gl/rawmemchr.c

   1 /* Searching in a string.
   2    Copyright (C) 2008-2012 Free Software Foundation, Inc.
   3
   4    This program is free software: you can redistribute it and/or modify
   5    it under the terms of the GNU General Public License as published by
   6    the Free Software Foundation; either version 3 of the License, or
   7    (at your option) any later version.
   8
   9    This program is distributed in the hope that it will be useful,
  10    but WITHOUT ANY WARRANTY; without even the implied warranty of
  11    MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
  12    GNU General Public License for more details.
  13
  14    You should have received a copy of the GNU General Public License
  15    along with this program.  If not, see <http://www.gnu.org/licenses/>.  */
  16
  17 #include <config.h>
  18
  19 /* Specification.  */
  20 #include <string.h>
  21
  22 /* Find the first occurrence of C in S.  */
  23 void *
  24 rawmemchr (const void *s, int c_in)
  25 {
  26   /* On 32-bit hardware, choosing longword to be a 32-bit unsigned
  27      long instead of a 64-bit uintmax_t tends to give better
  28      performance.  On 64-bit hardware, unsigned long is generally 64
  29      bits already.  Change this typedef to experiment with
  30      performance.  */
  31   typedef unsigned long int longword;
  32
  33   const unsigned char *char_ptr;
  34   const longword *longword_ptr;
  35   longword repeated_one;
  36   longword repeated_c;
  37   unsigned char c;
  38
  39   c = (unsigned char) c_in;
  40
  41   /* Handle the first few bytes by reading one byte at a time.
  42      Do this until CHAR_PTR is aligned on a longword boundary.  */
  43   for (char_ptr = (const unsigned char *) s;
  44        (size_t) char_ptr % sizeof (longword) != 0;
  45        ++char_ptr)
  46     if (*char_ptr == c)
  47       return (void *) char_ptr;
  48
  49   longword_ptr = (const longword *) char_ptr;
  50
  51   /* All these elucidatory comments refer to 4-byte longwords,
  52      but the theory applies equally well to any size longwords.  */
  53
  54   /* Compute auxiliary longword values:
  55      repeated_one is a value which has a 1 in every byte.
  56      repeated_c has c in every byte.  */
  57   repeated_one = 0x01010101;
  58   repeated_c = c | (c << 8);
  59   repeated_c |= repeated_c << 16;
  60   if (0xffffffffU < (longword) -1)
  61     {
  62       repeated_one |= repeated_one << 31 << 1;
  63       repeated_c |= repeated_c << 31 << 1;
  64       if (8 < sizeof (longword))
  65         {
  66           size_t i;
  67
  68           for (i = 64; i < sizeof (longword) * 8; i *= 2)
  69             {
  70               repeated_one |= repeated_one << i;
  71               repeated_c |= repeated_c << i;
  72             }
  73         }
  74     }
  75
  76   /* Instead of the traditional loop which tests each byte, we will
  77      test a longword at a time.  The tricky part is testing if *any of
  78      the four* bytes in the longword in question are equal to NUL or
  79      c.  We first use an xor with repeated_c.  This reduces the task
  80      to testing whether *any of the four* bytes in longword1 is zero.
  81
  82      We compute tmp =
  83        ((longword1 - repeated_one) & ~longword1) & (repeated_one << 7).
  84      That is, we perform the following operations:
  85        1. Subtract repeated_one.
  86        2. & ~longword1.
  87        3. & a mask consisting of 0x80 in every byte.
  88      Consider what happens in each byte:
  89        - If a byte of longword1 is zero, step 1 and 2 transform it into 0xff,
  90          and step 3 transforms it into 0x80.  A carry can also be propagated
  91          to more significant bytes.
  92        - If a byte of longword1 is nonzero, let its lowest 1 bit be at
  93          position k (0 <= k <= 7); so the lowest k bits are 0.  After step 1,
  94          the byte ends in a single bit of value 0 and k bits of value 1.
  95          After step 2, the result is just k bits of value 1: 2^k - 1.  After
  96          step 3, the result is 0.  And no carry is produced.
  97      So, if longword1 has only non-zero bytes, tmp is zero.
  98      Whereas if longword1 has a zero byte, call j the position of the least
  99      significant zero byte.  Then the result has a zero at positions 0, ...,
 100      j-1 and a 0x80 at position j.  We cannot predict the result at the more
 101      significant bytes (positions j+1..3), but it does not matter since we
 102      already have a non-zero bit at position 8*j+7.
 103
 104      The test whether any byte in longword1 is zero is equivalent
 105      to testing whether tmp is nonzero.
 106
 107      This test can read beyond the end of a string, depending on where
 108      C_IN is encountered.  However, this is considered safe since the
 109      initialization phase ensured that the read will be aligned,
 110      therefore, the read will not cross page boundaries and will not
 111      cause a fault.  */
 112
 113   while (1)
 114     {
 115       longword longword1 = *longword_ptr ^ repeated_c;
 116
 117       if ((((longword1 - repeated_one) & ~longword1)
 118            & (repeated_one << 7)) != 0)
 119         break;
 120       longword_ptr++;
 121     }
 122
 123   char_ptr = (const unsigned char *) longword_ptr;
 124
 125   /* At this point, we know that one of the sizeof (longword) bytes
 126      starting at char_ptr is == c.  On little-endian machines, we
 127      could determine the first such byte without any further memory
 128      accesses, just by looking at the tmp result from the last loop
 129      iteration.  But this does not work on big-endian machines.
 130      Choose code that works in both cases.  */
 131
 132   char_ptr = (unsigned char *) longword_ptr;
 133   while (*char_ptr != c)
 134     char_ptr++;
 135   return (void *) char_ptr;
 136 }