+++ /dev/null
-/* 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
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