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