/*- * SPDX-License-Identifier: BSD-2-Clause * * Copyright (c) 2009, 2010 Xin LI * * 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 #include #include #include /* * Portable strlen() for 32-bit and 64-bit systems. * * Rationale: it is generally much more efficient to do word length * operations and avoid branches on modern computer systems, as * compared to byte-length operations with a lot of branches. * * 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. * * On multi-issue processors, we can divide the above expression into: * a) (x - 0x01....01) * b) (~x & 0x80....80) * c) a & b * * Where, a) and b) can be partially computed in parallel. * * The algorithm above is found on "Hacker's Delight" by * Henry S. Warren, Jr. */ /* 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); }