S390: Ifunc resolver macro for vector instructions.
[glibc.git] / string / strchrnul.c
blobdaf0b3f659d6d0e9bfb3c79448f3d9593bfbe8ba
1 /* Copyright (C) 1991-2015 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/>. */
23 #include <string.h>
24 #include <memcopy.h>
25 #include <stdlib.h>
27 #undef __strchrnul
28 #undef strchrnul
30 #ifndef STRCHRNUL
31 # define STRCHRNUL __strchrnul
32 #endif
34 /* Find the first occurrence of C in S or the final NUL byte. */
35 char *
36 STRCHRNUL (s, c_in)
37 const char *s;
38 int c_in;
40 const unsigned char *char_ptr;
41 const unsigned long int *longword_ptr;
42 unsigned long int longword, magic_bits, charmask;
43 unsigned char c;
45 c = (unsigned char) c_in;
47 /* Handle the first few characters by reading one character at a time.
48 Do this until CHAR_PTR is aligned on a longword boundary. */
49 for (char_ptr = (const unsigned char *) s;
50 ((unsigned long int) char_ptr & (sizeof (longword) - 1)) != 0;
51 ++char_ptr)
52 if (*char_ptr == c || *char_ptr == '\0')
53 return (void *) char_ptr;
55 /* All these elucidatory comments refer to 4-byte longwords,
56 but the theory applies equally well to 8-byte longwords. */
58 longword_ptr = (unsigned long int *) char_ptr;
60 /* Bits 31, 24, 16, and 8 of this number are zero. Call these bits
61 the "holes." Note that there is a hole just to the left of
62 each byte, with an extra at the end:
64 bits: 01111110 11111110 11111110 11111111
65 bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD
67 The 1-bits make sure that carries propagate to the next 0-bit.
68 The 0-bits provide holes for carries to fall into. */
69 magic_bits = -1;
70 magic_bits = magic_bits / 0xff * 0xfe << 1 >> 1 | 1;
72 /* Set up a longword, each of whose bytes is C. */
73 charmask = c | (c << 8);
74 charmask |= charmask << 16;
75 if (sizeof (longword) > 4)
76 /* Do the shift in two steps to avoid a warning if long has 32 bits. */
77 charmask |= (charmask << 16) << 16;
78 if (sizeof (longword) > 8)
79 abort ();
81 /* Instead of the traditional loop which tests each character,
82 we will test a longword at a time. The tricky part is testing
83 if *any of the four* bytes in the longword in question are zero. */
84 for (;;)
86 /* We tentatively exit the loop if adding MAGIC_BITS to
87 LONGWORD fails to change any of the hole bits of LONGWORD.
89 1) Is this safe? Will it catch all the zero bytes?
90 Suppose there is a byte with all zeros. Any carry bits
91 propagating from its left will fall into the hole at its
92 least significant bit and stop. Since there will be no
93 carry from its most significant bit, the LSB of the
94 byte to the left will be unchanged, and the zero will be
95 detected.
97 2) Is this worthwhile? Will it ignore everything except
98 zero bytes? Suppose every byte of LONGWORD has a bit set
99 somewhere. There will be a carry into bit 8. If bit 8
100 is set, this will carry into bit 16. If bit 8 is clear,
101 one of bits 9-15 must be set, so there will be a carry
102 into bit 16. Similarly, there will be a carry into bit
103 24. If one of bits 24-30 is set, there will be a carry
104 into bit 31, so all of the hole bits will be changed.
106 The one misfire occurs when bits 24-30 are clear and bit
107 31 is set; in this case, the hole at bit 31 is not
108 changed. If we had access to the processor carry flag,
109 we could close this loophole by putting the fourth hole
110 at bit 32!
112 So it ignores everything except 128's, when they're aligned
113 properly.
115 3) But wait! Aren't we looking for C as well as zero?
116 Good point. So what we do is XOR LONGWORD with a longword,
117 each of whose bytes is C. This turns each byte that is C
118 into a zero. */
120 longword = *longword_ptr++;
122 /* Add MAGIC_BITS to LONGWORD. */
123 if ((((longword + magic_bits)
125 /* Set those bits that were unchanged by the addition. */
126 ^ ~longword)
128 /* Look at only the hole bits. If any of the hole bits
129 are unchanged, most likely one of the bytes was a
130 zero. */
131 & ~magic_bits) != 0 ||
133 /* That caught zeroes. Now test for C. */
134 ((((longword ^ charmask) + magic_bits) ^ ~(longword ^ charmask))
135 & ~magic_bits) != 0)
137 /* Which of the bytes was C or zero?
138 If none of them were, it was a misfire; continue the search. */
140 const unsigned char *cp = (const unsigned char *) (longword_ptr - 1);
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 (*++cp == c || *cp == '\0')
149 return (char *) cp;
150 if (sizeof (longword) > 4)
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 if (*++cp == c || *cp == '\0')
159 return (char *) cp;
164 /* This should never happen. */
165 return NULL;
168 weak_alias (__strchrnul, strchrnul)