1
/* Searching in a string.
2
Copyright (C) 2008, 2009, 2010 Free Software Foundation, Inc.
4
This program is free software: you can redistribute it and/or modify
5
it under the terms of the GNU General Public License as published by
6
the Free Software Foundation; either version 3 of the License, or
7
(at your option) any later version.
9
This program is distributed in the hope that it will be useful,
10
but WITHOUT ANY WARRANTY; without even the implied warranty of
11
MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
12
GNU General Public License for more details.
14
You should have received a copy of the GNU General Public License
15
along with this program. If not, see <http://www.gnu.org/licenses/>. */
22
/* Find the first occurrence of C in S. */
24
rawmemchr (const void *s, int c_in)
26
/* On 32-bit hardware, choosing longword to be a 32-bit unsigned
27
long instead of a 64-bit uintmax_t tends to give better
28
performance. On 64-bit hardware, unsigned long is generally 64
29
bits already. Change this typedef to experiment with
31
typedef unsigned long int longword;
33
const unsigned char *char_ptr;
34
const longword *longword_ptr;
35
longword repeated_one;
39
c = (unsigned char) c_in;
41
/* Handle the first few bytes by reading one byte at a time.
42
Do this until CHAR_PTR is aligned on a longword boundary. */
43
for (char_ptr = (const unsigned char *) s;
44
(size_t) char_ptr % sizeof (longword) != 0;
47
return (void *) char_ptr;
49
longword_ptr = (const longword *) char_ptr;
51
/* All these elucidatory comments refer to 4-byte longwords,
52
but the theory applies equally well to any size longwords. */
54
/* Compute auxiliary longword values:
55
repeated_one is a value which has a 1 in every byte.
56
repeated_c has c in every byte. */
57
repeated_one = 0x01010101;
58
repeated_c = c | (c << 8);
59
repeated_c |= repeated_c << 16;
60
if (0xffffffffU < (longword) -1)
62
repeated_one |= repeated_one << 31 << 1;
63
repeated_c |= repeated_c << 31 << 1;
64
if (8 < sizeof (longword))
68
for (i = 64; i < sizeof (longword) * 8; i *= 2)
70
repeated_one |= repeated_one << i;
71
repeated_c |= repeated_c << i;
76
/* Instead of the traditional loop which tests each byte, we will
77
test a longword at a time. The tricky part is testing if *any of
78
the four* bytes in the longword in question are equal to NUL or
79
c. We first use an xor with repeated_c. This reduces the task
80
to testing whether *any of the four* bytes in longword1 is zero.
83
((longword1 - repeated_one) & ~longword1) & (repeated_one << 7).
84
That is, we perform the following operations:
85
1. Subtract repeated_one.
87
3. & a mask consisting of 0x80 in every byte.
88
Consider what happens in each byte:
89
- If a byte of longword1 is zero, step 1 and 2 transform it into 0xff,
90
and step 3 transforms it into 0x80. A carry can also be propagated
91
to more significant bytes.
92
- If a byte of longword1 is nonzero, let its lowest 1 bit be at
93
position k (0 <= k <= 7); so the lowest k bits are 0. After step 1,
94
the byte ends in a single bit of value 0 and k bits of value 1.
95
After step 2, the result is just k bits of value 1: 2^k - 1. After
96
step 3, the result is 0. And no carry is produced.
97
So, if longword1 has only non-zero bytes, tmp is zero.
98
Whereas if longword1 has a zero byte, call j the position of the least
99
significant zero byte. Then the result has a zero at positions 0, ...,
100
j-1 and a 0x80 at position j. We cannot predict the result at the more
101
significant bytes (positions j+1..3), but it does not matter since we
102
already have a non-zero bit at position 8*j+7.
104
The test whether any byte in longword1 is zero is equivalent
105
to testing whether tmp is nonzero.
107
This test can read beyond the end of a string, depending on where
108
C_IN is encountered. However, this is considered safe since the
109
initialization phase ensured that the read will be aligned,
110
therefore, the read will not cross page boundaries and will not
115
longword longword1 = *longword_ptr ^ repeated_c;
117
if ((((longword1 - repeated_one) & ~longword1)
118
& (repeated_one << 7)) != 0)
123
char_ptr = (const unsigned char *) longword_ptr;
125
/* At this point, we know that one of the sizeof (longword) bytes
126
starting at char_ptr is == c. On little-endian machines, we
127
could determine the first such byte without any further memory
128
accesses, just by looking at the tmp result from the last loop
129
iteration. But this does not work on big-endian machines.
130
Choose code that works in both cases. */
132
char_ptr = (unsigned char *) longword_ptr;
133
while (*char_ptr != c)
135
return (void *) char_ptr;