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1/* Find the length of STRING, but scan at most MAXLEN characters.
2 Copyright (C) 1991-2019 Free Software Foundation, Inc.
3 Contributed by Jakub Jelinek <jakub@redhat.com>.
4
5 Based on strlen written by Torbjorn Granlund (tege@sics.se),
6 with help from Dan Sahlin (dan@sics.se);
7 commentary by Jim Blandy (jimb@ai.mit.edu).
8
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 License as
11 published by the Free Software Foundation; either version 2.1 of the
12 License, or (at your option) any later version.
13
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.
18
19 You should have received a copy of the GNU Lesser General Public
20 License along with the GNU C Library; see the file COPYING.LIB. If
21 not, see <http://www.gnu.org/licenses/>. */
22
23#include <string.h>
24#include <stdlib.h>
25
26/* Find the length of S, but scan at most MAXLEN characters. If no
27 '\0' terminator is found in that many characters, return MAXLEN. */
28
29#ifdef STRNLEN
30# define __strnlen STRNLEN
31#endif
32
33size_t
34__strnlen (const char *str, size_t maxlen)
35{
36 const char *char_ptr, *end_ptr = str + maxlen;
37 const unsigned long int *longword_ptr;
38 unsigned long int longword, himagic, lomagic;
39
40 if (maxlen == 0)
41 return 0;
42
43 if (__glibc_unlikely (end_ptr < str))
44 end_ptr = (const char *) ~0UL;
45
46 /* Handle the first few characters by reading one character at a time.
47 Do this until CHAR_PTR is aligned on a longword boundary. */
48 for (char_ptr = str; ((unsigned long int) char_ptr
49 & (sizeof (longword) - 1)) != 0;
50 ++char_ptr)
51 if (*char_ptr == '\0')
52 {
53 if (char_ptr > end_ptr)
54 char_ptr = end_ptr;
55 return char_ptr - str;
56 }
57
58 /* All these elucidatory comments refer to 4-byte longwords,
59 but the theory applies equally well to 8-byte longwords. */
60
61 longword_ptr = (unsigned long int *) char_ptr;
62
63 /* Bits 31, 24, 16, and 8 of this number are zero. Call these bits
64 the "holes." Note that there is a hole just to the left of
65 each byte, with an extra at the end:
66
67 bits: 01111110 11111110 11111110 11111111
68 bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD
69
70 The 1-bits make sure that carries propagate to the next 0-bit.
71 The 0-bits provide holes for carries to fall into. */
72 himagic = 0x80808080L;
73 lomagic = 0x01010101L;
74 if (sizeof (longword) > 4)
75 {
76 /* 64-bit version of the magic. */
77 /* Do the shift in two steps to avoid a warning if long has 32 bits. */
78 himagic = ((himagic << 16) << 16) | himagic;
79 lomagic = ((lomagic << 16) << 16) | lomagic;
80 }
81 if (sizeof (longword) > 8)
82 abort ();
83
84 /* Instead of the traditional loop which tests each character,
85 we will test a longword at a time. The tricky part is testing
86 if *any of the four* bytes in the longword in question are zero. */
87 while (longword_ptr < (unsigned long int *) end_ptr)
88 {
89 /* We tentatively exit the loop if adding MAGIC_BITS to
90 LONGWORD fails to change any of the hole bits of LONGWORD.
91
92 1) Is this safe? Will it catch all the zero bytes?
93 Suppose there is a byte with all zeros. Any carry bits
94 propagating from its left will fall into the hole at its
95 least significant bit and stop. Since there will be no
96 carry from its most significant bit, the LSB of the
97 byte to the left will be unchanged, and the zero will be
98 detected.
99
100 2) Is this worthwhile? Will it ignore everything except
101 zero bytes? Suppose every byte of LONGWORD has a bit set
102 somewhere. There will be a carry into bit 8. If bit 8
103 is set, this will carry into bit 16. If bit 8 is clear,
104 one of bits 9-15 must be set, so there will be a carry
105 into bit 16. Similarly, there will be a carry into bit
106 24. If one of bits 24-30 is set, there will be a carry
107 into bit 31, so all of the hole bits will be changed.
108
109 The one misfire occurs when bits 24-30 are clear and bit
110 31 is set; in this case, the hole at bit 31 is not
111 changed. If we had access to the processor carry flag,
112 we could close this loophole by putting the fourth hole
113 at bit 32!
114
115 So it ignores everything except 128's, when they're aligned
116 properly. */
117
118 longword = *longword_ptr++;
119
120 if ((longword - lomagic) & himagic)
121 {
122 /* Which of the bytes was the zero? If none of them were, it was
123 a misfire; continue the search. */
124
125 const char *cp = (const char *) (longword_ptr - 1);
126
127 char_ptr = cp;
128 if (cp[0] == 0)
129 break;
130 char_ptr = cp + 1;
131 if (cp[1] == 0)
132 break;
133 char_ptr = cp + 2;
134 if (cp[2] == 0)
135 break;
136 char_ptr = cp + 3;
137 if (cp[3] == 0)
138 break;
139 if (sizeof (longword) > 4)
140 {
141 char_ptr = cp + 4;
142 if (cp[4] == 0)
143 break;
144 char_ptr = cp + 5;
145 if (cp[5] == 0)
146 break;
147 char_ptr = cp + 6;
148 if (cp[6] == 0)
149 break;
150 char_ptr = cp + 7;
151 if (cp[7] == 0)
152 break;
153 }
154 }
155 char_ptr = end_ptr;
156 }
157
158 if (char_ptr > end_ptr)
159 char_ptr = end_ptr;
160 return char_ptr - str;
161}
162#ifndef STRNLEN
163libc_hidden_def (__strnlen)
164weak_alias (__strnlen, strnlen)
165#endif
166libc_hidden_def (strnlen)
167

Warning: That file was not part of the compilation database. It may have many parsing errors.