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122 lines
3.5 KiB
C
122 lines
3.5 KiB
C
/*-
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* SPDX-License-Identifier: BSD-2-Clause
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*
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* Copyright (c) 2009, 2010 Xin LI <delphij@FreeBSD.org>
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*
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* Redistribution and use in source and binary forms, with or without
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* modification, are permitted provided that the following conditions
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* are met:
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* 1. Redistributions of source code must retain the above copyright
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* notice, this list of conditions and the following disclaimer.
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* 2. Redistributions in binary form must reproduce the above copyright
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* notice, this list of conditions and the following disclaimer in the
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* documentation and/or other materials provided with the distribution.
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*
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* THIS SOFTWARE IS PROVIDED BY THE AUTHOR AND CONTRIBUTORS ``AS IS'' AND
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* ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
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* IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE
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* ARE DISCLAIMED. IN NO EVENT SHALL THE AUTHOR OR CONTRIBUTORS BE LIABLE
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* FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL
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* DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS
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* OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION)
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* HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT
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* LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY
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* OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF
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* SUCH DAMAGE.
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*/
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#include <sys/cdefs.h>
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#include <sys/libkern.h>
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#include <sys/limits.h>
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/*
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* Portable strlen() for 32-bit and 64-bit systems.
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*
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* The expression:
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*
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* ((x - 0x01....01) & ~x & 0x80....80)
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*
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* would evaluate to a non-zero value iff any of the bytes in the
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* original word is zero.
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*
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* The algorithm above is found on "Hacker's Delight" by
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* Henry S. Warren, Jr.
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*
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* Note: this leaves performance on the table and each architecture
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* would be best served with a tailor made routine instead, even if
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* using the same trick.
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*/
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/* Magic numbers for the algorithm */
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#if LONG_BIT == 32
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static const unsigned long mask01 = 0x01010101;
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static const unsigned long mask80 = 0x80808080;
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#elif LONG_BIT == 64
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static const unsigned long mask01 = 0x0101010101010101;
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static const unsigned long mask80 = 0x8080808080808080;
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#else
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#error Unsupported word size
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#endif
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#define LONGPTR_MASK (sizeof(long) - 1)
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/*
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* Helper macro to return string length if we caught the zero
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* byte.
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*/
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#define testbyte(x) \
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do { \
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if (p[x] == '\0') \
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return (p - str + x); \
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} while (0)
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size_t
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(strlen)(const char *str)
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{
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const char *p;
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const unsigned long *lp;
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long va, vb;
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/*
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* Before trying the hard (unaligned byte-by-byte access) way
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* to figure out whether there is a nul character, try to see
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* if there is a nul character is within this accessible word
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* first.
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*
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* p and (p & ~LONGPTR_MASK) must be equally accessible since
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* they always fall in the same memory page, as long as page
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* boundaries is integral multiple of word size.
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*/
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lp = (const unsigned long *)((uintptr_t)str & ~LONGPTR_MASK);
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va = (*lp - mask01);
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vb = ((~*lp) & mask80);
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lp++;
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if (va & vb)
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/* Check if we have \0 in the first part */
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for (p = str; p < (const char *)lp; p++)
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if (*p == '\0')
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return (p - str);
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/* Scan the rest of the string using word sized operation */
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for (; ; lp++) {
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va = (*lp - mask01);
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vb = ((~*lp) & mask80);
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if (va & vb) {
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p = (const char *)(lp);
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testbyte(0);
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testbyte(1);
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testbyte(2);
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testbyte(3);
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#if (LONG_BIT >= 64)
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testbyte(4);
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testbyte(5);
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testbyte(6);
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testbyte(7);
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#endif
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}
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}
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/* NOTREACHED */
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return (0);
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}
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