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/* Copyright (C) 1991, 1993, 1996-1997, 1999-2000, 2003-2004, 2006, 2008-2013
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;
}
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