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| 1 /* Copyright (C) 1991, 1993, 1996-1997, 1999-2000, 2003-2004, 2006, 2008-2011 | |
| 2 Free Software Foundation, Inc. | |
| 3 | |
| 4 Based on strlen implementation by Torbjorn Granlund (tege@sics.se), | |
| 5 with help from Dan Sahlin (dan@sics.se) and | |
| 6 commentary by Jim Blandy (jimb@ai.mit.edu); | |
| 7 adaptation to memchr suggested by Dick Karpinski (dick@cca.ucsf.edu), | |
| 8 and implemented by Roland McGrath (roland@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@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 | |
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