1 /* Copyright (C) 1991, 1993 Free Software Foundation, Inc.
2 Based on strlen implemention by Torbjorn Granlund (tege@sics.se),
3 with help from Dan Sahlin (dan@sics.se) and
4 commentary by Jim Blandy (jimb@ai.mit.edu);
5 adaptation to memchr suggested by Dick Karpinski (dick@cca.ucsf.edu),
6 and implemented by Roland McGrath (roland@ai.mit.edu).
8 NOTE: The canonical source of this file is maintained with the GNU C Library.
9 Bugs can be reported to bug-glibc@prep.ai.mit.edu.
11 This program is free software; you can redistribute it and/or modify it
12 under the terms of the GNU General Public License as published by the
13 Free Software Foundation; either version 2, or (at your option) any
16 This program is distributed in the hope that it will be useful,
17 but WITHOUT ANY WARRANTY; without even the implied warranty of
18 MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
19 GNU General Public License for more details.
21 You should have received a copy of the GNU General Public License
22 along with this program; if not, write to the Free Software
23 Foundation, Inc., 59 Temple Place - Suite 330, Boston, MA 02111-1307,
31 #if defined (__cplusplus) || (defined (__STDC__) && __STDC__)
32 # define __ptr_t void *
33 #else /* Not C++ or ANSI C. */
34 # define __ptr_t char *
35 #endif /* C++ or ANSI C. */
41 #if defined (HAVE_LIMITS_H) || defined (_LIBC)
45 #define LONG_MAX_32_BITS 2147483647
48 #define LONG_MAX LONG_MAX_32_BITS
51 #include <sys/types.h>
54 /* Search no more than N bytes of S for C. */
62 const unsigned char *char_ptr
;
63 const unsigned long int *longword_ptr
;
64 unsigned long int longword
, magic_bits
, charmask
;
66 c
= (unsigned char) c
;
68 /* Handle the first few characters by reading one character at a time.
69 Do this until CHAR_PTR is aligned on a longword boundary. */
70 for (char_ptr
= (const unsigned char *) s
;
71 n
> 0 && ((unsigned long int) char_ptr
72 & (sizeof (longword
) - 1)) != 0;
75 return (__ptr_t
) char_ptr
;
77 /* All these elucidatory comments refer to 4-byte longwords,
78 but the theory applies equally well to 8-byte longwords. */
80 longword_ptr
= (unsigned long int *) char_ptr
;
82 /* Bits 31, 24, 16, and 8 of this number are zero. Call these bits
83 the "holes." Note that there is a hole just to the left of
84 each byte, with an extra at the end:
86 bits: 01111110 11111110 11111110 11111111
87 bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD
89 The 1-bits make sure that carries propagate to the next 0-bit.
90 The 0-bits provide holes for carries to fall into. */
92 if (sizeof (longword
) != 4 && sizeof (longword
) != 8)
95 #if LONG_MAX <= LONG_MAX_32_BITS
96 magic_bits
= 0x7efefeff;
98 magic_bits
= ((unsigned long int) 0x7efefefe << 32) | 0xfefefeff;
101 /* Set up a longword, each of whose bytes is C. */
102 charmask
= c
| (c
<< 8);
103 charmask
|= charmask
<< 16;
104 #if LONG_MAX > LONG_MAX_32_BITS
105 charmask
|= charmask
<< 32;
108 /* Instead of the traditional loop which tests each character,
109 we will test a longword at a time. The tricky part is testing
110 if *any of the four* bytes in the longword in question are zero. */
111 while (n
>= sizeof (longword
))
113 /* We tentatively exit the loop if adding MAGIC_BITS to
114 LONGWORD fails to change any of the hole bits of LONGWORD.
116 1) Is this safe? Will it catch all the zero bytes?
117 Suppose there is a byte with all zeros. Any carry bits
118 propagating from its left will fall into the hole at its
119 least significant bit and stop. Since there will be no
120 carry from its most significant bit, the LSB of the
121 byte to the left will be unchanged, and the zero will be
124 2) Is this worthwhile? Will it ignore everything except
125 zero bytes? Suppose every byte of LONGWORD has a bit set
126 somewhere. There will be a carry into bit 8. If bit 8
127 is set, this will carry into bit 16. If bit 8 is clear,
128 one of bits 9-15 must be set, so there will be a carry
129 into bit 16. Similarly, there will be a carry into bit
130 24. If one of bits 24-30 is set, there will be a carry
131 into bit 31, so all of the hole bits will be changed.
133 The one misfire occurs when bits 24-30 are clear and bit
134 31 is set; in this case, the hole at bit 31 is not
135 changed. If we had access to the processor carry flag,
136 we could close this loophole by putting the fourth hole
139 So it ignores everything except 128's, when they're aligned
142 3) But wait! Aren't we looking for C, not zero?
143 Good point. So what we do is XOR LONGWORD with a longword,
144 each of whose bytes is C. This turns each byte that is C
147 longword
= *longword_ptr
++ ^ charmask
;
149 /* Add MAGIC_BITS to LONGWORD. */
150 if ((((longword
+ magic_bits
)
152 /* Set those bits that were unchanged by the addition. */
155 /* Look at only the hole bits. If any of the hole bits
156 are unchanged, most likely one of the bytes was a
160 /* Which of the bytes was C? If none of them were, it was
161 a misfire; continue the search. */
163 const unsigned char *cp
= (const unsigned char *) (longword_ptr
- 1);
168 return (__ptr_t
) &cp
[1];
170 return (__ptr_t
) &cp
[2];
172 return (__ptr_t
) &cp
[3];
173 #if LONG_MAX > 2147483647
175 return (__ptr_t
) &cp
[4];
177 return (__ptr_t
) &cp
[5];
179 return (__ptr_t
) &cp
[6];
181 return (__ptr_t
) &cp
[7];
185 n
-= sizeof (longword
);
188 char_ptr
= (const unsigned char *) longword_ptr
;
193 return (__ptr_t
) char_ptr
;