2010-05-23 02:57:01 +02:00
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/* Copyright (C) 1991, 1993, 1996-1997, 1999-2000, 2003-2004, 2006, 2008-2010
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Free Software Foundation, Inc.
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2008-04-14 20:05:12 +02:00
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Based on strlen implementation by Torbjorn Granlund (tege@sics.se),
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with help from Dan Sahlin (dan@sics.se) and
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commentary by Jim Blandy (jimb@ai.mit.edu);
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adaptation to memchr suggested by Dick Karpinski (dick@cca.ucsf.edu),
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and implemented by Roland McGrath (roland@ai.mit.edu).
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NOTE: The canonical source of this file is maintained with the GNU C Library.
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Bugs can be reported to bug-glibc@prep.ai.mit.edu.
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This program is free software: you can redistribute it and/or modify it
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under the terms of the GNU General Public License as published by the
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Free Software Foundation; either version 3 of the License, or any
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later version.
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This program is distributed in the hope that it will be useful,
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but WITHOUT ANY WARRANTY; without even the implied warranty of
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MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
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GNU General Public License for more details.
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You should have received a copy of the GNU General Public License
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along with this program. If not, see <http://www.gnu.org/licenses/>. */
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#ifndef _LIBC
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# include <config.h>
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#endif
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#include <string.h>
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#include <stddef.h>
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#if defined _LIBC
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# include <memcopy.h>
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#else
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# define reg_char char
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#endif
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#include <limits.h>
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#if HAVE_BP_SYM_H || defined _LIBC
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# include <bp-sym.h>
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#else
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# define BP_SYM(sym) sym
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#endif
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#undef __memchr
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2010-05-23 02:57:01 +02:00
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#ifdef _LIBC
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# undef memchr
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#endif
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#ifndef weak_alias
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# define __memchr memchr
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#endif
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2008-04-14 20:05:12 +02:00
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/* Search no more than N bytes of S for C. */
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void *
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__memchr (void const *s, int c_in, size_t n)
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{
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2010-05-23 02:57:01 +02:00
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/* On 32-bit hardware, choosing longword to be a 32-bit unsigned
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long instead of a 64-bit uintmax_t tends to give better
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performance. On 64-bit hardware, unsigned long is generally 64
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bits already. Change this typedef to experiment with
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performance. */
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typedef unsigned long int longword;
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const unsigned char *char_ptr;
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const longword *longword_ptr;
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longword repeated_one;
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longword repeated_c;
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unsigned reg_char c;
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c = (unsigned char) c_in;
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2010-05-23 02:57:01 +02:00
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/* Handle the first few bytes by reading one byte at a time.
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Do this until CHAR_PTR is aligned on a longword boundary. */
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for (char_ptr = (const unsigned char *) s;
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2010-05-23 02:57:01 +02:00
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n > 0 && (size_t) char_ptr % sizeof (longword) != 0;
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--n, ++char_ptr)
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if (*char_ptr == c)
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return (void *) char_ptr;
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longword_ptr = (const longword *) char_ptr;
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2008-04-14 20:05:12 +02:00
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/* All these elucidatory comments refer to 4-byte longwords,
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but the theory applies equally well to any size longwords. */
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2010-05-23 02:57:01 +02:00
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/* Compute auxiliary longword values:
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repeated_one is a value which has a 1 in every byte.
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repeated_c has c in every byte. */
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repeated_one = 0x01010101;
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repeated_c = c | (c << 8);
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repeated_c |= repeated_c << 16;
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if (0xffffffffU < (longword) -1)
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{
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repeated_one |= repeated_one << 31 << 1;
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repeated_c |= repeated_c << 31 << 1;
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if (8 < sizeof (longword))
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{
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size_t i;
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for (i = 64; i < sizeof (longword) * 8; i *= 2)
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{
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repeated_one |= repeated_one << i;
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repeated_c |= repeated_c << i;
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}
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}
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}
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2010-05-23 02:57:01 +02:00
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/* Instead of the traditional loop which tests each byte, we will test a
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longword at a time. The tricky part is testing if *any of the four*
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bytes in the longword in question are equal to c. We first use an xor
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with repeated_c. This reduces the task to testing whether *any of the
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four* bytes in longword1 is zero.
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We compute tmp =
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((longword1 - repeated_one) & ~longword1) & (repeated_one << 7).
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That is, we perform the following operations:
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1. Subtract repeated_one.
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2. & ~longword1.
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3. & a mask consisting of 0x80 in every byte.
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Consider what happens in each byte:
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- If a byte of longword1 is zero, step 1 and 2 transform it into 0xff,
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and step 3 transforms it into 0x80. A carry can also be propagated
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to more significant bytes.
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- If a byte of longword1 is nonzero, let its lowest 1 bit be at
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position k (0 <= k <= 7); so the lowest k bits are 0. After step 1,
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the byte ends in a single bit of value 0 and k bits of value 1.
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After step 2, the result is just k bits of value 1: 2^k - 1. After
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step 3, the result is 0. And no carry is produced.
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So, if longword1 has only non-zero bytes, tmp is zero.
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Whereas if longword1 has a zero byte, call j the position of the least
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significant zero byte. Then the result has a zero at positions 0, ...,
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j-1 and a 0x80 at position j. We cannot predict the result at the more
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significant bytes (positions j+1..3), but it does not matter since we
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already have a non-zero bit at position 8*j+7.
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So, the test whether any byte in longword1 is zero is equivalent to
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testing whether tmp is nonzero. */
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while (n >= sizeof (longword))
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{
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longword longword1 = *longword_ptr ^ repeated_c;
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if ((((longword1 - repeated_one) & ~longword1)
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& (repeated_one << 7)) != 0)
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break;
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longword_ptr++;
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n -= sizeof (longword);
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}
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char_ptr = (const unsigned char *) longword_ptr;
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2010-05-23 02:57:01 +02:00
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/* At this point, we know that either n < sizeof (longword), or one of the
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sizeof (longword) bytes starting at char_ptr is == c. On little-endian
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machines, we could determine the first such byte without any further
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memory accesses, just by looking at the tmp result from the last loop
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iteration. But this does not work on big-endian machines. Choose code
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that works in both cases. */
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for (; n > 0; --n, ++char_ptr)
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{
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if (*char_ptr == c)
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return (void *) char_ptr;
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}
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2010-05-23 02:57:01 +02:00
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return NULL;
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2008-04-14 20:05:12 +02:00
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}
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#ifdef weak_alias
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weak_alias (__memchr, BP_SYM (memchr))
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#endif
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