142 lines
5.5 KiB
C
142 lines
5.5 KiB
C
/* Searching in a string.
|
|
Copyright (C) 2003, 2007-2013 Free Software Foundation, Inc.
|
|
|
|
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/>. */
|
|
|
|
#include <config.h>
|
|
|
|
/* Specification. */
|
|
#include <string.h>
|
|
|
|
/* Find the first occurrence of C in S or the final NUL byte. */
|
|
char *
|
|
strchrnul (const char *s, int c_in)
|
|
{
|
|
/* 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 char c;
|
|
|
|
c = (unsigned char) c_in;
|
|
if (!c)
|
|
return rawmemchr (s, 0);
|
|
|
|
/* Handle the first 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;
|
|
(size_t) char_ptr % sizeof (longword) != 0;
|
|
++char_ptr)
|
|
if (!*char_ptr || *char_ptr == c)
|
|
return (char *) 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 NUL or
|
|
c. We first use an xor with repeated_c. This reduces the task
|
|
to testing whether *any of the four* bytes in longword1 or
|
|
longword2 is zero.
|
|
|
|
Let's consider longword1. 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.
|
|
|
|
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.
|
|
|
|
This test can read more than one byte beyond the end of a string,
|
|
depending on where the terminating NUL is encountered. However,
|
|
this is considered safe since the initialization phase ensured
|
|
that the read will be aligned, therefore, the read will not cross
|
|
page boundaries and will not cause a fault. */
|
|
|
|
while (1)
|
|
{
|
|
longword longword1 = *longword_ptr ^ repeated_c;
|
|
longword longword2 = *longword_ptr;
|
|
|
|
if (((((longword1 - repeated_one) & ~longword1)
|
|
| ((longword2 - repeated_one) & ~longword2))
|
|
& (repeated_one << 7)) != 0)
|
|
break;
|
|
longword_ptr++;
|
|
}
|
|
|
|
char_ptr = (const unsigned char *) longword_ptr;
|
|
|
|
/* At this point, we know that one of the sizeof (longword) bytes
|
|
starting at char_ptr is == 0 or == 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. */
|
|
|
|
char_ptr = (unsigned char *) longword_ptr;
|
|
while (*char_ptr && (*char_ptr != c))
|
|
char_ptr++;
|
|
return (char *) char_ptr;
|
|
}
|