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Diffstat (limited to 'data/_apt7/memrchr.c')
| -rw-r--r-- | data/_apt7/memrchr.c | 161 |
1 files changed, 0 insertions, 161 deletions
diff --git a/data/_apt7/memrchr.c b/data/_apt7/memrchr.c deleted file mode 100644 index da93ca0ba..000000000 --- a/data/_apt7/memrchr.c +++ /dev/null @@ -1,161 +0,0 @@ -/* memrchr -- find the last occurrence of a byte in a memory block - - Copyright (C) 1991, 1993, 1996, 1997, 1999, 2000, 2003, 2004, 2005, - 2006, 2007, 2008 Free Software Foundation, Inc. - - Based on strlen implementation by Torbjorn Granlund (tege@sics.se), - with help from Dan Sahlin (dan@sics.se) and - commentary by Jim Blandy (jimb@ai.mit.edu); - adaptation to memchr suggested by Dick Karpinski (dick@cca.ucsf.edu), - and implemented by Roland McGrath (roland@ai.mit.edu). - - This program is free software: you can redistribute it and/or modify - it under the terms of the GNU General Public License as published by - the Free Software Foundation; either version 3 of the License, or - (at your option) any later version. - - This program is distributed in the hope that it will be useful, - but WITHOUT ANY WARRANTY; without even the implied warranty of - MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the - GNU General Public License for more details. - - You should have received a copy of the GNU General Public License - along with this program. If not, see <http://www.gnu.org/licenses/>. */ - -#if defined _LIBC -# include <memcopy.h> -#else -# include <config.h> -# define reg_char char -#endif - -#include <string.h> -#include <limits.h> - -#undef __memrchr -#ifdef _LIBC -# undef memrchr -#endif - -#ifndef weak_alias -# define __memrchr memrchr -#endif - -/* Search no more than N bytes of S for C. */ -void * -__memrchr (void const *s, int c_in, size_t n) -{ - /* On 32-bit hardware, choosing longword to be a 32-bit unsigned - long instead of a 64-bit uintmax_t tends to give better - performance. On 64-bit hardware, unsigned long is generally 64 - bits already. Change this typedef to experiment with - performance. */ - typedef unsigned long int longword; - - const unsigned char *char_ptr; - const longword *longword_ptr; - longword repeated_one; - longword repeated_c; - unsigned reg_char c; - - c = (unsigned char) c_in; - - /* Handle the last few bytes by reading one byte at a time. - Do this until CHAR_PTR is aligned on a longword boundary. */ - for (char_ptr = (const unsigned char *) s + n; - n > 0 && (size_t) char_ptr % sizeof (longword) != 0; - --n) - if (*--char_ptr == c) - return (void *) char_ptr; - - longword_ptr = (const longword *) char_ptr; - - /* All these elucidatory comments refer to 4-byte longwords, - but the theory applies equally well to any size longwords. */ - - /* Compute auxiliary longword values: - repeated_one is a value which has a 1 in every byte. - repeated_c has c in every byte. */ - repeated_one = 0x01010101; - repeated_c = c | (c << 8); - repeated_c |= repeated_c << 16; - if (0xffffffffU < (longword) -1) - { - repeated_one |= repeated_one << 31 << 1; - repeated_c |= repeated_c << 31 << 1; - if (8 < sizeof (longword)) - { - size_t i; - - for (i = 64; i < sizeof (longword) * 8; i *= 2) - { - repeated_one |= repeated_one << i; - repeated_c |= repeated_c << i; - } - } - } - - /* Instead of the traditional loop which tests each byte, we will test a - longword at a time. The tricky part is testing if *any of the four* - bytes in the longword in question are equal to c. We first use an xor - with repeated_c. This reduces the task to testing whether *any of the - four* bytes in longword1 is zero. - - We compute tmp = - ((longword1 - repeated_one) & ~longword1) & (repeated_one << 7). - That is, we perform the following operations: - 1. Subtract repeated_one. - 2. & ~longword1. - 3. & a mask consisting of 0x80 in every byte. - Consider what happens in each byte: - - If a byte of longword1 is zero, step 1 and 2 transform it into 0xff, - and step 3 transforms it into 0x80. A carry can also be propagated - to more significant bytes. - - If a byte of longword1 is nonzero, let its lowest 1 bit be at - position k (0 <= k <= 7); so the lowest k bits are 0. After step 1, - the byte ends in a single bit of value 0 and k bits of value 1. - After step 2, the result is just k bits of value 1: 2^k - 1. After - step 3, the result is 0. And no carry is produced. - So, if longword1 has only non-zero bytes, tmp is zero. - Whereas if longword1 has a zero byte, call j the position of the least - significant zero byte. Then the result has a zero at positions 0, ..., - j-1 and a 0x80 at position j. We cannot predict the result at the more - significant bytes (positions j+1..3), but it does not matter since we - already have a non-zero bit at position 8*j+7. - - So, the test whether any byte in longword1 is zero is equivalent to - testing whether tmp is nonzero. */ - - while (n >= sizeof (longword)) - { - longword longword1 = *--longword_ptr ^ repeated_c; - - if ((((longword1 - repeated_one) & ~longword1) - & (repeated_one << 7)) != 0) - { - longword_ptr++; - break; - } - n -= sizeof (longword); - } - - char_ptr = (const unsigned char *) longword_ptr; - - /* At this point, we know that either n < sizeof (longword), or one of the - sizeof (longword) bytes starting at char_ptr is == c. On little-endian - machines, we could determine the first such byte without any further - memory accesses, just by looking at the tmp result from the last loop - iteration. But this does not work on big-endian machines. Choose code - that works in both cases. */ - - while (n-- > 0) - { - if (*--char_ptr == c) - return (void *) char_ptr; - } - - return NULL; -} -#ifdef weak_alias -weak_alias (__memrchr, memrchr) -#endif |