1 /* Copyright (C) 1991-2016 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).
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.
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.
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/>. */
31 # define STRCHRNUL __strchrnul
34 /* Find the first occurrence of C in S or the final NUL byte. */
36 STRCHRNUL (const char *s
, int c_in
)
38 const unsigned char *char_ptr
;
39 const unsigned long int *longword_ptr
;
40 unsigned long int longword
, magic_bits
, charmask
;
43 c
= (unsigned char) c_in
;
45 /* Handle the first few characters by reading one character at a time.
46 Do this until CHAR_PTR is aligned on a longword boundary. */
47 for (char_ptr
= (const unsigned char *) s
;
48 ((unsigned long int) char_ptr
& (sizeof (longword
) - 1)) != 0;
50 if (*char_ptr
== c
|| *char_ptr
== '\0')
51 return (void *) char_ptr
;
53 /* All these elucidatory comments refer to 4-byte longwords,
54 but the theory applies equally well to 8-byte longwords. */
56 longword_ptr
= (unsigned long int *) char_ptr
;
58 /* Bits 31, 24, 16, and 8 of this number are zero. Call these bits
59 the "holes." Note that there is a hole just to the left of
60 each byte, with an extra at the end:
62 bits: 01111110 11111110 11111110 11111111
63 bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD
65 The 1-bits make sure that carries propagate to the next 0-bit.
66 The 0-bits provide holes for carries to fall into. */
68 magic_bits
= magic_bits
/ 0xff * 0xfe << 1 >> 1 | 1;
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)
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. */
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
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
110 So it ignores everything except 128's, when they're aligned
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
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. */
126 /* Look at only the hole bits. If any of the hole bits
127 are unchanged, most likely one of the bytes was a
129 & ~magic_bits
) != 0 ||
131 /* That caught zeroes. Now test for C. */
132 ((((longword
^ charmask
) + magic_bits
) ^ ~(longword
^ charmask
))
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
|| *cp
== '\0')
142 if (*++cp
== c
|| *cp
== '\0')
144 if (*++cp
== c
|| *cp
== '\0')
146 if (*++cp
== c
|| *cp
== '\0')
148 if (sizeof (longword
) > 4)
150 if (*++cp
== c
|| *cp
== '\0')
152 if (*++cp
== c
|| *cp
== '\0')
154 if (*++cp
== c
|| *cp
== '\0')
156 if (*++cp
== c
|| *cp
== '\0')
162 /* This should never happen. */
166 weak_alias (__strchrnul
, strchrnul
)