1 /*-
2 * SPDX-License-Identifier: BSD-2-Clause
3 *
4 * Copyright (c) 2009, 2010 Xin LI <delphij@FreeBSD.org>
5 *
6 * Redistribution and use in source and binary forms, with or without
7 * modification, are permitted provided that the following conditions
8 * are met:
9 * 1. Redistributions of source code must retain the above copyright
10 * notice, this list of conditions and the following disclaimer.
11 * 2. Redistributions in binary form must reproduce the above copyright
12 * notice, this list of conditions and the following disclaimer in the
13 * documentation and/or other materials provided with the distribution.
14 *
15 * THIS SOFTWARE IS PROVIDED BY THE AUTHOR AND CONTRIBUTORS ``AS IS'' AND
16 * ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
17 * IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE
18 * ARE DISCLAIMED. IN NO EVENT SHALL THE AUTHOR OR CONTRIBUTORS BE LIABLE
19 * FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL
20 * DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS
21 * OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION)
22 * HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT
23 * LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY
24 * OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF
25 * SUCH DAMAGE.
26 */
27
28 #include <sys/cdefs.h>
29 #include <sys/libkern.h>
30 #include <sys/limits.h>
31
32 /*
33 * Portable strlen() for 32-bit and 64-bit systems.
34 *
35 * The expression:
36 *
37 * ((x - 0x01....01) & ~x & 0x80....80)
38 *
39 * would evaluate to a non-zero value iff any of the bytes in the
40 * original word is zero.
41 *
42 * The algorithm above is found on "Hacker's Delight" by
43 * Henry S. Warren, Jr.
44 *
45 * Note: this leaves performance on the table and each architecture
46 * would be best served with a tailor made routine instead, even if
47 * using the same trick.
48 */
49
50 /* Magic numbers for the algorithm */
51 #if LONG_BIT == 32
52 static const unsigned long mask01 = 0x01010101;
53 static const unsigned long mask80 = 0x80808080;
54 #elif LONG_BIT == 64
55 static const unsigned long mask01 = 0x0101010101010101;
56 static const unsigned long mask80 = 0x8080808080808080;
57 #else
58 #error Unsupported word size
59 #endif
60
61 #define LONGPTR_MASK (sizeof(long) - 1)
62
63 /*
64 * Helper macro to return string length if we caught the zero
65 * byte.
66 */
67 #define testbyte(x) \
68 do { \
69 if (p[x] == '\0') \
70 return (p - str + x); \
71 } while (0)
72
size_t(strlen)73 size_t
74 (strlen)(const char *str)
75 {
76 const char *p;
77 const unsigned long *lp;
78 long va, vb;
79
80 /*
81 * Before trying the hard (unaligned byte-by-byte access) way
82 * to figure out whether there is a nul character, try to see
83 * if there is a nul character is within this accessible word
84 * first.
85 *
86 * p and (p & ~LONGPTR_MASK) must be equally accessible since
87 * they always fall in the same memory page, as long as page
88 * boundaries is integral multiple of word size.
89 */
90 lp = (const unsigned long *)((uintptr_t)str & ~LONGPTR_MASK);
91 va = (*lp - mask01);
92 vb = ((~*lp) & mask80);
93 lp++;
94 if (va & vb)
95 /* Check if we have \0 in the first part */
96 for (p = str; p < (const char *)lp; p++)
97 if (*p == '\0')
98 return (p - str);
99
100 /* Scan the rest of the string using word sized operation */
101 for (; ; lp++) {
102 va = (*lp - mask01);
103 vb = ((~*lp) & mask80);
104 if (va & vb) {
105 p = (const char *)(lp);
106 testbyte(0);
107 testbyte(1);
108 testbyte(2);
109 testbyte(3);
110 #if (LONG_BIT >= 64)
111 testbyte(4);
112 testbyte(5);
113 testbyte(6);
114 testbyte(7);
115 #endif
116 }
117 }
118
119 /* NOTREACHED */
120 return (0);
121 }
122