1
/* Copyright (C) 1991, 1993, 1996-1997, 1999-2000, 2003-2004, 2006, 2008-2010
2
Free Software Foundation, Inc.
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).
10
NOTE: The canonical source of this file is maintained with the GNU C Library.
11
Bugs can be reported to bug-glibc@prep.ai.mit.edu.
13
This program is free software: you can redistribute it and/or modify it
14
under the terms of the GNU General Public License as published by the
15
Free Software Foundation; either version 3 of the License, or any
18
This program is distributed in the hope that it will be useful,
19
but WITHOUT ANY WARRANTY; without even the implied warranty of
20
MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
21
GNU General Public License for more details.
23
You should have received a copy of the GNU General Public License
24
along with this program. If not, see <http://www.gnu.org/licenses/>. */
37
# define reg_char char
42
#if HAVE_BP_SYM_H || defined _LIBC
45
# define BP_SYM(sym) sym
54
# define __memchr memchr
57
/* Search no more than N bytes of S for C. */
59
__memchr (void const *s, int c_in, size_t n)
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
66
typedef unsigned long int longword;
68
const unsigned char *char_ptr;
69
const longword *longword_ptr;
70
longword repeated_one;
74
c = (unsigned char) c_in;
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;
82
return (void *) char_ptr;
84
longword_ptr = (const longword *) char_ptr;
86
/* All these elucidatory comments refer to 4-byte longwords,
87
but the theory applies equally well to any size longwords. */
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)
97
repeated_one |= repeated_one << 31 << 1;
98
repeated_c |= repeated_c << 31 << 1;
99
if (8 < sizeof (longword))
103
for (i = 64; i < sizeof (longword) * 8; i *= 2)
105
repeated_one |= repeated_one << i;
106
repeated_c |= repeated_c << i;
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.
118
((longword1 - repeated_one) & ~longword1) & (repeated_one << 7).
119
That is, we perform the following operations:
120
1. Subtract repeated_one.
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.
139
So, the test whether any byte in longword1 is zero is equivalent to
140
testing whether tmp is nonzero. */
142
while (n >= sizeof (longword))
144
longword longword1 = *longword_ptr ^ repeated_c;
146
if ((((longword1 - repeated_one) & ~longword1)
147
& (repeated_one << 7)) != 0)
150
n -= sizeof (longword);
153
char_ptr = (const unsigned char *) longword_ptr;
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. */
162
for (; n > 0; --n, ++char_ptr)
165
return (void *) char_ptr;
171
weak_alias (__memchr, BP_SYM (memchr))
added
removed
lib/memchr.c
