abort ();

  magic_bits = 0x7efefeff;
  magic_bits = ((unsigned long int) 0x7efefefe << 32) | 0xfefefeff;

  /* Set up a longword, each of whose bytes is C.  */
  charmask = c | (c << 8);
  charmask |= charmask << 16;
  charmask |= charmask << 32;

  /* Instead of the traditional loop which tests each character,
     we will test a longword at a time.  The tricky part is testing
     if *any of the four* bytes in the longword in question are zero.  */
  while (n >= sizeof (longword))
      /* We tentatively exit the loop if adding MAGIC_BITS to
       LONGWORD fails to change any of the hole bits of LONGWORD.

       1) Is this safe?  Will it catch all the zero bytes?
       Suppose there is a byte with all zeros.  Any carry bits
       propagating from its left will fall into the hole at its
       least significant bit and stop.  Since there will be no
       carry from its most significant bit, the LSB of the
       byte to the left will be unchanged, and the zero will be

       2) Is this worthwhile?  Will it ignore everything except
       zero bytes?  Suppose every byte of LONGWORD has a bit set
       somewhere.  There will be a carry into bit 8.  If bit 8
       is set, this will carry into bit 16.  If bit 8 is clear,
       one of bits 9-15 must be set, so there will be a carry
       into bit 16.  Similarly, there will be a carry into bit
       24.  If one of bits 24-30 is set, there will be a carry
       into bit 31, so all of the hole bits will be changed.

       The one misfire occurs when bits 24-30 are clear and bit
       31 is set; in this case, the hole at bit 31 is not
       changed.  If we had access to the processor carry flag,
       we could close this loophole by putting the fourth hole
       at bit 32!

       So it ignores everything except 128's, when they're aligned

       3) But wait!  Aren't we looking for C, not zero?
       Good point.  So what we do is XOR LONGWORD with a longword,
       each of whose bytes is C.  This turns each byte that is C
       into a zero.  */

      longword = *--longword_ptr ^ charmask;

      /* Add MAGIC_BITS to LONGWORD.  */
      if ((((longword + magic_bits)

          /* Set those bits that were unchanged by the addition.  */
          ^ ~longword)

         /* Look at only the hole bits.  If any of the hole bits
            are unchanged, most likely one of the bytes was a
            zero.  */
         & ~magic_bits) != 0)
        /* Which of the bytes was C?  If none of them were, it was
           a misfire; continue the search.  */

        const unsigned char *cp = (const unsigned char *) longword_ptr;

#if LONG_MAX > 2147483647
        if (cp[7] == c)
          return (void *) &cp[7];
        if (cp[6] == c)
          return (void *) &cp[6];
        if (cp[5] == c)
          return (void *) &cp[5];
        if (cp[4] == c)
          return (void *) &cp[4];
        if (cp[3] == c)
          return (void *) &cp[3];
        if (cp[2] == c)
          return (void *) &cp[2];
        if (cp[1] == c)
          return (void *) &cp[1];
        if (cp[0] == c)
          return (void *) cp;

      n -= sizeof (longword);

  char_ptr = (const unsigned char *) longword_ptr;

  while (n-- > 0)
      if (*--char_ptr == c)
      return (void *) char_ptr;

  return 0;
#ifdef weak_alias
weak_alias (__memrchr, memrchr)

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