memchr2.c 6.3 KB
Newer Older
Paul Eggert's avatar
Paul Eggert committed
1
/* Copyright (C) 1991, 1993, 1996-1997, 1999-2000, 2003-2004, 2006, 2008-2016
2
   Free Software Foundation, Inc.
Eric Blake's avatar
Eric Blake committed
3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34

   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.  */
35
void *
Eric Blake's avatar
Eric Blake committed
36 37
memchr2 (void const *s, int c1_in, int c2_in, size_t n)
{
38 39 40 41 42
  /* 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.  */
Eric Blake's avatar
Eric Blake committed
43
  typedef unsigned long int longword;
44

Eric Blake's avatar
Eric Blake committed
45
  const unsigned char *char_ptr;
46
  void const *void_ptr;
47
  const longword *longword_ptr;
48 49 50
  longword repeated_one;
  longword repeated_c1;
  longword repeated_c2;
Eric Blake's avatar
Eric Blake committed
51 52 53 54 55 56 57 58 59
  unsigned char c1;
  unsigned char c2;

  c1 = (unsigned char) c1_in;
  c2 = (unsigned char) c2_in;

  if (c1 == c2)
    return memchr (s, c1, n);

60
  /* Handle the first few bytes by reading one byte at a time.
61 62 63 64 65 66 67 68 69 70 71 72
     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;
73

Eric Blake's avatar
Eric Blake committed
74 75 76
  /* All these elucidatory comments refer to 4-byte longwords,
     but the theory applies equally well to any size longwords.  */

77 78 79 80 81 82 83 84 85
  /* 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;
Eric Blake's avatar
Eric Blake committed
86
  if (0xffffffffU < (longword) -1)
87
    {
88 89 90 91
      repeated_one |= repeated_one << 31 << 1;
      repeated_c1 |= repeated_c1 << 31 << 1;
      repeated_c2 |= repeated_c2 << 31 << 1;
      if (8 < sizeof (longword))
92 93 94 95 96 97 98 99 100 101
        {
          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;
            }
        }
102
    }
Eric Blake's avatar
Eric Blake committed
103

104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132
  /* 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.

Paul Eggert's avatar
Paul Eggert committed
133
     Similarly, we compute tmp2 =
134 135 136 137 138 139 140
       ((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))
Eric Blake's avatar
Eric Blake committed
141
    {
142 143
      longword longword1 = *longword_ptr ^ repeated_c1;
      longword longword2 = *longword_ptr ^ repeated_c2;
Eric Blake's avatar
Eric Blake committed
144

145
      if (((((longword1 - repeated_one) & ~longword1)
146 147 148
            | ((longword2 - repeated_one) & ~longword2))
           & (repeated_one << 7)) != 0)
        break;
149
      longword_ptr++;
150
      n -= sizeof (longword);
Eric Blake's avatar
Eric Blake committed
151 152 153 154
    }

  char_ptr = (const unsigned char *) longword_ptr;

155 156 157 158 159 160 161 162
  /* 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)
Eric Blake's avatar
Eric Blake committed
163 164
    {
      if (*char_ptr == c1 || *char_ptr == c2)
165
        return (void *) char_ptr;
Eric Blake's avatar
Eric Blake committed
166 167
    }

168
  return NULL;
Eric Blake's avatar
Eric Blake committed
169
}