1 /*- 2 * SPDX-License-Identifier: BSD-2-Clause-FreeBSD 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 __FBSDID("$FreeBSD$"); 30 31 #include <sys/limits.h> 32 #include <sys/types.h> 33 #include <string.h> 34 35 /* 36 * Portable strlen() for 32-bit and 64-bit systems. 37 * 38 * The expression: 39 * 40 * ((x - 0x01....01) & ~x & 0x80....80) 41 * 42 * would evaluate to a non-zero value iff any of the bytes in the 43 * original word is zero. 44 * 45 * The algorithm above is found on "Hacker's Delight" by 46 * Henry S. Warren, Jr. 47 * 48 * Note: this leaves performance on the table and each architecture 49 * would be best served with a tailor made routine instead, even if 50 * using the same trick. 51 */ 52 53 /* Magic numbers for the algorithm */ 54 #if LONG_BIT == 32 55 static const unsigned long mask01 = 0x01010101; 56 static const unsigned long mask80 = 0x80808080; 57 #elif LONG_BIT == 64 58 static const unsigned long mask01 = 0x0101010101010101; 59 static const unsigned long mask80 = 0x8080808080808080; 60 #else 61 #error Unsupported word size 62 #endif 63 64 #define LONGPTR_MASK (sizeof(long) - 1) 65 66 /* 67 * Helper macro to return string length if we caught the zero 68 * byte. 69 */ 70 #define testbyte(x) \ 71 do { \ 72 if (p[x] == '\0') \ 73 return (p - str + x); \ 74 } while (0) 75 76 size_t 77 strlen(const char *str) 78 { 79 const char *p; 80 const unsigned long *lp; 81 long va, vb; 82 83 /* 84 * Before trying the hard (unaligned byte-by-byte access) way 85 * to figure out whether there is a nul character, try to see 86 * if there is a nul character is within this accessible word 87 * first. 88 * 89 * p and (p & ~LONGPTR_MASK) must be equally accessible since 90 * they always fall in the same memory page, as long as page 91 * boundaries is integral multiple of word size. 92 */ 93 lp = (const unsigned long *)((uintptr_t)str & ~LONGPTR_MASK); 94 va = (*lp - mask01); 95 vb = ((~*lp) & mask80); 96 lp++; 97 if (va & vb) 98 /* Check if we have \0 in the first part */ 99 for (p = str; p < (const char *)lp; p++) 100 if (*p == '\0') 101 return (p - str); 102 103 /* Scan the rest of the string using word sized operation */ 104 for (; ; lp++) { 105 va = (*lp - mask01); 106 vb = ((~*lp) & mask80); 107 if (va & vb) { 108 p = (const char *)(lp); 109 testbyte(0); 110 testbyte(1); 111 testbyte(2); 112 testbyte(3); 113 #if (LONG_BIT >= 64) 114 testbyte(4); 115 testbyte(5); 116 testbyte(6); 117 testbyte(7); 118 #endif 119 } 120 } 121 122 /* NOTREACHED */ 123 return (0); 124 } 125