1 /* Copyright (C) 1991,93,96,97,99,2000,2002 Free Software Foundation, Inc.
2 This file is part of the GNU C Library.
3 Based on strlen implementation by Torbjorn Granlund (tege@sics.se),
4 with help from Dan Sahlin (dan@sics.se) and
5 commentary by Jim Blandy (jimb@ai.mit.edu);
6 adaptation to memchr suggested by Dick Karpinski (dick@cca.ucsf.edu),
7 and implemented by Roland McGrath (roland@ai.mit.edu).
8
9 The GNU C Library is free software; you can redistribute it and/or
10 modify it under the terms of the GNU Lesser General Public
11 License as published by the Free Software Foundation; either
12 version 2.1 of the License, or (at your option) any later version.
13
14 The GNU C Library is distributed in the hope that it will be useful,
15 but WITHOUT ANY WARRANTY; without even the implied warranty of
16 MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU
17 Lesser General Public License for more details.
18
19 You should have received a copy of the GNU Lesser General Public
20 License along with the GNU C Library; if not, see
21 <http://www.gnu.org/licenses/>. */
22
23 #include <string.h>
24 #include <stdlib.h>
25 #include <limits.h>
26
27 #ifdef __USE_GNU
28
29 #include "memcopy.h"
30
31 #define LONG_MAX_32_BITS 2147483647
32
33 /* Find the first occurrence of C in S. */
rawmemchr(const void * s,int c_in)34 void *rawmemchr (const void * s, int c_in)
35 {
36 const unsigned char *char_ptr;
37 const unsigned long int *longword_ptr;
38 unsigned long int longword, magic_bits, charmask;
39 unsigned reg_char c;
40
41 c = (unsigned char) c_in;
42
43 /* Handle the first few characters by reading one character at a time.
44 Do this until CHAR_PTR is aligned on a longword boundary. */
45 for (char_ptr = (const unsigned char *) s;
46 ((unsigned long int) char_ptr & (sizeof (longword) - 1)) != 0;
47 ++char_ptr)
48 if (*char_ptr == c)
49 return (void *) char_ptr;
50
51 /* All these elucidatory comments refer to 4-byte longwords,
52 but the theory applies equally well to 8-byte longwords. */
53
54 longword_ptr = (unsigned long int *) char_ptr;
55
56 /* Bits 31, 24, 16, and 8 of this number are zero. Call these bits
57 the "holes." Note that there is a hole just to the left of
58 each byte, with an extra at the end:
59
60 bits: 01111110 11111110 11111110 11111111
61 bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD
62
63 The 1-bits make sure that carries propagate to the next 0-bit.
64 The 0-bits provide holes for carries to fall into. */
65
66 if (sizeof (longword) != 4 && sizeof (longword) != 8)
67 abort ();
68
69 #if LONG_MAX <= LONG_MAX_32_BITS
70 magic_bits = 0x7efefeff;
71 #else
72 magic_bits = ((unsigned long int) 0x7efefefe << 32) | 0xfefefeff;
73 #endif
74
75 /* Set up a longword, each of whose bytes is C. */
76 charmask = c | (c << 8);
77 charmask |= charmask << 16;
78 #if LONG_MAX > LONG_MAX_32_BITS
79 charmask |= charmask << 32;
80 #endif
81
82 /* Instead of the traditional loop which tests each character,
83 we will test a longword at a time. The tricky part is testing
84 if *any of the four* bytes in the longword in question are zero. */
85 while (1)
86 {
87 /* We tentatively exit the loop if adding MAGIC_BITS to
88 LONGWORD fails to change any of the hole bits of LONGWORD.
89
90 1) Is this safe? Will it catch all the zero bytes?
91 Suppose there is a byte with all zeros. Any carry bits
92 propagating from its left will fall into the hole at its
93 least significant bit and stop. Since there will be no
94 carry from its most significant bit, the LSB of the
95 byte to the left will be unchanged, and the zero will be
96 detected.
97
98 2) Is this worthwhile? Will it ignore everything except
99 zero bytes? Suppose every byte of LONGWORD has a bit set
100 somewhere. There will be a carry into bit 8. If bit 8
101 is set, this will carry into bit 16. If bit 8 is clear,
102 one of bits 9-15 must be set, so there will be a carry
103 into bit 16. Similarly, there will be a carry into bit
104 24. If one of bits 24-30 is set, there will be a carry
105 into bit 31, so all of the hole bits will be changed.
106
107 The one misfire occurs when bits 24-30 are clear and bit
108 31 is set; in this case, the hole at bit 31 is not
109 changed. If we had access to the processor carry flag,
110 we could close this loophole by putting the fourth hole
111 at bit 32!
112
113 So it ignores everything except 128's, when they're aligned
114 properly.
115
116 3) But wait! Aren't we looking for C, not zero?
117 Good point. So what we do is XOR LONGWORD with a longword,
118 each of whose bytes is C. This turns each byte that is C
119 into a zero. */
120
121 longword = *longword_ptr++ ^ charmask;
122
123 /* Add MAGIC_BITS to LONGWORD. */
124 if ((((longword + magic_bits)
125
126 /* Set those bits that were unchanged by the addition. */
127 ^ ~longword)
128
129 /* Look at only the hole bits. If any of the hole bits
130 are unchanged, most likely one of the bytes was a
131 zero. */
132 & ~magic_bits) != 0)
133 {
134 /* Which of the bytes was C? If none of them were, it was
135 a misfire; continue the search. */
136
137 const unsigned char *cp = (const unsigned char *) (longword_ptr - 1);
138
139 if (cp[0] == c)
140 return (void *) cp;
141 if (cp[1] == c)
142 return (void *) &cp[1];
143 if (cp[2] == c)
144 return (void *) &cp[2];
145 if (cp[3] == c)
146 return (void *) &cp[3];
147 #if LONG_MAX > 2147483647
148 if (cp[4] == c)
149 return (void *) &cp[4];
150 if (cp[5] == c)
151 return (void *) &cp[5];
152 if (cp[6] == c)
153 return (void *) &cp[6];
154 if (cp[7] == c)
155 return (void *) &cp[7];
156 #endif
157 }
158 }
159 }
160 libc_hidden_def(rawmemchr)
161 #endif
162