/* Searching in a string.

   Copyright (C) 2008-2019 Free Software Foundation, Inc.

   This program is free software: you can redistribute it and/or modify

   it under the terms of the GNU Lesser General Public License as published by

   the Free Software Foundation; either version 2.1 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 Lesser General Public License for more details.

   You should have received a copy of the GNU Lesser General Public License

   along with this program.  If not, see <https://www.gnu.org/licenses/>.  */

#include <config.h>

/* Specification.  */

#include <string.h>

/* Find the first occurrence of C in S.  */

void *

rawmemchr (const void *s, int c_in)

{

  /* On 32-bit hardware, choosing longword to be a 32-bit unsigned

     long instead of a 64-bit uintmax_t tends to give better

     performance.  On 64-bit hardware, unsigned long is generally 64

     bits already.  Change this typedef to experiment with

     performance.  */

  typedef unsigned long int longword;

  const unsigned char *char_ptr;

  const longword *longword_ptr;

  longword repeated_one;

  longword repeated_c;

  unsigned char c;

  c = (unsigned char) c_in;

  /* Handle the first few bytes by reading one byte at a time.

     Do this until CHAR_PTR is aligned on a longword boundary.  */

  for (char_ptr = (const unsigned char *) s;

       (size_t) char_ptr % sizeof (longword) != 0;

       ++char_ptr)

    if (*char_ptr == c)

      return (void *) char_ptr;

  longword_ptr = (const longword *) char_ptr;

  /* All these elucidatory comments refer to 4-byte longwords,

     but the theory applies equally well to any size longwords.  */

  /* Compute auxiliary longword values:

     repeated_one is a value which has a 1 in every byte.

     repeated_c has c in every byte.  */

  repeated_one = 0x01010101;

  repeated_c = c | (c << 8);

  repeated_c |= repeated_c << 16;

  if (0xffffffffU < (longword) -1)

    {

      repeated_one |= repeated_one << 31 << 1;

      repeated_c |= repeated_c << 31 << 1;

      if (8 < sizeof (longword))

        {

          size_t i;

          for (i = 64; i < sizeof (longword) * 8; i *= 2)

            {

              repeated_one |= repeated_one << i;

              repeated_c |= repeated_c << i;

            }

        }

    }

  /* Instead of the traditional loop which tests each byte, 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 equal to NUL or

     c.  We first use an xor with repeated_c.  This reduces the task

     to testing whether *any of the four* bytes in longword1 is zero.

     We compute tmp =

       ((longword1 - repeated_one) & ~longword1) & (repeated_one << 7).

     That is, we perform the following operations:

       1. Subtract repeated_one.

       2. & ~longword1.

       3. & a mask consisting of 0x80 in every byte.

     Consider what happens in each byte:

       - If a byte of longword1 is zero, step 1 and 2 transform it into 0xff,

         and step 3 transforms it into 0x80.  A carry can also be propagated

         to more significant bytes.

       - If a byte of longword1 is nonzero, let its lowest 1 bit be at

         position k (0 <= k <= 7); so the lowest k bits are 0.  After step 1,

         the byte ends in a single bit of value 0 and k bits of value 1.

         After step 2, the result is just k bits of value 1: 2^k - 1.  After

         step 3, the result is 0.  And no carry is produced.

     So, if longword1 has only non-zero bytes, tmp is zero.

     Whereas if longword1 has a zero byte, call j the position of the least

     significant zero byte.  Then the result has a zero at positions 0, ...,

     j-1 and a 0x80 at position j.  We cannot predict the result at the more

     significant bytes (positions j+1..3), but it does not matter since we

     already have a non-zero bit at position 8*j+7.

     The test whether any byte in longword1 is zero is equivalent

     to testing whether tmp is nonzero.

     This test can read beyond the end of a string, depending on where

     C_IN is encountered.  However, this is considered safe since the

     initialization phase ensured that the read will be aligned,

     therefore, the read will not cross page boundaries and will not

     cause a fault.  */

  while (1)

    {

      longword longword1 = *longword_ptr ^ repeated_c;

      if ((((longword1 - repeated_one) & ~longword1)

           & (repeated_one << 7)) != 0)

        break;

      longword_ptr++;

    }

  char_ptr = (const unsigned char *) longword_ptr;

  /* At this point, we know that one of the sizeof (longword) bytes

     starting at char_ptr is == c.  On little-endian machines, we

     could determine the first such byte without any further memory

     accesses, just by looking at the tmp result from the last loop

     iteration.  But this does not work on big-endian machines.

     Choose code that works in both cases.  */

  char_ptr = (unsigned char *) longword_ptr;

  while (*char_ptr != c)

    char_ptr++;

  return (void *) char_ptr;

}