1 /* Copyright (C) 1991, 1993, 1996, 1997, 1999, 2000, 2003, 2004, 2006 Free
2 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 2, or (at your option) 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, write to the Free Software Foundation,
25 Inc., 51 Franklin Street, Fifth Floor, Boston, MA 02110-1301, USA. */
38 # define reg_char char
43 #if HAVE_BP_SYM_H || defined _LIBC
46 # define BP_SYM(sym) sym
52 /* Search no more than N bytes of S for C. */
54 __memchr (void const *s
, int c_in
, size_t n
)
56 const unsigned char *char_ptr
;
57 const unsigned long int *longword_ptr
;
58 unsigned long int longword
, magic_bits
, charmask
;
62 c
= (unsigned char) c_in
;
64 /* Handle the first few characters by reading one character at a time.
65 Do this until CHAR_PTR is aligned on a longword boundary. */
66 for (char_ptr
= (const unsigned char *) s
;
67 n
> 0 && (size_t) char_ptr
% sizeof longword
!= 0;
70 return (void *) char_ptr
;
72 /* All these elucidatory comments refer to 4-byte longwords,
73 but the theory applies equally well to any size longwords. */
75 longword_ptr
= (const unsigned long int *) char_ptr
;
77 /* Bits 31, 24, 16, and 8 of this number are zero. Call these bits
78 the "holes." Note that there is a hole just to the left of
79 each byte, with an extra at the end:
81 bits: 01111110 11111110 11111110 11111111
82 bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD
84 The 1-bits make sure that carries propagate to the next 0-bit.
85 The 0-bits provide holes for carries to fall into. */
87 /* Set MAGIC_BITS to be this pattern of 1 and 0 bits.
88 Set CHARMASK to be a longword, each of whose bytes is C. */
90 magic_bits
= 0xfefefefe;
91 charmask
= c
| (c
<< 8);
92 charmask
|= charmask
<< 16;
93 #if 0xffffffffU < ULONG_MAX
94 magic_bits
|= magic_bits
<< 32;
95 charmask
|= charmask
<< 32;
96 if (8 < sizeof longword
)
97 for (i
= 64; i
< sizeof longword
* 8; i
*= 2)
99 magic_bits
|= magic_bits
<< i
;
100 charmask
|= charmask
<< i
;
103 magic_bits
= (ULONG_MAX
>> 1) & (magic_bits
| 1);
105 /* Instead of the traditional loop which tests each character,
106 we will test a longword at a time. The tricky part is testing
107 if *any of the four* bytes in the longword in question are zero. */
108 while (n
>= sizeof longword
)
110 /* We tentatively exit the loop if adding MAGIC_BITS to
111 LONGWORD fails to change any of the hole bits of LONGWORD.
113 1) Is this safe? Will it catch all the zero bytes?
114 Suppose there is a byte with all zeros. Any carry bits
115 propagating from its left will fall into the hole at its
116 least significant bit and stop. Since there will be no
117 carry from its most significant bit, the LSB of the
118 byte to the left will be unchanged, and the zero will be
121 2) Is this worthwhile? Will it ignore everything except
122 zero bytes? Suppose every byte of LONGWORD has a bit set
123 somewhere. There will be a carry into bit 8. If bit 8
124 is set, this will carry into bit 16. If bit 8 is clear,
125 one of bits 9-15 must be set, so there will be a carry
126 into bit 16. Similarly, there will be a carry into bit
127 24. If one of bits 24-30 is set, there will be a carry
128 into bit 31, so all of the hole bits will be changed.
130 The one misfire occurs when bits 24-30 are clear and bit
131 31 is set; in this case, the hole at bit 31 is not
132 changed. If we had access to the processor carry flag,
133 we could close this loophole by putting the fourth hole
136 So it ignores everything except 128's, when they're aligned
139 3) But wait! Aren't we looking for C, not zero?
140 Good point. So what we do is XOR LONGWORD with a longword,
141 each of whose bytes is C. This turns each byte that is C
144 longword
= *longword_ptr
++ ^ charmask
;
146 /* Add MAGIC_BITS to LONGWORD. */
147 if ((((longword
+ magic_bits
)
149 /* Set those bits that were unchanged by the addition. */
152 /* Look at only the hole bits. If any of the hole bits
153 are unchanged, most likely one of the bytes was a
157 /* Which of the bytes was C? If none of them were, it was
158 a misfire; continue the search. */
160 const unsigned char *cp
= (const unsigned char *) (longword_ptr
- 1);
165 return (void *) &cp
[1];
167 return (void *) &cp
[2];
169 return (void *) &cp
[3];
170 if (4 < sizeof longword
&& cp
[4] == c
)
171 return (void *) &cp
[4];
172 if (5 < sizeof longword
&& cp
[5] == c
)
173 return (void *) &cp
[5];
174 if (6 < sizeof longword
&& cp
[6] == c
)
175 return (void *) &cp
[6];
176 if (7 < sizeof longword
&& cp
[7] == c
)
177 return (void *) &cp
[7];
178 if (8 < sizeof longword
)
179 for (i
= 8; i
< sizeof longword
; i
++)
181 return (void *) &cp
[i
];
184 n
-= sizeof longword
;
187 char_ptr
= (const unsigned char *) longword_ptr
;
192 return (void *) char_ptr
;
200 weak_alias (__memchr
, BP_SYM (memchr
))