/* Copyright (C) 1991, 1993, 1996-1997, 1999-2000, 2003-2004, 2006, 2008-2017

   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 in glibc by Roland McGrath (roland@ai.mit.edu).

   Extension to memchr2 implemented by Eric Blake (ebb9@byu.net).

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 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 <http://www.gnu.org/licenses/>.  */

#include <config.h>

#include "memchr2.h"

#include <limits.h>

#include <stdint.h>

#include <string.h>

/* Return the first address of either C1 or C2 (treated as unsigned

   char) that occurs within N bytes of the memory region S.  If

   neither byte appears, return NULL.  */

void *

memchr2 (void const *s, int c1_in, int c2_in, size_t n)

{

  /* 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;

  void const *void_ptr;

  const longword *longword_ptr;

  longword repeated_one;

  longword repeated_c1;

  longword repeated_c2;

  unsigned char c1;

  unsigned char c2;

  c1 = (unsigned char) c1_in;

  c2 = (unsigned char) c2_in;

  if (c1 == c2)

    return memchr (s, c1, n);

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

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

  for (void_ptr = s;

       n > 0 && (uintptr_t) void_ptr % sizeof (longword) != 0;

       --n)

    {

      char_ptr = void_ptr;

      if (*char_ptr == c1 || *char_ptr == c2)

        return (void *) void_ptr;

      void_ptr = char_ptr + 1;

    }

  longword_ptr = void_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_c1 has c1 in every byte.

     repeated_c2 has c2 in every byte.  */

  repeated_one = 0x01010101;

  repeated_c1 = c1 | (c1 << 8);

  repeated_c2 = c2 | (c2 << 8);

  repeated_c1 |= repeated_c1 << 16;

  repeated_c2 |= repeated_c2 << 16;

  if (0xffffffffU < (longword) -1)

    {

      repeated_one |= repeated_one << 31 << 1;

      repeated_c1 |= repeated_c1 << 31 << 1;

      repeated_c2 |= repeated_c2 << 31 << 1;

      if (8 < sizeof (longword))

        {

          size_t i;

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

            {

              repeated_one |= repeated_one << i;

              repeated_c1 |= repeated_c1 << i;

              repeated_c2 |= repeated_c2 << 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 c1 or c2.  We first use

     an xor with repeated_c1 and repeated_c2, respectively.  This reduces

     the task to testing whether *any of the four* bytes in longword1 or

     longword2 is zero.

     Let's consider longword1.  We compute tmp1 =

       ((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, tmp1 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.

     Similarly, we compute tmp2 =

       ((longword2 - repeated_one) & ~longword2) & (repeated_one << 7).

     The test whether any byte in longword1 or longword2 is zero is equivalent

     to testing whether tmp1 is nonzero or tmp2 is nonzero.  We can combine

     this into a single test, whether (tmp1 | tmp2) is nonzero.  */

  while (n >= sizeof (longword))

    {

      longword longword1 = *longword_ptr ^ repeated_c1;

      longword longword2 = *longword_ptr ^ repeated_c2;

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

            | ((longword2 - repeated_one) & ~longword2))

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

        break;

      longword_ptr++;

      n -= sizeof (longword);

    }

  char_ptr = (const unsigned char *) longword_ptr;

  /* At this point, we know that either n < sizeof (longword), or one of the

     sizeof (longword) bytes starting at char_ptr is == c1 or == c2.  On

     little-endian machines, we could determine the first such byte without

     any further memory accesses, just by looking at the (tmp1 | tmp2) result

     from the last loop iteration.  But this does not work on big-endian

     machines.  Choose code that works in both cases.  */

  for (; n > 0; --n, ++char_ptr)

    {

      if (*char_ptr == c1 || *char_ptr == c2)

        return (void *) char_ptr;

    }

  return NULL;

}