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Diffstat (limited to 'apt-pkg/memrchr.cc')
| -rw-r--r-- | apt-pkg/memrchr.cc | 157 |
1 files changed, 157 insertions, 0 deletions
diff --git a/apt-pkg/memrchr.cc b/apt-pkg/memrchr.cc new file mode 100644 index 000000000..edf8f346a --- /dev/null +++ b/apt-pkg/memrchr.cc @@ -0,0 +1,157 @@ +/* memrchr -- find the last occurrence of a byte in a memory block + + Copyright (C) 1991, 1993, 1996-1997, 1999-2000, 2003-2015 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/>. */ + +#include <config.h> + +#ifndef HAVE_MEMRCHR +#define reg_char char + +#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. */ +extern "C" void * +memrchr (const void *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; +} +#endif |