/* memrchr -- find the last occurrence of a byte in a memory block
- Copyright (C) 1991, 93, 96, 97, 99, 2000 Free Software Foundation, Inc.
+
+ Copyright (C) 1991, 1993, 1996-1997, 1999-2000, 2003-2014 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
+ 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 2, or (at your option)
- any later version.
+ 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, write to the Free Software Foundation,
- Inc., 59 Temple Place - Suite 330, Boston, MA 02111-1307, USA. */
-
-#ifdef HAVE_CONFIG_H
-# include <config.h>
-#endif
-
-#include <stdlib.h>
+ You should have received a copy of the GNU General Public License
+ along with this program. If not, see <http://www.gnu.org/licenses/>. */
-#undef __ptr_t
-#if defined (__cplusplus) || (defined (__STDC__) && __STDC__)
-# define __ptr_t void *
-#else /* Not C++ or ANSI C. */
-# define __ptr_t char *
-#endif /* C++ or ANSI C. */
-
-#if defined (_LIBC)
-# include <string.h>
+#if defined _LIBC
# include <memcopy.h>
#else
+# include <config.h>
# define reg_char char
#endif
-#if defined (HAVE_LIMITS_H) || defined (_LIBC)
-# include <limits.h>
-#endif
-
-#define LONG_MAX_32_BITS 2147483647
-
-#ifndef LONG_MAX
-# define LONG_MAX LONG_MAX_32_BITS
-#endif
-
-#include <sys/types.h>
+#include <string.h>
+#include <limits.h>
#undef __memrchr
-#undef memrchr
+#ifdef _LIBC
+# undef memrchr
+#endif
#ifndef weak_alias
# define __memrchr memrchr
#endif
/* Search no more than N bytes of S for C. */
-__ptr_t
-__memrchr (s, c_in, n)
- const __ptr_t s;
- int c_in;
- size_t n;
+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 unsigned long int *longword_ptr;
- unsigned long int longword, magic_bits, charmask;
+ const longword *longword_ptr;
+ longword repeated_one;
+ longword repeated_c;
unsigned reg_char c;
c = (unsigned char) c_in;
- /* Handle the last few characters by reading one character at a time.
+ /* 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 && ((unsigned long int) char_ptr
- & (sizeof (longword) - 1)) != 0;
+ n > 0 && (size_t) char_ptr % sizeof (longword) != 0;
--n)
if (*--char_ptr == c)
- return (__ptr_t) char_ptr;
+ return (void *) char_ptr;
- /* All these elucidatory comments refer to 4-byte longwords,
- but the theory applies equally well to 8-byte longwords. */
-
- longword_ptr = (unsigned long int *) char_ptr;
-
- /* Bits 31, 24, 16, and 8 of this number are zero. Call these bits
- the "holes." Note that there is a hole just to the left of
- each byte, with an extra at the end:
-
- bits: 01111110 11111110 11111110 11111111
- bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD
+ longword_ptr = (const longword *) char_ptr;
- The 1-bits make sure that carries propagate to the next 0-bit.
- The 0-bits provide holes for carries to fall into. */
-
- if (sizeof (longword) != 4 && sizeof (longword) != 8)
- abort ();
-
-#if LONG_MAX <= LONG_MAX_32_BITS
- magic_bits = 0x7efefeff;
-#else
- magic_bits = ((unsigned long int) 0x7efefefe << 32) | 0xfefefeff;
-#endif
+ /* 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;
+ }
+ }
+ }
- /* Set up a longword, each of whose bytes is C. */
- charmask = c | (c << 8);
- charmask |= charmask << 16;
-#if LONG_MAX > LONG_MAX_32_BITS
- charmask |= charmask << 32;
-#endif
+ /* 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. */
- /* Instead of the traditional loop which tests each character,
- 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 zero. */
while (n >= sizeof (longword))
{
- /* We tentatively exit the loop if adding MAGIC_BITS to
- LONGWORD fails to change any of the hole bits of LONGWORD.
-
- 1) Is this safe? Will it catch all the zero bytes?
- Suppose there is a byte with all zeros. Any carry bits
- propagating from its left will fall into the hole at its
- least significant bit and stop. Since there will be no
- carry from its most significant bit, the LSB of the
- byte to the left will be unchanged, and the zero will be
- detected.
-
- 2) Is this worthwhile? Will it ignore everything except
- zero bytes? Suppose every byte of LONGWORD has a bit set
- somewhere. There will be a carry into bit 8. If bit 8
- is set, this will carry into bit 16. If bit 8 is clear,
- one of bits 9-15 must be set, so there will be a carry
- into bit 16. Similarly, there will be a carry into bit
- 24. If one of bits 24-30 is set, there will be a carry
- into bit 31, so all of the hole bits will be changed.
-
- The one misfire occurs when bits 24-30 are clear and bit
- 31 is set; in this case, the hole at bit 31 is not
- changed. If we had access to the processor carry flag,
- we could close this loophole by putting the fourth hole
- at bit 32!
-
- So it ignores everything except 128's, when they're aligned
- properly.
-
- 3) But wait! Aren't we looking for C, not zero?
- Good point. So what we do is XOR LONGWORD with a longword,
- each of whose bytes is C. This turns each byte that is C
- into a zero. */
-
- longword = *--longword_ptr ^ charmask;
-
- /* Add MAGIC_BITS to LONGWORD. */
- if ((((longword + magic_bits)
-
- /* Set those bits that were unchanged by the addition. */
- ^ ~longword)
-
- /* Look at only the hole bits. If any of the hole bits
- are unchanged, most likely one of the bytes was a
- zero. */
- & ~magic_bits) != 0)
- {
- /* Which of the bytes was C? If none of them were, it was
- a misfire; continue the search. */
-
- const unsigned char *cp = (const unsigned char *) longword_ptr;
-
-#if LONG_MAX > 2147483647
- if (cp[7] == c)
- return (__ptr_t) &cp[7];
- if (cp[6] == c)
- return (__ptr_t) &cp[6];
- if (cp[5] == c)
- return (__ptr_t) &cp[5];
- if (cp[4] == c)
- return (__ptr_t) &cp[4];
-#endif
- if (cp[3] == c)
- return (__ptr_t) &cp[3];
- if (cp[2] == c)
- return (__ptr_t) &cp[2];
- if (cp[1] == c)
- return (__ptr_t) &cp[1];
- if (cp[0] == c)
- return (__ptr_t) cp;
- }
-
+ 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 (__ptr_t) char_ptr;
+ return (void *) char_ptr;
}
- return 0;
+ return NULL;
}
#ifdef weak_alias
weak_alias (__memrchr, memrchr)
projects / gnulib.git / blobdiff