2 * SPDX-License-Identifier: BSD-2-Clause-FreeBSD
4 * Copyright (c) 2009, 2010 Xin LI <delphij@FreeBSD.org>
7 * Redistribution and use in source and binary forms, with or without
8 * modification, are permitted provided that the following conditions
10 * 1. Redistributions of source code must retain the above copyright
11 * notice, this list of conditions and the following disclaimer.
12 * 2. Redistributions in binary form must reproduce the above copyright
13 * notice, this list of conditions and the following disclaimer in the
14 * documentation and/or other materials provided with the distribution.
16 * THIS SOFTWARE IS PROVIDED BY THE AUTHOR AND CONTRIBUTORS ``AS IS'' AND
17 * ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
18 * IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE
19 * ARE DISCLAIMED. IN NO EVENT SHALL THE AUTHOR OR CONTRIBUTORS BE LIABLE
20 * FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL
21 * DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS
22 * OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION)
23 * HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT
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29 #include <sys/cdefs.h>
30 __FBSDID("$FreeBSD$");
32 #include <sys/libkern.h>
33 #include <sys/limits.h>
36 * Portable strlen() for 32-bit and 64-bit systems.
38 * Rationale: it is generally much more efficient to do word length
39 * operations and avoid branches on modern computer systems, as
40 * compared to byte-length operations with a lot of branches.
44 * ((x - 0x01....01) & ~x & 0x80....80)
46 * would evaluate to a non-zero value iff any of the bytes in the
47 * original word is zero.
49 * On multi-issue processors, we can divide the above expression into:
51 * b) (~x & 0x80....80)
54 * Where, a) and b) can be partially computed in parallel.
56 * The algorithm above is found on "Hacker's Delight" by
57 * Henry S. Warren, Jr.
60 /* Magic numbers for the algorithm */
62 static const unsigned long mask01 = 0x01010101;
63 static const unsigned long mask80 = 0x80808080;
65 static const unsigned long mask01 = 0x0101010101010101;
66 static const unsigned long mask80 = 0x8080808080808080;
68 #error Unsupported word size
71 #define LONGPTR_MASK (sizeof(long) - 1)
74 * Helper macro to return string length if we caught the zero
80 return (p - str + x); \
84 strlen(const char *str)
87 const unsigned long *lp;
91 * Before trying the hard (unaligned byte-by-byte access) way
92 * to figure out whether there is a nul character, try to see
93 * if there is a nul character is within this accessible word
96 * p and (p & ~LONGPTR_MASK) must be equally accessible since
97 * they always fall in the same memory page, as long as page
98 * boundaries is integral multiple of word size.
100 lp = (const unsigned long *)((uintptr_t)str & ~LONGPTR_MASK);
102 vb = ((~*lp) & mask80);
105 /* Check if we have \0 in the first part */
106 for (p = str; p < (const char *)lp; p++)
110 /* Scan the rest of the string using word sized operation */
113 vb = ((~*lp) & mask80);
115 p = (const char *)(lp);