|
421
|
1 /* Searching in a string.
|
|
|
2 Copyright (C) 2003, 2007, 2008, 2009, 2010 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. */
|
|
|
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 }
|
