lib/memchr2.c

   1 /* Copyright (C) 1991, 1993, 1996-1997, 1999-2000, 2003-2004, 2006, 2008-2024
   2    Free Software Foundation, Inc.
   3
   4    Based on strlen implementation by Torbjorn Granlund (tege@sics.se),
   5    with help from Dan Sahlin (dan@sics.se) and
   6    commentary by Jim Blandy (jimb@ai.mit.edu);
   7    adaptation to memchr suggested by Dick Karpinski (dick@cca.ucsf.edu),
   8    and implemented in glibc by Roland McGrath (roland@ai.mit.edu).
   9    Extension to memchr2 implemented by Eric Blake (ebb9@byu.net).
  10
  11    This file is free software: you can redistribute it and/or modify
  12    it under the terms of the GNU Lesser General Public License as
  13    published by the Free Software Foundation; either version 2.1 of the
  14    License, or (at your option) any later version.
  15
  16    This file is distributed in the hope that it will be useful,
  17    but WITHOUT ANY WARRANTY; without even the implied warranty of
  18    MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
  19    GNU Lesser General Public License for more details.
  20
  21    You should have received a copy of the GNU Lesser General Public License
  22    along with this program.  If not, see <https://www.gnu.org/licenses/>.  */
  23
  24 #include <config.h>
  25
  26 #include "memchr2.h"
  27
  28 #include <limits.h>
  29 #include <stdint.h>
  30 #include <string.h>
  31
  32 /* Return the first address of either C1 or C2 (treated as unsigned
  33    char) that occurs within N bytes of the memory region S.  If
  34    neither byte appears, return NULL.  */
  35 void *
  36 memchr2 (void const *s, int c1_in, int c2_in, size_t n)
  37 {
  38   /* On 32-bit hardware, choosing longword to be a 32-bit unsigned
  39      long instead of a 64-bit uintmax_t tends to give better
  40      performance.  On 64-bit hardware, unsigned long is generally 64
  41      bits already.  Change this typedef to experiment with
  42      performance.  */
  43   typedef unsigned long int longword;
  44
  45   const unsigned char *char_ptr;
  46   void const *void_ptr;
  47   const longword *longword_ptr;
  48   longword repeated_one;
  49   longword repeated_c1;
  50   longword repeated_c2;
  51   unsigned char c1;
  52   unsigned char c2;
  53
  54   c1 = (unsigned char) c1_in;
  55   c2 = (unsigned char) c2_in;
  56
  57   if (c1 == c2)
  58     return memchr (s, c1, n);
  59
  60   /* Handle the first few bytes by reading one byte at a time.
  61      Do this until VOID_PTR is aligned on a longword boundary.  */
  62   for (void_ptr = s;
  63        n > 0 && (uintptr_t) void_ptr % sizeof (longword) != 0;
  64        --n)
  65     {
  66       char_ptr = void_ptr;
  67       if (*char_ptr == c1 || *char_ptr == c2)
  68         return (void *) void_ptr;
  69       void_ptr = char_ptr + 1;
  70     }
  71
  72   longword_ptr = void_ptr;
  73
  74   /* All these elucidatory comments refer to 4-byte longwords,
  75      but the theory applies equally well to any size longwords.  */
  76
  77   /* Compute auxiliary longword values:
  78      repeated_one is a value which has a 1 in every byte.
  79      repeated_c1 has c1 in every byte.
  80      repeated_c2 has c2 in every byte.  */
  81   repeated_one = 0x01010101;
  82   repeated_c1 = c1 | (c1 << 8);
  83   repeated_c2 = c2 | (c2 << 8);
  84   repeated_c1 |= repeated_c1 << 16;
  85   repeated_c2 |= repeated_c2 << 16;
  86   if (0xffffffffU < (longword) -1)
  87     {
  88       repeated_one |= repeated_one << 31 << 1;
  89       repeated_c1 |= repeated_c1 << 31 << 1;
  90       repeated_c2 |= repeated_c2 << 31 << 1;
  91       if (8 < sizeof (longword))
  92         {
  93           size_t i;
  94
  95           for (i = 64; i < sizeof (longword) * 8; i *= 2)
  96             {
  97               repeated_one |= repeated_one << i;
  98               repeated_c1 |= repeated_c1 << i;
  99               repeated_c2 |= repeated_c2 << i;
 100             }
 101         }
 102     }
 103
 104   /* Instead of the traditional loop which tests each byte, we will test a
 105      longword at a time.  The tricky part is testing if *any of the four*
 106      bytes in the longword in question are equal to c1 or c2.  We first use
 107      an xor with repeated_c1 and repeated_c2, respectively.  This reduces
 108      the task to testing whether *any of the four* bytes in longword1 or
 109      longword2 is zero.
 110
 111      Let's consider longword1.  We compute tmp1 =
 112        ((longword1 - repeated_one) & ~longword1) & (repeated_one << 7).
 113      That is, we perform the following operations:
 114        1. Subtract repeated_one.
 115        2. & ~longword1.
 116        3. & a mask consisting of 0x80 in every byte.
 117      Consider what happens in each byte:
 118        - If a byte of longword1 is zero, step 1 and 2 transform it into 0xff,
 119          and step 3 transforms it into 0x80.  A carry can also be propagated
 120          to more significant bytes.
 121        - If a byte of longword1 is nonzero, let its lowest 1 bit be at
 122          position k (0 <= k <= 7); so the lowest k bits are 0.  After step 1,
 123          the byte ends in a single bit of value 0 and k bits of value 1.
 124          After step 2, the result is just k bits of value 1: 2^k - 1.  After
 125          step 3, the result is 0.  And no carry is produced.
 126      So, if longword1 has only non-zero bytes, tmp1 is zero.
 127      Whereas if longword1 has a zero byte, call j the position of the least
 128      significant zero byte.  Then the result has a zero at positions 0, ...,
 129      j-1 and a 0x80 at position j.  We cannot predict the result at the more
 130      significant bytes (positions j+1..3), but it does not matter since we
 131      already have a non-zero bit at position 8*j+7.
 132
 133      Similarly, we compute tmp2 =
 134        ((longword2 - repeated_one) & ~longword2) & (repeated_one << 7).
 135
 136      The test whether any byte in longword1 or longword2 is zero is equivalent
 137      to testing whether tmp1 is nonzero or tmp2 is nonzero.  We can combine
 138      this into a single test, whether (tmp1 | tmp2) is nonzero.  */
 139
 140   while (n >= sizeof (longword))
 141     {
 142       longword longword1 = *longword_ptr ^ repeated_c1;
 143       longword longword2 = *longword_ptr ^ repeated_c2;
 144
 145       if (((((longword1 - repeated_one) & ~longword1)
 146             | ((longword2 - repeated_one) & ~longword2))
 147            & (repeated_one << 7)) != 0)
 148         break;
 149       longword_ptr++;
 150       n -= sizeof (longword);
 151     }
 152
 153   char_ptr = (const unsigned char *) longword_ptr;
 154
 155   /* At this point, we know that either n < sizeof (longword), or one of the
 156      sizeof (longword) bytes starting at char_ptr is == c1 or == c2.  On
 157      little-endian machines, we could determine the first such byte without
 158      any further memory accesses, just by looking at the (tmp1 | tmp2) result
 159      from the last loop iteration.  But this does not work on big-endian
 160      machines.  Choose code that works in both cases.  */
 161
 162   for (; n > 0; --n, ++char_ptr)
 163     {
 164       if (*char_ptr == c1 || *char_ptr == c2)
 165         return (void *) char_ptr;
 166     }
 167
 168   return NULL;
 169 }