1 /* -*- buffer-read-only: t -*- vi: set ro: */
2 /* DO NOT EDIT! GENERATED AUTOMATICALLY! */
3 /* Copyright (C) 1991, 1993, 1996-1997, 1999-2000, 2003-2004, 2006, 2008-2011
4 Free Software Foundation, Inc.
6 Based on strlen implementation by Torbjorn Granlund (tege@sics.se),
7 with help from Dan Sahlin (dan@sics.se) and
8 commentary by Jim Blandy (jimb@ai.mit.edu);
9 adaptation to memchr suggested by Dick Karpinski (dick@cca.ucsf.edu),
10 and implemented by Roland McGrath (roland@ai.mit.edu).
12 NOTE: The canonical source of this file is maintained with the GNU C Library.
13 Bugs can be reported to bug-glibc@prep.ai.mit.edu.
15 This program is free software: you can redistribute it and/or modify it
16 under the terms of the GNU General Public License as published by the
17 Free Software Foundation; either version 3 of the License, or any
20 This program is distributed in the hope that it will be useful,
21 but WITHOUT ANY WARRANTY; without even the implied warranty of
22 MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
23 GNU General Public License for more details.
25 You should have received a copy of the GNU General Public License
26 along with this program. If not, see <http://www.gnu.org/licenses/>. */
39 # define reg_char char
44 #if HAVE_BP_SYM_H || defined _LIBC
47 # define BP_SYM(sym) sym
56 # define __memchr memchr
59 /* Search no more than N bytes of S for C. */
61 __memchr (void const *s
, int c_in
, size_t n
)
63 /* On 32-bit hardware, choosing longword to be a 32-bit unsigned
64 long instead of a 64-bit uintmax_t tends to give better
65 performance. On 64-bit hardware, unsigned long is generally 64
66 bits already. Change this typedef to experiment with
68 typedef unsigned long int longword
;
70 const unsigned char *char_ptr
;
71 const longword
*longword_ptr
;
72 longword repeated_one
;
76 c
= (unsigned char) c_in
;
78 /* Handle the first few bytes by reading one byte at a time.
79 Do this until CHAR_PTR is aligned on a longword boundary. */
80 for (char_ptr
= (const unsigned char *) s
;
81 n
> 0 && (size_t) char_ptr
% sizeof (longword
) != 0;
84 return (void *) char_ptr
;
86 longword_ptr
= (const longword
*) char_ptr
;
88 /* All these elucidatory comments refer to 4-byte longwords,
89 but the theory applies equally well to any size longwords. */
91 /* Compute auxiliary longword values:
92 repeated_one is a value which has a 1 in every byte.
93 repeated_c has c in every byte. */
94 repeated_one
= 0x01010101;
95 repeated_c
= c
| (c
<< 8);
96 repeated_c
|= repeated_c
<< 16;
97 if (0xffffffffU
< (longword
) -1)
99 repeated_one
|= repeated_one
<< 31 << 1;
100 repeated_c
|= repeated_c
<< 31 << 1;
101 if (8 < sizeof (longword
))
105 for (i
= 64; i
< sizeof (longword
) * 8; i
*= 2)
107 repeated_one
|= repeated_one
<< i
;
108 repeated_c
|= repeated_c
<< i
;
113 /* Instead of the traditional loop which tests each byte, we will test a
114 longword at a time. The tricky part is testing if *any of the four*
115 bytes in the longword in question are equal to c. We first use an xor
116 with repeated_c. This reduces the task to testing whether *any of the
117 four* bytes in longword1 is zero.
120 ((longword1 - repeated_one) & ~longword1) & (repeated_one << 7).
121 That is, we perform the following operations:
122 1. Subtract repeated_one.
124 3. & a mask consisting of 0x80 in every byte.
125 Consider what happens in each byte:
126 - If a byte of longword1 is zero, step 1 and 2 transform it into 0xff,
127 and step 3 transforms it into 0x80. A carry can also be propagated
128 to more significant bytes.
129 - If a byte of longword1 is nonzero, let its lowest 1 bit be at
130 position k (0 <= k <= 7); so the lowest k bits are 0. After step 1,
131 the byte ends in a single bit of value 0 and k bits of value 1.
132 After step 2, the result is just k bits of value 1: 2^k - 1. After
133 step 3, the result is 0. And no carry is produced.
134 So, if longword1 has only non-zero bytes, tmp is zero.
135 Whereas if longword1 has a zero byte, call j the position of the least
136 significant zero byte. Then the result has a zero at positions 0, ...,
137 j-1 and a 0x80 at position j. We cannot predict the result at the more
138 significant bytes (positions j+1..3), but it does not matter since we
139 already have a non-zero bit at position 8*j+7.
141 So, the test whether any byte in longword1 is zero is equivalent to
142 testing whether tmp is nonzero. */
144 while (n
>= sizeof (longword
))
146 longword longword1
= *longword_ptr
^ repeated_c
;
148 if ((((longword1
- repeated_one
) & ~longword1
)
149 & (repeated_one
<< 7)) != 0)
152 n
-= sizeof (longword
);
155 char_ptr
= (const unsigned char *) longword_ptr
;
157 /* At this point, we know that either n < sizeof (longword), or one of the
158 sizeof (longword) bytes starting at char_ptr is == c. On little-endian
159 machines, we could determine the first such byte without any further
160 memory accesses, just by looking at the tmp result from the last loop
161 iteration. But this does not work on big-endian machines. Choose code
162 that works in both cases. */
164 for (; n
> 0; --n
, ++char_ptr
)
167 return (void *) char_ptr
;
173 weak_alias (__memchr
, BP_SYM (memchr
))