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 switch (sizeof (longword))
63 {
64 case 4: magic_bits = 0x7efefeffL; break;
65 case 8: magic_bits = ((0x7efefefeL << 16) << 16) | 0xfefefeffL; break;
66 default:
67 abort ();
68 }
69
70 /* Set up a longword, each of whose bytes is C. */
71 charmask = c | (c << 8);
72 charmask |= charmask << 16;
73 if (sizeof (longword) > 4)
74 /* Do the shift in two steps to avoid a warning if long has 32 bits. */
75 charmask |= (charmask << 16) << 16;
76 if (sizeof (longword) > 8)
77 abort ();
78
79 /* Instead of the traditional loop which tests each character,
80 we will test a longword at a time. The tricky part is testing
81 if *any of the four* bytes in the longword in question are zero. */
82 for (;;)
83 {
84 /* We tentatively exit the loop if adding MAGIC_BITS to
85 LONGWORD fails to change any of the hole bits of LONGWORD.
86
87 1) Is this safe? Will it catch all the zero bytes?
88 Suppose there is a byte with all zeros. Any carry bits
89 propagating from its left will fall into the hole at its
90 least significant bit and stop. Since there will be no
91 carry from its most significant bit, the LSB of the
92 byte to the left will be unchanged, and the zero will be
93 detected.
94
95 2) Is this worthwhile? Will it ignore everything except
96 zero bytes? Suppose every byte of LONGWORD has a bit set
97 somewhere. There will be a carry into bit 8. If bit 8
98 is set, this will carry into bit 16. If bit 8 is clear,
99 one of bits 9-15 must be set, so there will be a carry
100 into bit 16. Similarly, there will be a carry into bit
101 24. If one of bits 24-30 is set, there will be a carry
102 into bit 31, so all of the hole bits will be changed.
103
104 The one misfire occurs when bits 24-30 are clear and bit
105 31 is set; in this case, the hole at bit 31 is not
106 changed. If we had access to the processor carry flag,
107 we could close this loophole by putting the fourth hole
108 at bit 32!
109
110 So it ignores everything except 128's, when they're aligned
111 properly.
112
113 3) But wait! Aren't we looking for C as well as zero?
114 Good point. So what we do is XOR LONGWORD with a longword,
115 each of whose bytes is C. This turns each byte that is C
116 into a zero. */
117
118 longword = *longword_ptr++;
119
120 /* Add MAGIC_BITS to LONGWORD. */
121 if ((((longword + magic_bits)
122
123 /* Set those bits that were unchanged by the addition. */
124 ^ ~longword)
125
126 /* Look at only the hole bits. If any of the hole bits
127 are unchanged, most likely one of the bytes was a
128 zero. */
129 & ~magic_bits) != 0 ||
130
131 /* That caught zeroes. Now test for C. */
132 ((((longword ^ charmask) + magic_bits) ^ ~(longword ^ charmask))
133 & ~magic_bits) != 0)
134 {
135 /* Which of the bytes was C or zero?
136 If none of them were, it was a misfire; continue the search. */
137
138 const unsigned char *cp = (const unsigned char *) (longword_ptr - 1);
139
140 if (*cp == c || *cp == '\0')
141 return (char *) cp;
142 if (*++cp == c || *cp == '\0')
143 return (char *) cp;
144 if (*++cp == c || *cp == '\0')
145 return (char *) cp;
146 if (*++cp == c || *cp == '\0')
147 return (char *) cp;
148 if (sizeof (longword) > 4)
149 {
150 if (*++cp == c || *cp == '\0')
151 return (char *) cp;
152 if (*++cp == c || *cp == '\0')
153 return (char *) cp;
154 if (*++cp == c || *cp == '\0')
155 return (char *) cp;
156 if (*++cp == c || *cp == '\0')
157 return (char *) cp;
158 }
159 }
160 }
161
162 /* This should never happen. */
163 return NULL;
164 }
165 libc_hidden_def(strchrnul)
166 #endif
167