freebsd-skq/sys/libkern/strlen.c
Mateusz Guzik 3acea07c18 Restore the augmented strlen commentary
... lost in revert
2021-02-08 19:15:21 +00:00

124 lines
3.5 KiB
C

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