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 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
20 not, write to the Free Software Foundation, Inc., 675 Mass Ave,
21 Cambridge, MA 02139, 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. */
38 #if defined (HAVE_LIMITS_H) || defined (_LIBC)
42 #define LONG_MAX_32_BITS 2147483647
45 #define LONG_MAX LONG_MAX_32_BITS
48 #include <sys/types.h>
51 /* Search no more than N bytes of S for C. */
59 const unsigned char *char_ptr
;
60 const unsigned long int *longword_ptr
;
61 unsigned long int longword
, magic_bits
, charmask
;
63 c
= (unsigned char) c
;
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 n
> 0 && ((unsigned long int) char_ptr
69 & (sizeof (longword
) - 1)) != 0;
72 return (__ptr_t
) char_ptr
;
74 /* All these elucidatory comments refer to 4-byte longwords,
75 but the theory applies equally well to 8-byte longwords. */
77 longword_ptr
= (unsigned long int *) char_ptr
;
79 /* Bits 31, 24, 16, and 8 of this number are zero. Call these bits
80 the "holes." Note that there is a hole just to the left of
81 each byte, with an extra at the end:
83 bits: 01111110 11111110 11111110 11111111
84 bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD
86 The 1-bits make sure that carries propagate to the next 0-bit.
87 The 0-bits provide holes for carries to fall into. */
89 if (sizeof (longword
) != 4 && sizeof (longword
) != 8)
92 #if LONG_MAX <= LONG_MAX_32_BITS
93 magic_bits
= 0x7efefeff;
95 magic_bits
= ((unsigned long int) 0x7efefefe << 32) | 0xfefefeff;
98 /* Set up a longword, each of whose bytes is C. */
99 charmask
= c
| (c
<< 8);
100 charmask
|= charmask
<< 16;
101 #if LONG_MAX > LONG_MAX_32_BITS
102 charmask
|= charmask
<< 32;
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 (__ptr_t
) &cp
[1];
167 return (__ptr_t
) &cp
[2];
169 return (__ptr_t
) &cp
[3];
170 #if LONG_MAX > 2147483647
172 return (__ptr_t
) &cp
[4];
174 return (__ptr_t
) &cp
[5];
176 return (__ptr_t
) &cp
[6];
178 return (__ptr_t
) &cp
[7];
182 n
-= sizeof (longword
);
185 char_ptr
= (const unsigned char *) longword_ptr
;
190 return (__ptr_t
) char_ptr
;