u-tools/apps/tar/gnu/rawmemchr.c

   1 /* -*- buffer-read-only: t -*- vi: set ro: */
   2 /* DO NOT EDIT! GENERATED AUTOMATICALLY! */
   3 /* Searching in a string.
   4    Copyright (C) 2008-2011 Free Software Foundation, Inc.
   5
   6    This program is free software: you can redistribute it and/or modify
   7    it under the terms of the GNU General Public License as published by
   8    the Free Software Foundation; either version 3 of the License, or
   9    (at your option) any later version.
  10
  11    This program is distributed in the hope that it will be useful,
  12    but WITHOUT ANY WARRANTY; without even the implied warranty of
  13    MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
  14    GNU General Public License for more details.
  15
  16    You should have received a copy of the GNU General Public License
  17    along with this program.  If not, see <http://www.gnu.org/licenses/>.  */
  18
  19 #include <config.h>
  20
  21 /* Specification.  */
  22 #include <string.h>
  23
  24 /* Find the first occurrence of C in S.  */
  25 void *
  26 rawmemchr (const void *s, int c_in)
  27 {
  28   /* On 32-bit hardware, choosing longword to be a 32-bit unsigned
  29      long instead of a 64-bit uintmax_t tends to give better
  30      performance.  On 64-bit hardware, unsigned long is generally 64
  31      bits already.  Change this typedef to experiment with
  32      performance.  */
  33   typedef unsigned long int longword;
  34
  35   const unsigned char *char_ptr;
  36   const longword *longword_ptr;
  37   longword repeated_one;
  38   longword repeated_c;
  39   unsigned char c;
  40
  41   c = (unsigned char) c_in;
  42
  43   /* Handle the first few bytes by reading one byte at a time.
  44      Do this until CHAR_PTR is aligned on a longword boundary.  */
  45   for (char_ptr = (const unsigned char *) s;
  46        (size_t) char_ptr % sizeof (longword) != 0;
  47        ++char_ptr)
  48     if (*char_ptr == c)
  49       return (void *) char_ptr;
  50
  51   longword_ptr = (const longword *) char_ptr;
  52
  53   /* All these elucidatory comments refer to 4-byte longwords,
  54      but the theory applies equally well to any size longwords.  */
  55
  56   /* Compute auxiliary longword values:
  57      repeated_one is a value which has a 1 in every byte.
  58      repeated_c has c in every byte.  */
  59   repeated_one = 0x01010101;
  60   repeated_c = c | (c << 8);
  61   repeated_c |= repeated_c << 16;
  62   if (0xffffffffU < (longword) -1)
  63     {
  64       repeated_one |= repeated_one << 31 << 1;
  65       repeated_c |= repeated_c << 31 << 1;
  66       if (8 < sizeof (longword))
  67         {
  68           size_t i;
  69
  70           for (i = 64; i < sizeof (longword) * 8; i *= 2)
  71             {
  72               repeated_one |= repeated_one << i;
  73               repeated_c |= repeated_c << i;
  74             }
  75         }
  76     }
  77
  78   /* Instead of the traditional loop which tests each byte, we will
  79      test a longword at a time.  The tricky part is testing if *any of
  80      the four* bytes in the longword in question are equal to NUL or
  81      c.  We first use an xor with repeated_c.  This reduces the task
  82      to testing whether *any of the four* bytes in longword1 is zero.
  83
  84      We compute tmp =
  85        ((longword1 - repeated_one) & ~longword1) & (repeated_one << 7).
  86      That is, we perform the following operations:
  87        1. Subtract repeated_one.
  88        2. & ~longword1.
  89        3. & a mask consisting of 0x80 in every byte.
  90      Consider what happens in each byte:
  91        - If a byte of longword1 is zero, step 1 and 2 transform it into 0xff,
  92          and step 3 transforms it into 0x80.  A carry can also be propagated
  93          to more significant bytes.
  94        - If a byte of longword1 is nonzero, let its lowest 1 bit be at
  95          position k (0 <= k <= 7); so the lowest k bits are 0.  After step 1,
  96          the byte ends in a single bit of value 0 and k bits of value 1.
  97          After step 2, the result is just k bits of value 1: 2^k - 1.  After
  98          step 3, the result is 0.  And no carry is produced.
  99      So, if longword1 has only non-zero bytes, tmp is zero.
 100      Whereas if longword1 has a zero byte, call j the position of the least
 101      significant zero byte.  Then the result has a zero at positions 0, ...,
 102      j-1 and a 0x80 at position j.  We cannot predict the result at the more
 103      significant bytes (positions j+1..3), but it does not matter since we
 104      already have a non-zero bit at position 8*j+7.
 105
 106      The test whether any byte in longword1 is zero is equivalent
 107      to testing whether tmp is nonzero.
 108
 109      This test can read beyond the end of a string, depending on where
 110      C_IN is encountered.  However, this is considered safe since the
 111      initialization phase ensured that the read will be aligned,
 112      therefore, the read will not cross page boundaries and will not
 113      cause a fault.  */
 114
 115   while (1)
 116     {
 117       longword longword1 = *longword_ptr ^ repeated_c;
 118
 119       if ((((longword1 - repeated_one) & ~longword1)
 120            & (repeated_one << 7)) != 0)
 121         break;
 122       longword_ptr++;
 123     }
 124
 125   char_ptr = (const unsigned char *) longword_ptr;
 126
 127   /* At this point, we know that one of the sizeof (longword) bytes
 128      starting at char_ptr is == c.  On little-endian machines, we
 129      could determine the first such byte without any further memory
 130      accesses, just by looking at the tmp result from the last loop
 131      iteration.  But this does not work on big-endian machines.
 132      Choose code that works in both cases.  */
 133
 134   char_ptr = (unsigned char *) longword_ptr;
 135   while (*char_ptr != c)
 136     char_ptr++;
 137   return (void *) char_ptr;
 138 }