root/lib/rawmemchr.c

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DEFINITIONS

This source file includes following definitions.
  1. rawmemchr

     1 /* Searching in a string.
     2    Copyright (C) 2008-2023 Free Software Foundation, Inc.
     3 
     4    This file is free software: you can redistribute it and/or modify
     5    it under the terms of the GNU Lesser General Public License as
     6    published by the Free Software Foundation; either version 2.1 of the
     7    License, or (at your option) any later version.
     8 
     9    This file 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 Lesser General Public License for more details.
    13 
    14    You should have received a copy of the GNU Lesser 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 /* A function definition is only needed if HAVE_RAWMEMCHR is not defined.  */
    23 #if !HAVE_RAWMEMCHR
    24 
    25 # include <limits.h>
    26 # include <stdint.h>
    27 
    28 
    29 /* Find the first occurrence of C in S.  */
    30 void *
    31 rawmemchr (const void *s, int c_in)
    32 {
    33   /* Change this typedef to experiment with performance.  */
    34   typedef uintptr_t longword;
    35   /* If you change the "uintptr_t", you should change UINTPTR_WIDTH to match.
    36      This verifies that the type does not have padding bits.  */
    37   static_assert (UINTPTR_WIDTH == UCHAR_WIDTH * sizeof (longword));
    38 
    39   const unsigned char *char_ptr;
    40   unsigned char c = c_in;
    41 
    42   /* Handle the first few bytes by reading one byte at a time.
    43      Do this until CHAR_PTR is aligned on a longword boundary.  */
    44   for (char_ptr = (const unsigned char *) s;
    45        (uintptr_t) char_ptr % alignof (longword) != 0;
    46        ++char_ptr)
    47     if (*char_ptr == c)
    48       return (void *) char_ptr;
    49 
    50   longword const *longword_ptr = s = char_ptr;
    51 
    52   /* Compute auxiliary longword values:
    53      repeated_one is a value which has a 1 in every byte.
    54      repeated_c has c in every byte.  */
    55   longword repeated_one = (longword) -1 / UCHAR_MAX;
    56   longword repeated_c = repeated_one * c;
    57   longword repeated_hibit = repeated_one * (UCHAR_MAX / 2 + 1);
    58 
    59   /* Instead of the traditional loop which tests each byte, we will
    60      test a longword at a time.  The tricky part is testing if any of
    61      the bytes in the longword in question are equal to
    62      c.  We first use an xor with repeated_c.  This reduces the task
    63      to testing whether any of the bytes in longword1 is zero.
    64 
    65      (The following comments assume 8-bit bytes, as POSIX requires;
    66      the code's use of UCHAR_MAX should work even if bytes have more
    67      than 8 bits.)
    68 
    69      We compute tmp =
    70        ((longword1 - repeated_one) & ~longword1) & (repeated_one * 0x80).
    71      That is, we perform the following operations:
    72        1. Subtract repeated_one.
    73        2. & ~longword1.
    74        3. & a mask consisting of 0x80 in every byte.
    75      Consider what happens in each byte:
    76        - If a byte of longword1 is zero, step 1 and 2 transform it into 0xff,
    77          and step 3 transforms it into 0x80.  A carry can also be propagated
    78          to more significant bytes.
    79        - If a byte of longword1 is nonzero, let its lowest 1 bit be at
    80          position k (0 <= k <= 7); so the lowest k bits are 0.  After step 1,
    81          the byte ends in a single bit of value 0 and k bits of value 1.
    82          After step 2, the result is just k bits of value 1: 2^k - 1.  After
    83          step 3, the result is 0.  And no carry is produced.
    84      So, if longword1 has only non-zero bytes, tmp is zero.
    85      Whereas if longword1 has a zero byte, call j the position of the least
    86      significant zero byte.  Then the result has a zero at positions 0, ...,
    87      j-1 and a 0x80 at position j.  We cannot predict the result at the more
    88      significant bytes (positions j+1..3), but it does not matter since we
    89      already have a non-zero bit at position 8*j+7.
    90 
    91      The test whether any byte in longword1 is zero is equivalent
    92      to testing whether tmp is nonzero.
    93 
    94      This test can read beyond the end of a string, depending on where
    95      C_IN is encountered.  However, this is considered safe since the
    96      initialization phase ensured that the read will be aligned,
    97      therefore, the read will not cross page boundaries and will not
    98      cause a fault.  */
    99 
   100   while (1)
   101     {
   102       longword longword1 = *longword_ptr ^ repeated_c;
   103 
   104       if ((((longword1 - repeated_one) & ~longword1) & repeated_hibit) != 0)
   105         break;
   106       longword_ptr++;
   107     }
   108 
   109   char_ptr = s = longword_ptr;
   110 
   111   /* At this point, we know that one of the sizeof (longword) bytes
   112      starting at char_ptr is == c.  If we knew endianness, we
   113      could determine the first such byte without any further memory
   114      accesses, just by looking at the tmp result from the last loop
   115      iteration.  However, the following simple and portable code does
   116      not attempt this potential optimization.  */
   117 
   118   while (*char_ptr != c)
   119     char_ptr++;
   120   return (void *) char_ptr;
   121 }
   122 
   123 #endif

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