1 /* Copyright (C) 1991,93,94,95,96,97,99,2000 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 #ifdef __USE_GNU
27
28 #include "memcopy.h"
29
30 /* Find the first occurrence of C in S or the final NUL byte. */
strchrnul(const char * s,int c_in)31 char *strchrnul (const char *s, int c_in)
32 {
33 const unsigned char *char_ptr;
34 const unsigned long int *longword_ptr;
35 unsigned long int longword, magic_bits, charmask;
36 unsigned reg_char c;
37
38 c = (unsigned char) c_in;
39
40 /* Handle the first few characters by reading one character at a time.
41 Do this until CHAR_PTR is aligned on a longword boundary. */
42 for (char_ptr = (const unsigned char *) s;
43 ((unsigned long int) char_ptr & (sizeof (longword) - 1)) != 0;
44 ++char_ptr)
45 if (*char_ptr == c || *char_ptr == '\0')
46 return (void *) char_ptr;
47
48 /* All these elucidatory comments refer to 4-byte longwords,
49 but the theory applies equally well to 8-byte longwords. */
50
51 longword_ptr = (unsigned long int *) char_ptr;
52
53 /* Bits 31, 24, 16, and 8 of this number are zero. Call these bits
54 the "holes." Note that there is a hole just to the left of
55 each byte, with an extra at the end:
56
57 bits: 01111110 11111110 11111110 11111111
58 bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD
59
60 The 1-bits make sure that carries propagate to the next 0-bit.
61 The 0-bits provide holes for carries to fall into. */
62
63 #if __WORDSIZE == 32
64 magic_bits = 0x7efefeffL;
65 charmask = c | (c << 8);
66 charmask |= charmask << 16;
67 #elif __WORDSIZE == 64
68 magic_bits = ((0x7efefefeL << 16) << 16) | 0xfefefeffL;
69 charmask = c | (c << 8);
70 charmask |= charmask << 16;
71 charmask |= (charmask << 16) << 16;
72 #else
73 #error unexpected integer size strchr()
74 #endif
75
76 /* Instead of the traditional loop which tests each character,
77 we will test a longword at a time. The tricky part is testing
78 if *any of the four* bytes in the longword in question are zero. */
79 for (;;)
80 {
81 /* We tentatively exit the loop if adding MAGIC_BITS to
82 LONGWORD fails to change any of the hole bits of LONGWORD.
83
84 1) Is this safe? Will it catch all the zero bytes?
85 Suppose there is a byte with all zeros. Any carry bits
86 propagating from its left will fall into the hole at its
87 least significant bit and stop. Since there will be no
88 carry from its most significant bit, the LSB of the
89 byte to the left will be unchanged, and the zero will be
90 detected.
91
92 2) Is this worthwhile? Will it ignore everything except
93 zero bytes? Suppose every byte of LONGWORD has a bit set
94 somewhere. There will be a carry into bit 8. If bit 8
95 is set, this will carry into bit 16. If bit 8 is clear,
96 one of bits 9-15 must be set, so there will be a carry
97 into bit 16. Similarly, there will be a carry into bit
98 24. If one of bits 24-30 is set, there will be a carry
99 into bit 31, so all of the hole bits will be changed.
100
101 The one misfire occurs when bits 24-30 are clear and bit
102 31 is set; in this case, the hole at bit 31 is not
103 changed. If we had access to the processor carry flag,
104 we could close this loophole by putting the fourth hole
105 at bit 32!
106
107 So it ignores everything except 128's, when they're aligned
108 properly.
109
110 3) But wait! Aren't we looking for C as well as zero?
111 Good point. So what we do is XOR LONGWORD with a longword,
112 each of whose bytes is C. This turns each byte that is C
113 into a zero. */
114
115 longword = *longword_ptr++;
116
117 /* Add MAGIC_BITS to LONGWORD. */
118 if ((((longword + magic_bits)
119
120 /* Set those bits that were unchanged by the addition. */
121 ^ ~longword)
122
123 /* Look at only the hole bits. If any of the hole bits
124 are unchanged, most likely one of the bytes was a
125 zero. */
126 & ~magic_bits) != 0 ||
127
128 /* That caught zeroes. Now test for C. */
129 ((((longword ^ charmask) + magic_bits) ^ ~(longword ^ charmask))
130 & ~magic_bits) != 0)
131 {
132 /* Which of the bytes was C or zero?
133 If none of them were, it was a misfire; continue the search. */
134
135 const unsigned char *cp = (const unsigned char *) (longword_ptr - 1);
136
137 if (*cp == c || *cp == '\0')
138 return (char *) cp;
139 if (*++cp == c || *cp == '\0')
140 return (char *) cp;
141 if (*++cp == c || *cp == '\0')
142 return (char *) cp;
143 if (*++cp == c || *cp == '\0')
144 return (char *) cp;
145 if (sizeof (longword) > 4)
146 {
147 if (*++cp == c || *cp == '\0')
148 return (char *) cp;
149 if (*++cp == c || *cp == '\0')
150 return (char *) cp;
151 if (*++cp == c || *cp == '\0')
152 return (char *) cp;
153 if (*++cp == c || *cp == '\0')
154 return (char *) cp;
155 }
156 }
157 }
158
159 /* This should never happen. */
160 return NULL;
161 }
162 libc_hidden_def(strchrnul)
163 #endif
164