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[glibc/history.git] / sysdeps / generic / strchr.c
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1 /* Copyright (C) 1991, 93, 94, 95, 96, 97 Free Software Foundation, Inc.
2 Based on strlen implementation by Torbjorn Granlund (tege@sics.se),
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).
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.
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.
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. */
23 #include <string.h>
26 /* Find the first occurrence of C in S. */
28 char *
29 strchr (s, c)
30 const char *s;
31 int c;
33 const unsigned char *char_ptr;
34 const unsigned long int *longword_ptr;
35 unsigned long int longword, magic_bits, charmask;
37 c = (unsigned char) c;
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)
45 return (void *) char_ptr;
46 else if (*char_ptr == '\0')
47 return NULL;
49 /* All these elucidatory comments refer to 4-byte longwords,
50 but the theory applies equally well to 8-byte longwords. */
52 longword_ptr = (unsigned long int *) char_ptr;
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:
58 bits: 01111110 11111110 11111110 11111111
59 bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD
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))
65 case 4: magic_bits = 0x7efefeffL; break;
66 case 8: magic_bits = (0x7efefefeL << 32) | 0xfefefeffL; break;
67 default:
68 abort ();
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 ();
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 (;;)
84 /* We tentatively exit the loop if adding MAGIC_BITS to
85 LONGWORD fails to change any of the hole bits of LONGWORD.
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.
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.
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!
110 So it ignores everything except 128's, when they're aligned
111 properly.
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. */
118 longword = *longword_ptr++;
120 /* Add MAGIC_BITS to LONGWORD. */
121 if ((((longword + magic_bits)
123 /* Set those bits that were unchanged by the addition. */
124 ^ ~longword)
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 ||
131 /* That caught zeroes. Now test for C. */
132 ((((longword ^ charmask) + magic_bits) ^ ~(longword ^ charmask))
133 & ~magic_bits) != 0)
135 /* Which of the bytes was C or zero?
136 If none of them were, it was a misfire; continue the search. */
138 const unsigned char *cp = (const unsigned char *) (longword_ptr - 1);
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)
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;
178 return NULL;
181 #ifdef weak_alias
182 #undef index
183 weak_alias (strchr, index)
184 #endif