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 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
51 /* Search no more than N bytes of S for C. */
53 __memchr (void const *s
, int c_in
, size_t n
)
55 const unsigned char *char_ptr
;
56 const unsigned long int *longword_ptr
;
57 unsigned long int longword
, magic_bits
, charmask
;
61 c
= (unsigned char) c_in
;
63 /* Handle the first few characters by reading one character at a time.
64 Do this until CHAR_PTR is aligned on a longword boundary. */
65 for (char_ptr
= (const unsigned char *) s
;
66 n
> 0 && (size_t) char_ptr
% sizeof longword
!= 0;
69 return (void *) char_ptr
;
71 /* All these elucidatory comments refer to 4-byte longwords,
72 but the theory applies equally well to any size longwords. */
74 longword_ptr
= (const unsigned long int *) char_ptr
;
76 /* Bits 31, 24, 16, and 8 of this number are zero. Call these bits
77 the "holes." Note that there is a hole just to the left of
78 each byte, with an extra at the end:
80 bits: 01111110 11111110 11111110 11111111
81 bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD
83 The 1-bits make sure that carries propagate to the next 0-bit.
84 The 0-bits provide holes for carries to fall into. */
86 /* Set MAGIC_BITS to be this pattern of 1 and 0 bits.
87 Set CHARMASK to be a longword, each of whose bytes is C. */
89 magic_bits
= 0xfefefefe;
90 charmask
= c
| (c
<< 8);
91 charmask
|= charmask
<< 16;
92 #if 0xffffffffU < ULONG_MAX
93 magic_bits
|= magic_bits
<< 32;
94 charmask
|= charmask
<< 32;
95 if (8 < sizeof longword
)
96 for (i
= 64; i
< sizeof longword
* 8; i
*= 2)
98 magic_bits
|= magic_bits
<< i
;
99 charmask
|= charmask
<< i
;
102 magic_bits
= (ULONG_MAX
>> 1) & (magic_bits
| 1);
104 /* Instead of the traditional loop which tests each character,
105 we will test a longword at a time. The tricky part is testing
106 if *any of the four* bytes in the longword in question are zero. */
107 while (n
>= sizeof longword
)
109 /* We tentatively exit the loop if adding MAGIC_BITS to
110 LONGWORD fails to change any of the hole bits of LONGWORD.
112 1) Is this safe? Will it catch all the zero bytes?
113 Suppose there is a byte with all zeros. Any carry bits
114 propagating from its left will fall into the hole at its
115 least significant bit and stop. Since there will be no
116 carry from its most significant bit, the LSB of the
117 byte to the left will be unchanged, and the zero will be
120 2) Is this worthwhile? Will it ignore everything except
121 zero bytes? Suppose every byte of LONGWORD has a bit set
122 somewhere. There will be a carry into bit 8. If bit 8
123 is set, this will carry into bit 16. If bit 8 is clear,
124 one of bits 9-15 must be set, so there will be a carry
125 into bit 16. Similarly, there will be a carry into bit
126 24. If one of bits 24-30 is set, there will be a carry
127 into bit 31, so all of the hole bits will be changed.
129 The one misfire occurs when bits 24-30 are clear and bit
130 31 is set; in this case, the hole at bit 31 is not
131 changed. If we had access to the processor carry flag,
132 we could close this loophole by putting the fourth hole
135 So it ignores everything except 128's, when they're aligned
138 3) But wait! Aren't we looking for C, not zero?
139 Good point. So what we do is XOR LONGWORD with a longword,
140 each of whose bytes is C. This turns each byte that is C
143 longword
= *longword_ptr
++ ^ charmask
;
145 /* Add MAGIC_BITS to LONGWORD. */
146 if ((((longword
+ magic_bits
)
148 /* Set those bits that were unchanged by the addition. */
151 /* Look at only the hole bits. If any of the hole bits
152 are unchanged, most likely one of the bytes was a
156 /* Which of the bytes was C? If none of them were, it was
157 a misfire; continue the search. */
159 const unsigned char *cp
= (const unsigned char *) (longword_ptr
- 1);
164 return (void *) &cp
[1];
166 return (void *) &cp
[2];
168 return (void *) &cp
[3];
169 if (4 < sizeof longword
&& cp
[4] == c
)
170 return (void *) &cp
[4];
171 if (5 < sizeof longword
&& cp
[5] == c
)
172 return (void *) &cp
[5];
173 if (6 < sizeof longword
&& cp
[6] == c
)
174 return (void *) &cp
[6];
175 if (7 < sizeof longword
&& cp
[7] == c
)
176 return (void *) &cp
[7];
177 if (8 < sizeof longword
)
178 for (i
= 8; i
< sizeof longword
; i
++)
180 return (void *) &cp
[i
];
183 n
-= sizeof longword
;
186 char_ptr
= (const unsigned char *) longword_ptr
;
191 return (void *) char_ptr
;
199 weak_alias (__memchr
, BP_SYM (memchr
))
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