1 /* Searching in a string.
2 Copyright (C) 2003, 2007, 2008 Free Software Foundation, Inc.
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.
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.
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/>. */
22 /* Find the first occurrence of C in S or the final NUL byte. */
24 strchrnul (const char *s, int c_in)
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
31 typedef unsigned long int longword;
33 const unsigned char *char_ptr;
34 const longword *longword_ptr;
35 longword repeated_one;
39 c = (unsigned char) c_in;
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;
46 if (!*char_ptr || *char_ptr == c)
47 return (char *) char_ptr;
49 longword_ptr = (const longword *) char_ptr;
51 /* All these elucidatory comments refer to 4-byte longwords,
52 but the theory applies equally well to any size longwords. */
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)
62 repeated_one |= repeated_one << 31 << 1;
63 repeated_c |= repeated_c << 31 << 1;
64 if (8 < sizeof (longword))
68 for (i = 64; i < sizeof (longword) * 8; i *= 2)
70 repeated_one |= repeated_one << i;
71 repeated_c |= repeated_c << i;
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 or
83 Let's consider longword1. We compute tmp =
84 ((longword1 - repeated_one) & ~longword1) & (repeated_one << 7).
85 That is, we perform the following operations:
86 1. Subtract repeated_one.
88 3. & a mask consisting of 0x80 in every byte.
89 Consider what happens in each byte:
90 - If a byte of longword1 is zero, step 1 and 2 transform it into 0xff,
91 and step 3 transforms it into 0x80. A carry can also be propagated
92 to more significant bytes.
93 - If a byte of longword1 is nonzero, let its lowest 1 bit be at
94 position k (0 <= k <= 7); so the lowest k bits are 0. After step 1,
95 the byte ends in a single bit of value 0 and k bits of value 1.
96 After step 2, the result is just k bits of value 1: 2^k - 1. After
97 step 3, the result is 0. And no carry is produced.
98 So, if longword1 has only non-zero bytes, tmp is zero.
99 Whereas if longword1 has a zero byte, call j the position of the least
100 significant zero byte. Then the result has a zero at positions 0, ...,
101 j-1 and a 0x80 at position j. We cannot predict the result at the more
102 significant bytes (positions j+1..3), but it does not matter since we
103 already have a non-zero bit at position 8*j+7.
105 The test whether any byte in longword1 or longword2 is zero is equivalent
106 to testing whether tmp1 is nonzero or tmp2 is nonzero. We can combine
107 this into a single test, whether (tmp1 | tmp2) is nonzero.
109 This test can read more than one byte beyond the end of a string,
110 depending on where the terminating NUL is encountered. However,
111 this is considered safe since the initialization phase ensured
112 that the read will be aligned, therefore, the read will not cross
113 page boundaries and will not cause a fault. */
117 longword longword1 = *longword_ptr ^ repeated_c;
118 longword longword2 = *longword_ptr;
120 if (((((longword1 - repeated_one) & ~longword1)
121 | ((longword2 - repeated_one) & ~longword2))
122 & (repeated_one << 7)) != 0)
127 char_ptr = (const unsigned char *) longword_ptr;
129 /* At this point, we know that one of the sizeof (longword) bytes
130 starting at char_ptr is == 0 or == c. On little-endian machines,
131 we could determine the first such byte without any further memory
132 accesses, just by looking at the tmp result from the last loop
133 iteration. But this does not work on big-endian machines.
134 Choose code that works in both cases. */
136 char_ptr = (unsigned char *) longword_ptr;
137 while (*char_ptr && (*char_ptr != c))
139 return (char *) char_ptr;