1 /* Copyright (C) 1991-2015 Free Software Foundation, Inc.
2 This file is part of the GNU C Library.
3 Based on strlen implementation by Torbjorn Granlund (tege@sics.se),
4 with help from Dan Sahlin (dan@sics.se) and
5 commentary by Jim Blandy (jimb@ai.mit.edu);
6 adaptation to memchr suggested by Dick Karpinski (dick@cca.ucsf.edu),
7 and implemented by Roland McGrath (roland@ai.mit.edu).
9 The GNU C Library is free software; you can redistribute it and/or
10 modify it under the terms of the GNU Lesser General Public
11 License as published by the Free Software Foundation; either
12 version 2.1 of the License, or (at your option) any later version.
14 The GNU C Library is distributed in the hope that it will be useful,
15 but WITHOUT ANY WARRANTY; without even the implied warranty of
16 MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU
17 Lesser General Public License for more details.
19 You should have received a copy of the GNU Lesser General Public
20 License along with the GNU C Library; if not, see
21 <http://www.gnu.org/licenses/>. */
28 #define __ptr_t void *
36 #if defined (HAVE_LIMITS_H) || defined (_LIBC)
40 #define LONG_MAX_32_BITS 2147483647
43 #define LONG_MAX LONG_MAX_32_BITS
46 #include <sys/types.h>
51 # define RAWMEMCHR __rawmemchr
54 /* Find the first occurrence of C in S. */
56 RAWMEMCHR (const __ptr_t s
, int c_in
)
58 const unsigned char *char_ptr
;
59 const unsigned long int *longword_ptr
;
60 unsigned long int longword
, magic_bits
, charmask
;
63 c
= (unsigned char) c_in
;
65 /* Handle the first few characters by reading one character at a time.
66 Do this until CHAR_PTR is aligned on a longword boundary. */
67 for (char_ptr
= (const unsigned char *) s
;
68 ((unsigned long int) char_ptr
& (sizeof (longword
) - 1)) != 0;
71 return (__ptr_t
) char_ptr
;
73 /* All these elucidatory comments refer to 4-byte longwords,
74 but the theory applies equally well to 8-byte longwords. */
76 longword_ptr
= (unsigned long int *) char_ptr
;
78 /* Bits 31, 24, 16, and 8 of this number are zero. Call these bits
79 the "holes." Note that there is a hole just to the left of
80 each byte, with an extra at the end:
82 bits: 01111110 11111110 11111110 11111111
83 bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD
85 The 1-bits make sure that carries propagate to the next 0-bit.
86 The 0-bits provide holes for carries to fall into. */
88 magic_bits
= magic_bits
/ 0xff * 0xfe << 1 >> 1 | 1;
90 /* Set up a longword, each of whose bytes is C. */
91 charmask
= c
| (c
<< 8);
92 charmask
|= charmask
<< 16;
93 #if LONG_MAX > LONG_MAX_32_BITS
94 charmask
|= charmask
<< 32;
97 /* Instead of the traditional loop which tests each character,
98 we will test a longword at a time. The tricky part is testing
99 if *any of the four* bytes in the longword in question are zero. */
102 /* We tentatively exit the loop if adding MAGIC_BITS to
103 LONGWORD fails to change any of the hole bits of LONGWORD.
105 1) Is this safe? Will it catch all the zero bytes?
106 Suppose there is a byte with all zeros. Any carry bits
107 propagating from its left will fall into the hole at its
108 least significant bit and stop. Since there will be no
109 carry from its most significant bit, the LSB of the
110 byte to the left will be unchanged, and the zero will be
113 2) Is this worthwhile? Will it ignore everything except
114 zero bytes? Suppose every byte of LONGWORD has a bit set
115 somewhere. There will be a carry into bit 8. If bit 8
116 is set, this will carry into bit 16. If bit 8 is clear,
117 one of bits 9-15 must be set, so there will be a carry
118 into bit 16. Similarly, there will be a carry into bit
119 24. If one of bits 24-30 is set, there will be a carry
120 into bit 31, so all of the hole bits will be changed.
122 The one misfire occurs when bits 24-30 are clear and bit
123 31 is set; in this case, the hole at bit 31 is not
124 changed. If we had access to the processor carry flag,
125 we could close this loophole by putting the fourth hole
128 So it ignores everything except 128's, when they're aligned
131 3) But wait! Aren't we looking for C, not zero?
132 Good point. So what we do is XOR LONGWORD with a longword,
133 each of whose bytes is C. This turns each byte that is C
136 longword
= *longword_ptr
++ ^ charmask
;
138 /* Add MAGIC_BITS to LONGWORD. */
139 if ((((longword
+ magic_bits
)
141 /* Set those bits that were unchanged by the addition. */
144 /* Look at only the hole bits. If any of the hole bits
145 are unchanged, most likely one of the bytes was a
149 /* Which of the bytes was C? If none of them were, it was
150 a misfire; continue the search. */
152 const unsigned char *cp
= (const unsigned char *) (longword_ptr
- 1);
157 return (__ptr_t
) &cp
[1];
159 return (__ptr_t
) &cp
[2];
161 return (__ptr_t
) &cp
[3];
162 #if LONG_MAX > 2147483647
164 return (__ptr_t
) &cp
[4];
166 return (__ptr_t
) &cp
[5];
168 return (__ptr_t
) &cp
[6];
170 return (__ptr_t
) &cp
[7];
175 libc_hidden_def (__rawmemchr
)
176 weak_alias (__rawmemchr
, rawmemchr
)