source: grub-pc/trunk/fuentes/grub-core/gnulib/rawmemchr.c @ 22

Last change on this file since 22 was 22, checked in by mabarracus, 4 years ago

updated version and apply net.ifnames=0 into debian/rules

File size: 5.1 KB
Line 
1/* Searching in a string.
2   Copyright (C) 2008-2013 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.  */
23void *
24rawmemchr (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}
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