1 /* Copyright (C) 1991,1993-1997,1999,2000,2003,2006
2 Free Software Foundation, Inc.
3 This file is part of the GNU C Library.
4 Based on strlen implementation by Torbjorn Granlund (tege@sics.se),
5 with help from Dan Sahlin (dan@sics.se) and
6 bug fix and commentary by Jim Blandy (jimb@ai.mit.edu);
7 adaptation to strchr suggested by Dick Karpinski (dick@cca.ucsf.edu),
8 and implemented by Roland McGrath (roland@ai.mit.edu).
10 The GNU C Library is free software; you can redistribute it and/or
11 modify it under the terms of the GNU Lesser General Public
12 License as published by the Free Software Foundation; either
13 version 2.1 of the License, or (at your option) any later version.
15 The GNU C Library is distributed in the hope that it will be useful,
16 but WITHOUT ANY WARRANTY; without even the implied warranty of
17 MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU
18 Lesser General Public License for more details.
20 You should have received a copy of the GNU Lesser General Public
21 License along with the GNU C Library; if not, write to the Free
22 Software Foundation, Inc., 59 Temple Place, Suite 330, Boston, MA
31 /* Find the first occurrence of C in S. */
37 const unsigned char *char_ptr
;
38 const unsigned long int *longword_ptr
;
39 unsigned long int longword
, magic_bits
, charmask
;
42 c
= (unsigned char) c_in
;
44 /* Handle the first few characters by reading one character at a time.
45 Do this until CHAR_PTR is aligned on a longword boundary. */
46 for (char_ptr
= (const unsigned char *) s
;
47 ((unsigned long int) char_ptr
& (sizeof (longword
) - 1)) != 0;
50 return (void *) char_ptr
;
51 else if (*char_ptr
== '\0')
54 /* All these elucidatory comments refer to 4-byte longwords,
55 but the theory applies equally well to 8-byte longwords. */
57 longword_ptr
= (unsigned long int *) char_ptr
;
59 /* Bits 31, 24, 16, and 8 of this number are zero. Call these bits
60 the "holes." Note that there is a hole just to the left of
61 each byte, with an extra at the end:
63 bits: 01111110 11111110 11111110 11111111
64 bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD
66 The 1-bits make sure that carries propagate to the next 0-bit.
67 The 0-bits provide holes for carries to fall into. */
68 switch (sizeof (longword
))
70 case 4: magic_bits
= 0x7efefeffL
; break;
71 case 8: magic_bits
= ((0x7efefefeL
<< 16) << 16) | 0xfefefeffL
; break;
76 /* Set up a longword, each of whose bytes is C. */
77 charmask
= c
| (c
<< 8);
78 charmask
|= charmask
<< 16;
79 if (sizeof (longword
) > 4)
80 /* Do the shift in two steps to avoid a warning if long has 32 bits. */
81 charmask
|= (charmask
<< 16) << 16;
82 if (sizeof (longword
) > 8)
85 /* Instead of the traditional loop which tests each character,
86 we will test a longword at a time. The tricky part is testing
87 if *any of the four* bytes in the longword in question are zero. */
90 /* We tentatively exit the loop if adding MAGIC_BITS to
91 LONGWORD fails to change any of the hole bits of LONGWORD.
93 1) Is this safe? Will it catch all the zero bytes?
94 Suppose there is a byte with all zeros. Any carry bits
95 propagating from its left will fall into the hole at its
96 least significant bit and stop. Since there will be no
97 carry from its most significant bit, the LSB of the
98 byte to the left will be unchanged, and the zero will be
101 2) Is this worthwhile? Will it ignore everything except
102 zero bytes? Suppose every byte of LONGWORD has a bit set
103 somewhere. There will be a carry into bit 8. If bit 8
104 is set, this will carry into bit 16. If bit 8 is clear,
105 one of bits 9-15 must be set, so there will be a carry
106 into bit 16. Similarly, there will be a carry into bit
107 24. If one of bits 24-30 is set, there will be a carry
108 into bit 31, so all of the hole bits will be changed.
110 The one misfire occurs when bits 24-30 are clear and bit
111 31 is set; in this case, the hole at bit 31 is not
112 changed. If we had access to the processor carry flag,
113 we could close this loophole by putting the fourth hole
116 So it ignores everything except 128's, when they're aligned
119 3) But wait! Aren't we looking for C as well as zero?
120 Good point. So what we do is XOR LONGWORD with a longword,
121 each of whose bytes is C. This turns each byte that is C
124 longword
= *longword_ptr
++;
126 /* Add MAGIC_BITS to LONGWORD. */
127 if ((((longword
+ magic_bits
)
129 /* Set those bits that were unchanged by the addition. */
132 /* Look at only the hole bits. If any of the hole bits
133 are unchanged, most likely one of the bytes was a
135 & ~magic_bits
) != 0 ||
137 /* That caught zeroes. Now test for C. */
138 ((((longword
^ charmask
) + magic_bits
) ^ ~(longword
^ charmask
))
141 /* Which of the bytes was C or zero?
142 If none of them were, it was a misfire; continue the search. */
144 const unsigned char *cp
= (const unsigned char *) (longword_ptr
- 1);
148 else if (*cp
== '\0')
152 else if (*cp
== '\0')
156 else if (*cp
== '\0')
160 else if (*cp
== '\0')
162 if (sizeof (longword
) > 4)
166 else if (*cp
== '\0')
170 else if (*cp
== '\0')
174 else if (*cp
== '\0')
178 else if (*cp
== '\0')
189 weak_alias (strchr
, index
)
191 libc_hidden_builtin_def (strchr
)