1 /* Copyright (C) 1991,93,96,97,99,2000 Free Software Foundation, Inc.
2 Based on strlen implementation 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 The GNU C Library is free software; you can redistribute it and/or
9 modify it under the terms of the GNU Library General Public License as
10 published by the Free Software Foundation; either version 2 of the
11 License, or (at your option) any later version.
13 The GNU C Library is distributed in the hope that it will be useful,
14 but WITHOUT ANY WARRANTY; without even the implied warranty of
15 MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU
16 Library General Public License for more details.
18 You should have received a copy of the GNU Library General Public
19 License along with the GNU C Library; see the file COPYING.LIB. If not,
20 write to the Free Software Foundation, Inc., 59 Temple Place - Suite 330,
21 Boston, MA 02111-1307, USA. */
28 #if defined (__cplusplus) || (defined (__STDC__) && __STDC__)
29 # define __ptr_t void *
30 #else /* Not C++ or ANSI C. */
31 # define __ptr_t char *
32 #endif /* C++ or ANSI C. */
39 # define reg_char char
42 #if defined (HAVE_LIMITS_H) || defined (_LIBC)
46 #define LONG_MAX_32_BITS 2147483647
49 #define LONG_MAX LONG_MAX_32_BITS
52 #include <sys/types.h>
57 /* Find the first occurrence of C in S. */
63 const unsigned char *char_ptr
;
64 const unsigned long int *longword_ptr
;
65 unsigned long int longword
, magic_bits
, charmask
;
68 c
= (unsigned char) c_in
;
70 /* Handle the first few characters by reading one character at a time.
71 Do this until CHAR_PTR is aligned on a longword boundary. */
72 for (char_ptr
= (const unsigned char *) s
;
73 ((unsigned long int) char_ptr
& (sizeof (longword
) - 1)) != 0;
76 return (__ptr_t
) char_ptr
;
78 /* All these elucidatory comments refer to 4-byte longwords,
79 but the theory applies equally well to 8-byte longwords. */
81 longword_ptr
= (unsigned long int *) char_ptr
;
83 /* Bits 31, 24, 16, and 8 of this number are zero. Call these bits
84 the "holes." Note that there is a hole just to the left of
85 each byte, with an extra at the end:
87 bits: 01111110 11111110 11111110 11111111
88 bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD
90 The 1-bits make sure that carries propagate to the next 0-bit.
91 The 0-bits provide holes for carries to fall into. */
93 if (sizeof (longword
) != 4 && sizeof (longword
) != 8)
96 #if LONG_MAX <= LONG_MAX_32_BITS
97 magic_bits
= 0x7efefeff;
99 magic_bits
= ((unsigned long int) 0x7efefefe << 32) | 0xfefefeff;
102 /* Set up a longword, each of whose bytes is C. */
103 charmask
= c
| (c
<< 8);
104 charmask
|= charmask
<< 16;
105 #if LONG_MAX > LONG_MAX_32_BITS
106 charmask
|= charmask
<< 32;
109 /* Instead of the traditional loop which tests each character,
110 we will test a longword at a time. The tricky part is testing
111 if *any of the four* bytes in the longword in question are zero. */
114 /* We tentatively exit the loop if adding MAGIC_BITS to
115 LONGWORD fails to change any of the hole bits of LONGWORD.
117 1) Is this safe? Will it catch all the zero bytes?
118 Suppose there is a byte with all zeros. Any carry bits
119 propagating from its left will fall into the hole at its
120 least significant bit and stop. Since there will be no
121 carry from its most significant bit, the LSB of the
122 byte to the left will be unchanged, and the zero will be
125 2) Is this worthwhile? Will it ignore everything except
126 zero bytes? Suppose every byte of LONGWORD has a bit set
127 somewhere. There will be a carry into bit 8. If bit 8
128 is set, this will carry into bit 16. If bit 8 is clear,
129 one of bits 9-15 must be set, so there will be a carry
130 into bit 16. Similarly, there will be a carry into bit
131 24. If one of bits 24-30 is set, there will be a carry
132 into bit 31, so all of the hole bits will be changed.
134 The one misfire occurs when bits 24-30 are clear and bit
135 31 is set; in this case, the hole at bit 31 is not
136 changed. If we had access to the processor carry flag,
137 we could close this loophole by putting the fourth hole
140 So it ignores everything except 128's, when they're aligned
143 3) But wait! Aren't we looking for C, not zero?
144 Good point. So what we do is XOR LONGWORD with a longword,
145 each of whose bytes is C. This turns each byte that is C
148 longword
= *longword_ptr
++ ^ charmask
;
150 /* Add MAGIC_BITS to LONGWORD. */
151 if ((((longword
+ magic_bits
)
153 /* Set those bits that were unchanged by the addition. */
156 /* Look at only the hole bits. If any of the hole bits
157 are unchanged, most likely one of the bytes was a
161 /* Which of the bytes was C? If none of them were, it was
162 a misfire; continue the search. */
164 const unsigned char *cp
= (const unsigned char *) (longword_ptr
- 1);
169 return (__ptr_t
) &cp
[1];
171 return (__ptr_t
) &cp
[2];
173 return (__ptr_t
) &cp
[3];
174 #if LONG_MAX > 2147483647
176 return (__ptr_t
) &cp
[4];
178 return (__ptr_t
) &cp
[5];
180 return (__ptr_t
) &cp
[6];
182 return (__ptr_t
) &cp
[7];
187 weak_alias (__rawmemchr
, rawmemchr
)