| 1 | /* Copyright (C) 1991, 1993, 1996-1997, 1999-2000, 2003-2004, 2006, 2008-2015 |
| 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 |