1 /* Copyright (C) 1991, 1993, 1996, 1997, 1999, 2000, 2003, 2004, 2006, 2008 2 Free Software Foundation, Inc. 3 4 Based on strlen implementation by Torbjorn Granlund (tege (at) sics.se), 5 with help from Dan Sahlin (dan (at) sics.se) and 6 commentary by Jim Blandy (jimb (at) ai.mit.edu); 7 adaptation to memchr suggested by Dick Karpinski (dick (at) cca.ucsf.edu), 8 and implemented by Roland McGrath (roland (at) ai.mit.edu). 9 10 NOTE: The canonical source of this file is maintained with the GNU C Library. 11 Bugs can be reported to bug-glibc (at) prep.ai.mit.edu. 12 13 This program is free software: you can redistribute it and/or modify it 14 under the terms of the GNU General Public License as published by the 15 Free Software Foundation; either version 3 of the License, or any 16 later version. 17 18 This program is distributed in the hope that it will be useful, 19 but WITHOUT ANY WARRANTY; without even the implied warranty of 20 MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the 21 GNU General Public License for more details. 22 23 You should have received a copy of the GNU General Public License 24 along with this program. If not, see <http://www.gnu.org/licenses/>. */ 25 26 #ifndef _LIBC 27 # include <config.h> 28 #endif 29 30 #include <string.h> 31 32 #include <stddef.h> 33 34 #if defined _LIBC 35 # include <memcopy.h> 36 #else 37 # define reg_char char 38 #endif 39 40 #include <limits.h> 41 42 #if HAVE_BP_SYM_H || defined _LIBC 43 # include <bp-sym.h> 44 #else 45 # define BP_SYM(sym) sym 46 #endif 47 48 #undef __memchr 49 #ifdef _LIBC 50 # undef memchr 51 #endif 52 53 #ifndef weak_alias 54 # define __memchr memchr 55 #endif 56 57 /* Search no more than N bytes of S for C. */ 58 void * 59 __memchr (void const *s, int c_in, size_t n) 60 { 61 /* On 32-bit hardware, choosing longword to be a 32-bit unsigned 62 long instead of a 64-bit uintmax_t tends to give better 63 performance. On 64-bit hardware, unsigned long is generally 64 64 bits already. Change this typedef to experiment with 65 performance. */ 66 typedef unsigned long int longword; 67 68 const unsigned char *char_ptr; 69 const longword *longword_ptr; 70 longword repeated_one; 71 longword repeated_c; 72 unsigned reg_char c; 73 74 c = (unsigned char) c_in; 75 76 /* Handle the first few bytes by reading one byte at a time. 77 Do this until CHAR_PTR is aligned on a longword boundary. */ 78 for (char_ptr = (const unsigned char *) s; 79 n > 0 && (size_t) char_ptr % sizeof (longword) != 0; 80 --n, ++char_ptr) 81 if (*char_ptr == c) 82 return (void *) char_ptr; 83 84 longword_ptr = (const longword *) char_ptr; 85 86 /* All these elucidatory comments refer to 4-byte longwords, 87 but the theory applies equally well to any size longwords. */ 88 89 /* Compute auxiliary longword values: 90 repeated_one is a value which has a 1 in every byte. 91 repeated_c has c in every byte. */ 92 repeated_one = 0x01010101; 93 repeated_c = c | (c << 8); 94 repeated_c |= repeated_c << 16; 95 if (0xffffffffU < (longword) -1) 96 { 97 repeated_one |= repeated_one << 31 << 1; 98 repeated_c |= repeated_c << 31 << 1; 99 if (8 < sizeof (longword)) 100 { 101 size_t i; 102 103 for (i = 64; i < sizeof (longword) * 8; i *= 2) 104 { 105 repeated_one |= repeated_one << i; 106 repeated_c |= repeated_c << i; 107 } 108 } 109 } 110 111 /* Instead of the traditional loop which tests each byte, we will test a 112 longword at a time. The tricky part is testing if *any of the four* 113 bytes in the longword in question are equal to c. We first use an xor 114 with repeated_c. This reduces the task to testing whether *any of the 115 four* bytes in longword1 is zero. 116 117 We compute tmp = 118 ((longword1 - repeated_one) & ~longword1) & (repeated_one << 7). 119 That is, we perform the following operations: 120 1. Subtract repeated_one. 121 2. & ~longword1. 122 3. & a mask consisting of 0x80 in every byte. 123 Consider what happens in each byte: 124 - If a byte of longword1 is zero, step 1 and 2 transform it into 0xff, 125 and step 3 transforms it into 0x80. A carry can also be propagated 126 to more significant bytes. 127 - If a byte of longword1 is nonzero, let its lowest 1 bit be at 128 position k (0 <= k <= 7); so the lowest k bits are 0. After step 1, 129 the byte ends in a single bit of value 0 and k bits of value 1. 130 After step 2, the result is just k bits of value 1: 2^k - 1. After 131 step 3, the result is 0. And no carry is produced. 132 So, if longword1 has only non-zero bytes, tmp is zero. 133 Whereas if longword1 has a zero byte, call j the position of the least 134 significant zero byte. Then the result has a zero at positions 0, ..., 135 j-1 and a 0x80 at position j. We cannot predict the result at the more 136 significant bytes (positions j+1..3), but it does not matter since we 137 already have a non-zero bit at position 8*j+7. 138 139 So, the test whether any byte in longword1 is zero is equivalent to 140 testing whether tmp is nonzero. */ 141 142 while (n >= sizeof (longword)) 143 { 144 longword longword1 = *longword_ptr ^ repeated_c; 145 146 if ((((longword1 - repeated_one) & ~longword1) 147 & (repeated_one << 7)) != 0) 148 break; 149 longword_ptr++; 150 n -= sizeof (longword); 151 } 152 153 char_ptr = (const unsigned char *) longword_ptr; 154 155 /* At this point, we know that either n < sizeof (longword), or one of the 156 sizeof (longword) bytes starting at char_ptr is == c. On little-endian 157 machines, we could determine the first such byte without any further 158 memory accesses, just by looking at the tmp result from the last loop 159 iteration. But this does not work on big-endian machines. Choose code 160 that works in both cases. */ 161 162 for (; n > 0; --n, ++char_ptr) 163 { 164 if (*char_ptr == c) 165 return (void *) char_ptr; 166 } 167 168 return NULL; 169 } 170 #ifdef weak_alias 171 weak_alias (__memchr, BP_SYM (memchr)) 172 #endif 173
