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Member "gnuastro-0.8/bootstrapped/lib/strchrnul.c" (28 Dec 2018, 5646 Bytes) of package /linux/privat/gnuastro-0.8.tar.lz:


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    1 /* Searching in a string.
    2    Copyright (C) 2003, 2007-2018 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 <https://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 or the final NUL byte.  */
   23 char *
   24 strchrnul (const char *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   if (!c)
   41     return rawmemchr (s, 0);
   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 || *char_ptr == c)
   49       return (char *) 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 or
   83      longword2 is zero.
   84 
   85      Let's consider longword1.  We compute tmp =
   86        ((longword1 - repeated_one) & ~longword1) & (repeated_one << 7).
   87      That is, we perform the following operations:
   88        1. Subtract repeated_one.
   89        2. & ~longword1.
   90        3. & a mask consisting of 0x80 in every byte.
   91      Consider what happens in each byte:
   92        - If a byte of longword1 is zero, step 1 and 2 transform it into 0xff,
   93          and step 3 transforms it into 0x80.  A carry can also be propagated
   94          to more significant bytes.
   95        - If a byte of longword1 is nonzero, let its lowest 1 bit be at
   96          position k (0 <= k <= 7); so the lowest k bits are 0.  After step 1,
   97          the byte ends in a single bit of value 0 and k bits of value 1.
   98          After step 2, the result is just k bits of value 1: 2^k - 1.  After
   99          step 3, the result is 0.  And no carry is produced.
  100      So, if longword1 has only non-zero bytes, tmp is zero.
  101      Whereas if longword1 has a zero byte, call j the position of the least
  102      significant zero byte.  Then the result has a zero at positions 0, ...,
  103      j-1 and a 0x80 at position j.  We cannot predict the result at the more
  104      significant bytes (positions j+1..3), but it does not matter since we
  105      already have a non-zero bit at position 8*j+7.
  106 
  107      The test whether any byte in longword1 or longword2 is zero is equivalent
  108      to testing whether tmp1 is nonzero or tmp2 is nonzero.  We can combine
  109      this into a single test, whether (tmp1 | tmp2) is nonzero.
  110 
  111      This test can read more than one byte beyond the end of a string,
  112      depending on where the terminating NUL is encountered.  However,
  113      this is considered safe since the initialization phase ensured
  114      that the read will be aligned, therefore, the read will not cross
  115      page boundaries and will not cause a fault.  */
  116 
  117   while (1)
  118     {
  119       longword longword1 = *longword_ptr ^ repeated_c;
  120       longword longword2 = *longword_ptr;
  121 
  122       if (((((longword1 - repeated_one) & ~longword1)
  123             | ((longword2 - repeated_one) & ~longword2))
  124            & (repeated_one << 7)) != 0)
  125         break;
  126       longword_ptr++;
  127     }
  128 
  129   char_ptr = (const unsigned char *) longword_ptr;
  130 
  131   /* At this point, we know that one of the sizeof (longword) bytes
  132      starting at char_ptr is == 0 or == c.  On little-endian machines,
  133      we could determine the first such byte without any further memory
  134      accesses, just by looking at the tmp result from the last loop
  135      iteration.  But this does not work on big-endian machines.
  136      Choose code that works in both cases.  */
  137 
  138   char_ptr = (unsigned char *) longword_ptr;
  139   while (*char_ptr && (*char_ptr != c))
  140     char_ptr++;
  141   return (char *) char_ptr;
  142 }