1 /* Copyright (C) 1991,93,94,95,96,97,99,2000,03 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 bug fix and commentary by Jim Blandy (jimb@ai.mit.edu);
6 adaptation to strchr 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
26
27 #include "memcopy.h"
28
29 /* Find the first occurrence of C in S. */
strchr(const char * s,int c_in)30 char *strchr (const char *s, int c_in)
31 {
32 const unsigned char *char_ptr;
33 const unsigned long int *longword_ptr;
34 unsigned long int longword, magic_bits, charmask;
35 unsigned reg_char c;
36
37 c = (unsigned char) c_in;
38
39 /* Handle the first few characters by reading one character at a time.
40 Do this until CHAR_PTR is aligned on a longword boundary. */
41 for (char_ptr = (const unsigned char *) s;
42 ((unsigned long int) char_ptr & (sizeof (longword) - 1)) != 0;
43 ++char_ptr)
44 if (*char_ptr == c)
45 return (void *) char_ptr;
46 else if (*char_ptr == '\0')
47 return NULL;
48
49 /* All these elucidatory comments refer to 4-byte longwords,
50 but the theory applies equally well to 8-byte longwords. */
51
52 longword_ptr = (unsigned long int *) char_ptr;
53
54 /* Bits 31, 24, 16, and 8 of this number are zero. Call these bits
55 the "holes." Note that there is a hole just to the left of
56 each byte, with an extra at the end:
57
58 bits: 01111110 11111110 11111110 11111111
59 bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD
60
61 The 1-bits make sure that carries propagate to the next 0-bit.
62 The 0-bits provide holes for carries to fall into. */
63 /* Set up a longword, each of whose bytes is C. */
64 #if __WORDSIZE == 32
65 magic_bits = 0x7efefeffL;
66 charmask = c | (c << 8);
67 charmask |= charmask << 16;
68 #elif __WORDSIZE == 64
69 magic_bits = ((0x7efefefeL << 16) << 16) | 0xfefefeffL;
70 charmask = c | (c << 8);
71 charmask |= charmask << 16;
72 charmask |= (charmask << 16) << 16;
73 #else
74 #error unexpected integer size strchr()
75 #endif
76
77 /* Instead of the traditional loop which tests each character,
78 we will test a longword at a time. The tricky part is testing
79 if *any of the four* bytes in the longword in question are zero. */
80 for (;;)
81 {
82 /* We tentatively exit the loop if adding MAGIC_BITS to
83 LONGWORD fails to change any of the hole bits of LONGWORD.
84
85 1) Is this safe? Will it catch all the zero bytes?
86 Suppose there is a byte with all zeros. Any carry bits
87 propagating from its left will fall into the hole at its
88 least significant bit and stop. Since there will be no
89 carry from its most significant bit, the LSB of the
90 byte to the left will be unchanged, and the zero will be
91 detected.
92
93 2) Is this worthwhile? Will it ignore everything except
94 zero bytes? Suppose every byte of LONGWORD has a bit set
95 somewhere. There will be a carry into bit 8. If bit 8
96 is set, this will carry into bit 16. If bit 8 is clear,
97 one of bits 9-15 must be set, so there will be a carry
98 into bit 16. Similarly, there will be a carry into bit
99 24. If one of bits 24-30 is set, there will be a carry
100 into bit 31, so all of the hole bits will be changed.
101
102 The one misfire occurs when bits 24-30 are clear and bit
103 31 is set; in this case, the hole at bit 31 is not
104 changed. If we had access to the processor carry flag,
105 we could close this loophole by putting the fourth hole
106 at bit 32!
107
108 So it ignores everything except 128's, when they're aligned
109 properly.
110
111 3) But wait! Aren't we looking for C as well as zero?
112 Good point. So what we do is XOR LONGWORD with a longword,
113 each of whose bytes is C. This turns each byte that is C
114 into a zero. */
115
116 longword = *longword_ptr++;
117
118 /* Add MAGIC_BITS to LONGWORD. */
119 if ((((longword + magic_bits)
120
121 /* Set those bits that were unchanged by the addition. */
122 ^ ~longword)
123
124 /* Look at only the hole bits. If any of the hole bits
125 are unchanged, most likely one of the bytes was a
126 zero. */
127 & ~magic_bits) != 0 ||
128
129 /* That caught zeroes. Now test for C. */
130 ((((longword ^ charmask) + magic_bits) ^ ~(longword ^ charmask))
131 & ~magic_bits) != 0)
132 {
133 /* Which of the bytes was C or zero?
134 If none of them were, it was a misfire; continue the search. */
135
136 const unsigned char *cp = (const unsigned char *) (longword_ptr - 1);
137
138 if (*cp == c)
139 return (char *) cp;
140 else if (*cp == '\0')
141 return NULL;
142 if (*++cp == c)
143 return (char *) cp;
144 else if (*cp == '\0')
145 return NULL;
146 if (*++cp == c)
147 return (char *) cp;
148 else if (*cp == '\0')
149 return NULL;
150 if (*++cp == c)
151 return (char *) cp;
152 else if (*cp == '\0')
153 return NULL;
154 if (sizeof (longword) > 4)
155 {
156 if (*++cp == c)
157 return (char *) cp;
158 else if (*cp == '\0')
159 return NULL;
160 if (*++cp == c)
161 return (char *) cp;
162 else if (*cp == '\0')
163 return NULL;
164 if (*++cp == c)
165 return (char *) cp;
166 else if (*cp == '\0')
167 return NULL;
168 if (*++cp == c)
169 return (char *) cp;
170 else if (*cp == '\0')
171 return NULL;
172 }
173 }
174 }
175
176 return NULL;
177 }
178 libc_hidden_weak(strchr)
179 #ifdef __UCLIBC_SUSV3_LEGACY__
180 weak_alias(strchr,index)
181 #endif
182