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c84142e8 | 1 | /* Copyright (C) 1991, 93, 94, 95, 96, 97 Free Software Foundation, Inc. |
6d52618b | 2 | Based on strlen implementation by Torbjorn Granlund (tege@sics.se), |
28f540f4 RM |
3 | with help from Dan Sahlin (dan@sics.se) and |
4 | bug fix and commentary by Jim Blandy (jimb@ai.mit.edu); | |
5 | adaptation to strchr suggested by Dick Karpinski (dick@cca.ucsf.edu), | |
6 | and implemented by Roland McGrath (roland@ai.mit.edu). | |
7 | ||
6d52618b UD |
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. | |
28f540f4 | 12 | |
6d52618b UD |
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. | |
28f540f4 | 17 | |
6d52618b UD |
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 not, | |
20 | write to the Free Software Foundation, Inc., 59 Temple Place - Suite 330, | |
21 | Boston, MA 02111-1307, USA. */ | |
28f540f4 | 22 | |
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23 | #include <string.h> |
24 | ||
25 | ||
6d52618b | 26 | /* Find the first occurrence of C in S. */ |
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27 | |
28 | char * | |
c84142e8 UD |
29 | strchr (s, c) |
30 | const char *s; | |
31 | int c; | |
28f540f4 | 32 | { |
c84142e8 UD |
33 | const unsigned char *char_ptr; |
34 | const unsigned long int *longword_ptr; | |
28f540f4 RM |
35 | unsigned long int longword, magic_bits, charmask; |
36 | ||
37 | c = (unsigned char) c; | |
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 = s; ((unsigned long int) char_ptr | |
42 | & (sizeof (longword) - 1)) != 0; | |
43 | ++char_ptr) | |
44 | if (*char_ptr == c) | |
c84142e8 | 45 | return (void *) char_ptr; |
28f540f4 RM |
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: | |
6d52618b | 57 | |
28f540f4 | 58 | bits: 01111110 11111110 11111110 11111111 |
6d52618b | 59 | bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD |
28f540f4 RM |
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 | switch (sizeof (longword)) | |
64 | { | |
65 | case 4: magic_bits = 0x7efefeffL; break; | |
66 | case 8: magic_bits = (0x7efefefeL << 32) | 0xfefefeffL; break; | |
67 | default: | |
68 | abort (); | |
69 | } | |
70 | ||
71 | /* Set up a longword, each of whose bytes is C. */ | |
72 | charmask = c | (c << 8); | |
73 | charmask |= charmask << 16; | |
74 | if (sizeof (longword) > 4) | |
75 | charmask |= charmask << 32; | |
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) | |
6d52618b | 122 | |
28f540f4 RM |
123 | /* Set those bits that were unchanged by the addition. */ |
124 | ^ ~longword) | |
6d52618b | 125 | |
28f540f4 RM |
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 | ||
c84142e8 | 138 | const unsigned char *cp = (const unsigned char *) (longword_ptr - 1); |
28f540f4 RM |
139 | |
140 | if (*cp == c) | |
141 | return (char *) cp; | |
142 | else if (*cp == '\0') | |
143 | return NULL; | |
144 | if (*++cp == c) | |
145 | return (char *) cp; | |
146 | else if (*cp == '\0') | |
147 | return NULL; | |
148 | if (*++cp == c) | |
149 | return (char *) cp; | |
150 | else if (*cp == '\0') | |
151 | return NULL; | |
152 | if (*++cp == c) | |
153 | return (char *) cp; | |
154 | else if (*cp == '\0') | |
155 | return NULL; | |
156 | if (sizeof (longword) > 4) | |
157 | { | |
158 | if (*++cp == c) | |
159 | return (char *) cp; | |
160 | else if (*cp == '\0') | |
161 | return NULL; | |
162 | if (*++cp == c) | |
163 | return (char *) cp; | |
164 | else if (*cp == '\0') | |
165 | return NULL; | |
166 | if (*++cp == c) | |
167 | return (char *) cp; | |
168 | else if (*cp == '\0') | |
169 | return NULL; | |
170 | if (*++cp == c) | |
171 | return (char *) cp; | |
172 | else if (*cp == '\0') | |
173 | return NULL; | |
174 | } | |
175 | } | |
176 | } | |
177 | ||
178 | return NULL; | |
179 | } | |
180 | ||
181 | #ifdef weak_alias | |
182 | #undef index | |
183 | weak_alias (strchr, index) | |
184 | #endif |