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