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authorLorry Tar Creator <lorry-tar-importer@lorry>2016-01-20 10:55:18 +0000
committerLorry Tar Creator <lorry-tar-importer@lorry>2016-01-20 10:55:18 +0000
commit70e9163c9c18e995515598085cb824e554eb7ae7 (patch)
treea42dc8b2a6c031354bf31472de888bfc8a060132 /lib/memchr2.c
parentcbf5993c43f49281173f185863577d86bfac6eae (diff)
downloadcoreutils-tarball-master.tar.gz
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+/* Copyright (C) 1991, 1993, 1996-1997, 1999-2000, 2003-2004, 2006, 2008-2016
+ 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 in glibc by Roland McGrath (roland@ai.mit.edu).
+ Extension to memchr2 implemented by Eric Blake (ebb9@byu.net).
+
+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 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>
+
+#include "memchr2.h"
+
+#include <limits.h>
+#include <stdint.h>
+#include <string.h>
+
+/* Return the first address of either C1 or C2 (treated as unsigned
+ char) that occurs within N bytes of the memory region S. If
+ neither byte appears, return NULL. */
+void *
+memchr2 (void const *s, int c1_in, int c2_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;
+ void const *void_ptr;
+ const longword *longword_ptr;
+ longword repeated_one;
+ longword repeated_c1;
+ longword repeated_c2;
+ unsigned char c1;
+ unsigned char c2;
+
+ c1 = (unsigned char) c1_in;
+ c2 = (unsigned char) c2_in;
+
+ if (c1 == c2)
+ return memchr (s, c1, n);
+
+ /* Handle the first few bytes by reading one byte at a time.
+ Do this until VOID_PTR is aligned on a longword boundary. */
+ for (void_ptr = s;
+ n > 0 && (uintptr_t) void_ptr % sizeof (longword) != 0;
+ --n)
+ {
+ char_ptr = void_ptr;
+ if (*char_ptr == c1 || *char_ptr == c2)
+ return (void *) void_ptr;
+ void_ptr = char_ptr + 1;
+ }
+
+ longword_ptr = void_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_c1 has c1 in every byte.
+ repeated_c2 has c2 in every byte. */
+ repeated_one = 0x01010101;
+ repeated_c1 = c1 | (c1 << 8);
+ repeated_c2 = c2 | (c2 << 8);
+ repeated_c1 |= repeated_c1 << 16;
+ repeated_c2 |= repeated_c2 << 16;
+ if (0xffffffffU < (longword) -1)
+ {
+ repeated_one |= repeated_one << 31 << 1;
+ repeated_c1 |= repeated_c1 << 31 << 1;
+ repeated_c2 |= repeated_c2 << 31 << 1;
+ if (8 < sizeof (longword))
+ {
+ size_t i;
+
+ for (i = 64; i < sizeof (longword) * 8; i *= 2)
+ {
+ repeated_one |= repeated_one << i;
+ repeated_c1 |= repeated_c1 << i;
+ repeated_c2 |= repeated_c2 << 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 c1 or c2. We first use
+ an xor with repeated_c1 and repeated_c2, respectively. This reduces
+ the task to testing whether *any of the four* bytes in longword1 or
+ longword2 is zero.
+
+ Let's consider longword1. We compute tmp1 =
+ ((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, tmp1 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.
+
+ Similarly, we compute tmp2 =
+ ((longword2 - repeated_one) & ~longword2) & (repeated_one << 7).
+
+ The test whether any byte in longword1 or longword2 is zero is equivalent
+ to testing whether tmp1 is nonzero or tmp2 is nonzero. We can combine
+ this into a single test, whether (tmp1 | tmp2) is nonzero. */
+
+ while (n >= sizeof (longword))
+ {
+ longword longword1 = *longword_ptr ^ repeated_c1;
+ longword longword2 = *longword_ptr ^ repeated_c2;
+
+ if (((((longword1 - repeated_one) & ~longword1)
+ | ((longword2 - repeated_one) & ~longword2))
+ & (repeated_one << 7)) != 0)
+ break;
+ longword_ptr++;
+ 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 == c1 or == c2. On
+ little-endian machines, we could determine the first such byte without
+ any further memory accesses, just by looking at the (tmp1 | tmp2) result
+ from the last loop iteration. But this does not work on big-endian
+ machines. Choose code that works in both cases. */
+
+ for (; n > 0; --n, ++char_ptr)
+ {
+ if (*char_ptr == c1 || *char_ptr == c2)
+ return (void *) char_ptr;
+ }
+
+ return NULL;
+}