1 /* Find the length of STRING, but scan at most MAXLEN characters.
2 Copyright (C) 1991-2022 Free Software Foundation, Inc.
3
4 Based on strlen written by Torbjorn Granlund (tege@sics.se),
5 with help from Dan Sahlin (dan@sics.se);
6 commentary by Jim Blandy (jimb@ai.mit.edu).
7
8 The GNU C Library is free software; you can redistribute it and/or
9 modify it under the terms of the GNU Lesser General Public License as
10 published by the Free Software Foundation; either version 2.1 of the
11 License, or (at your option) any later version.
12
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 Lesser General Public License for more details.
17
18 You should have received a copy of the GNU Lesser General Public
19 License along with the GNU C Library; see the file COPYING.LIB. If
20 not, see <https://www.gnu.org/licenses/>. */
21
22 #include <string.h>
23 #include <stdlib.h>
24
25 /* Find the length of S, but scan at most MAXLEN characters. If no
26 '\0' terminator is found in that many characters, return MAXLEN. */
27
28 #ifdef STRNLEN
29 # define __strnlen STRNLEN
30 #endif
31
32 size_t
__strnlen(const char * str,size_t maxlen)33 __strnlen (const char *str, size_t maxlen)
34 {
35 const char *char_ptr, *end_ptr = str + maxlen;
36 const unsigned long int *longword_ptr;
37 unsigned long int longword, himagic, lomagic;
38
39 if (maxlen == 0)
40 return 0;
41
42 if (__glibc_unlikely (end_ptr < str))
43 end_ptr = (const char *) ~0UL;
44
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 = str; ((unsigned long int) char_ptr
48 & (sizeof (longword) - 1)) != 0;
49 ++char_ptr)
50 if (*char_ptr == '\0')
51 {
52 if (char_ptr > end_ptr)
53 char_ptr = end_ptr;
54 return char_ptr - str;
55 }
56
57 /* All these elucidatory comments refer to 4-byte longwords,
58 but the theory applies equally well to 8-byte longwords. */
59
60 longword_ptr = (unsigned long int *) char_ptr;
61
62 /* Bits 31, 24, 16, and 8 of this number are zero. Call these bits
63 the "holes." Note that there is a hole just to the left of
64 each byte, with an extra at the end:
65
66 bits: 01111110 11111110 11111110 11111111
67 bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD
68
69 The 1-bits make sure that carries propagate to the next 0-bit.
70 The 0-bits provide holes for carries to fall into. */
71 himagic = 0x80808080L;
72 lomagic = 0x01010101L;
73 if (sizeof (longword) > 4)
74 {
75 /* 64-bit version of the magic. */
76 /* Do the shift in two steps to avoid a warning if long has 32 bits. */
77 himagic = ((himagic << 16) << 16) | himagic;
78 lomagic = ((lomagic << 16) << 16) | lomagic;
79 }
80 if (sizeof (longword) > 8)
81 abort ();
82
83 /* Instead of the traditional loop which tests each character,
84 we will test a longword at a time. The tricky part is testing
85 if *any of the four* bytes in the longword in question are zero. */
86 while (longword_ptr < (unsigned long int *) end_ptr)
87 {
88 /* We tentatively exit the loop if adding MAGIC_BITS to
89 LONGWORD fails to change any of the hole bits of LONGWORD.
90
91 1) Is this safe? Will it catch all the zero bytes?
92 Suppose there is a byte with all zeros. Any carry bits
93 propagating from its left will fall into the hole at its
94 least significant bit and stop. Since there will be no
95 carry from its most significant bit, the LSB of the
96 byte to the left will be unchanged, and the zero will be
97 detected.
98
99 2) Is this worthwhile? Will it ignore everything except
100 zero bytes? Suppose every byte of LONGWORD has a bit set
101 somewhere. There will be a carry into bit 8. If bit 8
102 is set, this will carry into bit 16. If bit 8 is clear,
103 one of bits 9-15 must be set, so there will be a carry
104 into bit 16. Similarly, there will be a carry into bit
105 24. If one of bits 24-30 is set, there will be a carry
106 into bit 31, so all of the hole bits will be changed.
107
108 The one misfire occurs when bits 24-30 are clear and bit
109 31 is set; in this case, the hole at bit 31 is not
110 changed. If we had access to the processor carry flag,
111 we could close this loophole by putting the fourth hole
112 at bit 32!
113
114 So it ignores everything except 128's, when they're aligned
115 properly. */
116
117 longword = *longword_ptr++;
118
119 if ((longword - lomagic) & himagic)
120 {
121 /* Which of the bytes was the zero? If none of them were, it was
122 a misfire; continue the search. */
123
124 const char *cp = (const char *) (longword_ptr - 1);
125
126 char_ptr = cp;
127 if (cp[0] == 0)
128 break;
129 char_ptr = cp + 1;
130 if (cp[1] == 0)
131 break;
132 char_ptr = cp + 2;
133 if (cp[2] == 0)
134 break;
135 char_ptr = cp + 3;
136 if (cp[3] == 0)
137 break;
138 if (sizeof (longword) > 4)
139 {
140 char_ptr = cp + 4;
141 if (cp[4] == 0)
142 break;
143 char_ptr = cp + 5;
144 if (cp[5] == 0)
145 break;
146 char_ptr = cp + 6;
147 if (cp[6] == 0)
148 break;
149 char_ptr = cp + 7;
150 if (cp[7] == 0)
151 break;
152 }
153 }
154 char_ptr = end_ptr;
155 }
156
157 if (char_ptr > end_ptr)
158 char_ptr = end_ptr;
159 return char_ptr - str;
160 }
161 #ifndef STRNLEN
162 libc_hidden_def (__strnlen)
163 weak_alias (__strnlen, strnlen)
164 #endif
165 libc_hidden_def (strnlen)
166