1/* Copyright (C) 1991, 1993, 1996-1997, 1999-2000, 2003-2004, 2006, 2008-2012
2 Free Software Foundation, Inc.
3
4 Based on strlen implementation by Torbjorn Granlund (tege@sics.se),
5 with help from Dan Sahlin (dan@sics.se) and
6 commentary by Jim Blandy (jimb@ai.mit.edu);
7 adaptation to memchr suggested by Dick Karpinski (dick@cca.ucsf.edu),
8 and implemented by Roland McGrath (roland@ai.mit.edu).
9
10NOTE: The canonical source of this file is maintained with the GNU C Library.
11Bugs can be reported to bug-glibc@prep.ai.mit.edu.
12
13This program is free software: you can redistribute it and/or modify it
14under the terms of the GNU Lesser General Public License as published by the
15Free Software Foundation; either version 2.1 of the License, or any
16later version.
17
18This program is distributed in the hope that it will be useful,
19but WITHOUT ANY WARRANTY; without even the implied warranty of
20MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
21GNU Lesser General Public License for more details.
22
23You should have received a copy of the GNU Lesser General Public License
24along with this program. If not, see <http://www.gnu.org/licenses/>. */
25
26#ifndef _LIBC
27# include <config.h>
28#endif
29
30#include <string.h>
31
32#include <stddef.h>
33
34#if defined _LIBC
35# include <memcopy.h>
36#else
37# define reg_char char
38#endif
39
40#include <limits.h>
41
42#if HAVE_BP_SYM_H || defined _LIBC
43# include <bp-sym.h>
44#else
45# define BP_SYM(sym) sym
46#endif
47
48#undef __memchr
49#ifdef _LIBC
50# undef memchr
51#endif
52
53#ifndef weak_alias
54# define __memchr memchr
55#endif
56
57/* Search no more than N bytes of S for C. */
58void *
59__memchr (void const *s, int c_in, size_t n)
60{
61 /* On 32-bit hardware, choosing longword to be a 32-bit unsigned
62 long instead of a 64-bit uintmax_t tends to give better
63 performance. On 64-bit hardware, unsigned long is generally 64
64 bits already. Change this typedef to experiment with
65 performance. */
66 typedef unsigned long int longword;
67
68 const unsigned char *char_ptr;
69 const longword *longword_ptr;
70 longword repeated_one;
71 longword repeated_c;
72 unsigned reg_char c;
73
74 c = (unsigned char) c_in;
75
76 /* Handle the first few bytes by reading one byte at a time.
77 Do this until CHAR_PTR is aligned on a longword boundary. */
78 for (char_ptr = (const unsigned char *) s;
79 n > 0 && (size_t) char_ptr % sizeof (longword) != 0;
80 --n, ++char_ptr)
81 if (*char_ptr == c)
82 return (void *) char_ptr;
83
84 longword_ptr = (const longword *) char_ptr;
85
86 /* All these elucidatory comments refer to 4-byte longwords,
87 but the theory applies equally well to any size longwords. */
88
89 /* Compute auxiliary longword values:
90 repeated_one is a value which has a 1 in every byte.
91 repeated_c has c in every byte. */
92 repeated_one = 0x01010101;
93 repeated_c = c | (c << 8);
94 repeated_c |= repeated_c << 16;
95 if (0xffffffffU < (longword) -1)
96 {
97 repeated_one |= repeated_one << 31 << 1;
98 repeated_c |= repeated_c << 31 << 1;
99 if (8 < sizeof (longword))
100 {
101 size_t i;
102
103 for (i = 64; i < sizeof (longword) * 8; i *= 2)
104 {
105 repeated_one |= repeated_one << i;
106 repeated_c |= repeated_c << i;
107 }
108 }
109 }
110
111 /* Instead of the traditional loop which tests each byte, we will test a
112 longword at a time. The tricky part is testing if *any of the four*
113 bytes in the longword in question are equal to c. We first use an xor
114 with repeated_c. This reduces the task to testing whether *any of the
115 four* bytes in longword1 is zero.
116
117 We compute tmp =
118 ((longword1 - repeated_one) & ~longword1) & (repeated_one << 7).
119 That is, we perform the following operations:
120 1. Subtract repeated_one.
121 2. & ~longword1.
122 3. & a mask consisting of 0x80 in every byte.
123 Consider what happens in each byte:
124 - If a byte of longword1 is zero, step 1 and 2 transform it into 0xff,
125 and step 3 transforms it into 0x80. A carry can also be propagated
126 to more significant bytes.
127 - If a byte of longword1 is nonzero, let its lowest 1 bit be at
128 position k (0 <= k <= 7); so the lowest k bits are 0. After step 1,
129 the byte ends in a single bit of value 0 and k bits of value 1.
130 After step 2, the result is just k bits of value 1: 2^k - 1. After
131 step 3, the result is 0. And no carry is produced.
132 So, if longword1 has only non-zero bytes, tmp is zero.
133 Whereas if longword1 has a zero byte, call j the position of the least
134 significant zero byte. Then the result has a zero at positions 0, ...,
135 j-1 and a 0x80 at position j. We cannot predict the result at the more
136 significant bytes (positions j+1..3), but it does not matter since we
137 already have a non-zero bit at position 8*j+7.
138
139 So, the test whether any byte in longword1 is zero is equivalent to
140 testing whether tmp is nonzero. */
141
142 while (n >= sizeof (longword))
143 {
144 longword longword1 = *longword_ptr ^ repeated_c;
145
146 if ((((longword1 - repeated_one) & ~longword1)
147 & (repeated_one << 7)) != 0)
148 break;
149 longword_ptr++;
150 n -= sizeof (longword);
151 }
152
153 char_ptr = (const unsigned char *) longword_ptr;
154
155 /* At this point, we know that either n < sizeof (longword), or one of the
156 sizeof (longword) bytes starting at char_ptr is == c. On little-endian
157 machines, we could determine the first such byte without any further
158 memory accesses, just by looking at the tmp result from the last loop
159 iteration. But this does not work on big-endian machines. Choose code
160 that works in both cases. */
161
162 for (; n > 0; --n, ++char_ptr)
163 {
164 if (*char_ptr == c)
165 return (void *) char_ptr;
166 }
167
168 return NULL;
169}
170#ifdef weak_alias
171weak_alias (__memchr, BP_SYM (memchr))
172#endif
173