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
52static const unsigned long mask01 = 0x01010101;
53static const unsigned long mask80 = 0x80808080;
54#elif LONG_BIT == 64
55static const unsigned long mask01 = 0x0101010101010101;
56static 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
73size_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