1/* memrchr -- find the last occurrence of a byte in a memory block 2 3 Copyright (C) 1991, 1993, 1996, 1997, 1999, 2000, 2003, 2004, 2005, 4 2006, 2007 Free Software Foundation, Inc. 5 6 Based on strlen implementation by Torbjorn Granlund (tege@sics.se), 7 with help from Dan Sahlin (dan@sics.se) and 8 commentary by Jim Blandy (jimb@ai.mit.edu); 9 adaptation to memchr suggested by Dick Karpinski (dick@cca.ucsf.edu), 10 and implemented by Roland McGrath (roland@ai.mit.edu). 11 12 This program is free software: you can redistribute it and/or modify 13 it under the terms of the GNU General Public License as published by 14 the Free Software Foundation; either version 3 of the License, or 15 (at your option) any later version. 16 17 This program is distributed in the hope that it will be useful, 18 but WITHOUT ANY WARRANTY; without even the implied warranty of 19 MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the 20 GNU General Public License for more details. 21 22 You should have received a copy of the GNU General Public License 23 along with this program. If not, see <http://www.gnu.org/licenses/>. */ 24 25#if defined _LIBC 26# include <memcopy.h> 27#else 28# include <config.h> 29# define reg_char char 30#endif 31 32#include <string.h> 33#include <limits.h> 34 35#undef __memrchr 36#undef memrchr 37 38#ifndef weak_alias 39# define __memrchr memrchr 40#endif 41 42/* Search no more than N bytes of S for C. */ 43void * 44__memrchr (void const *s, int c_in, size_t n) 45{ 46 const unsigned char *char_ptr; 47 const unsigned long int *longword_ptr; 48 unsigned long int longword, magic_bits, charmask; 49 unsigned reg_char c; 50 int i; 51 52 c = (unsigned char) c_in; 53 54 /* Handle the last few characters by reading one character at a time. 55 Do this until CHAR_PTR is aligned on a longword boundary. */ 56 for (char_ptr = (const unsigned char *) s + n; 57 n > 0 && (size_t) char_ptr % sizeof longword != 0; 58 --n) 59 if (*--char_ptr == c) 60 return (void *) char_ptr; 61 62 /* All these elucidatory comments refer to 4-byte longwords, 63 but the theory applies equally well to any size longwords. */ 64 65 longword_ptr = (const unsigned long int *) char_ptr; 66 67 /* Bits 31, 24, 16, and 8 of this number are zero. Call these bits 68 the "holes." Note that there is a hole just to the left of 69 each byte, with an extra at the end: 70 71 bits: 01111110 11111110 11111110 11111111 72 bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD 73 74 The 1-bits make sure that carries propagate to the next 0-bit. 75 The 0-bits provide holes for carries to fall into. */ 76 77 /* Set MAGIC_BITS to be this pattern of 1 and 0 bits. 78 Set CHARMASK to be a longword, each of whose bytes is C. */ 79 80 magic_bits = 0xfefefefe; 81 charmask = c | (c << 8); 82 charmask |= charmask << 16; 83#if 0xffffffffU < ULONG_MAX 84 magic_bits |= magic_bits << 32; 85 charmask |= charmask << 32; 86 if (8 < sizeof longword) 87 for (i = 64; i < sizeof longword * 8; i *= 2) 88 { 89 magic_bits |= magic_bits << i; 90 charmask |= charmask << i; 91 } 92#endif 93 magic_bits = (ULONG_MAX >> 1) & (magic_bits | 1); 94 95 /* Instead of the traditional loop which tests each character, 96 we will test a longword at a time. The tricky part is testing 97 if *any of the four* bytes in the longword in question are zero. */ 98 while (n >= sizeof longword) 99 { 100 /* We tentatively exit the loop if adding MAGIC_BITS to 101 LONGWORD fails to change any of the hole bits of LONGWORD. 102 103 1) Is this safe? Will it catch all the zero bytes? 104 Suppose there is a byte with all zeros. Any carry bits 105 propagating from its left will fall into the hole at its 106 least significant bit and stop. Since there will be no 107 carry from its most significant bit, the LSB of the 108 byte to the left will be unchanged, and the zero will be 109 detected. 110 111 2) Is this worthwhile? Will it ignore everything except 112 zero bytes? Suppose every byte of LONGWORD has a bit set 113 somewhere. There will be a carry into bit 8. If bit 8 114 is set, this will carry into bit 16. If bit 8 is clear, 115 one of bits 9-15 must be set, so there will be a carry 116 into bit 16. Similarly, there will be a carry into bit 117 24. If one of bits 24-30 is set, there will be a carry 118 into bit 31, so all of the hole bits will be changed. 119 120 The one misfire occurs when bits 24-30 are clear and bit 121 31 is set; in this case, the hole at bit 31 is not 122 changed. If we had access to the processor carry flag, 123 we could close this loophole by putting the fourth hole 124 at bit 32! 125 126 So it ignores everything except 128's, when they're aligned 127 properly. 128 129 3) But wait! Aren't we looking for C, not zero? 130 Good point. So what we do is XOR LONGWORD with a longword, 131 each of whose bytes is C. This turns each byte that is C 132 into a zero. */ 133 134 longword = *--longword_ptr ^ charmask; 135 136 /* Add MAGIC_BITS to LONGWORD. */ 137 if ((((longword + magic_bits) 138 139 /* Set those bits that were unchanged by the addition. */ 140 ^ ~longword) 141 142 /* Look at only the hole bits. If any of the hole bits 143 are unchanged, most likely one of the bytes was a 144 zero. */ 145 & ~magic_bits) != 0) 146 { 147 /* Which of the bytes was C? If none of them were, it was 148 a misfire; continue the search. */ 149 150 const unsigned char *cp = (const unsigned char *) longword_ptr; 151 152 if (8 < sizeof longword) 153 for (i = sizeof longword - 1; 8 <= i; i--) 154 if (cp[i] == c) 155 return (void *) &cp[i]; 156 if (7 < sizeof longword && cp[7] == c) 157 return (void *) &cp[7]; 158 if (6 < sizeof longword && cp[6] == c) 159 return (void *) &cp[6]; 160 if (5 < sizeof longword && cp[5] == c) 161 return (void *) &cp[5]; 162 if (4 < sizeof longword && cp[4] == c) 163 return (void *) &cp[4]; 164 if (cp[3] == c) 165 return (void *) &cp[3]; 166 if (cp[2] == c) 167 return (void *) &cp[2]; 168 if (cp[1] == c) 169 return (void *) &cp[1]; 170 if (cp[0] == c) 171 return (void *) cp; 172 } 173 174 n -= sizeof longword; 175 } 176 177 char_ptr = (const unsigned char *) longword_ptr; 178 179 while (n-- > 0) 180 { 181 if (*--char_ptr == c) 182 return (void *) char_ptr; 183 } 184 185 return 0; 186} 187#ifdef weak_alias 188weak_alias (__memrchr, memrchr) 189#endif 190