1/* Searching in a string.
2   Copyright (C) 2003, 2007-2012 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 or the final NUL byte.  */
23char *
24strchrnul (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}
143