1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
|
/* Copyright (c) 2007, 2011, Oracle and/or its affiliates.
Copyright (c) 2009-2011, Monty Program Ab
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; version 2 of the License.
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., 51 Franklin St, Fifth Floor, Boston, MA 02110-1335 USA */
#ifndef MY_BIT_INCLUDED
#define MY_BIT_INCLUDED
#include <my_global.h>
/*
Some useful bit functions
*/
C_MODE_START
extern const char _my_bits_nbits[256];
extern const uchar _my_bits_reverse_table[256];
/*
Find smallest X in 2^X >= value
This can be used to divide a number with value by doing a shift instead
*/
static inline uint my_bit_log2(ulong value)
{
uint bit;
for (bit=0 ; value > 1 ; value>>=1, bit++) ;
return bit;
}
static inline uint my_count_bits(ulonglong v)
{
#if SIZEOF_LONG_LONG > 4
/* The following code is a bit faster on 16 bit machines than if we would
only shift v */
ulong v2=(ulong) (v >> 32);
return (uint) (uchar) (_my_bits_nbits[(uchar) v] +
_my_bits_nbits[(uchar) (v >> 8)] +
_my_bits_nbits[(uchar) (v >> 16)] +
_my_bits_nbits[(uchar) (v >> 24)] +
_my_bits_nbits[(uchar) (v2)] +
_my_bits_nbits[(uchar) (v2 >> 8)] +
_my_bits_nbits[(uchar) (v2 >> 16)] +
_my_bits_nbits[(uchar) (v2 >> 24)]);
#else
return (uint) (uchar) (_my_bits_nbits[(uchar) v] +
_my_bits_nbits[(uchar) (v >> 8)] +
_my_bits_nbits[(uchar) (v >> 16)] +
_my_bits_nbits[(uchar) (v >> 24)]);
#endif
}
static inline uint my_count_bits_uint32(uint32 v)
{
return (uint) (uchar) (_my_bits_nbits[(uchar) v] +
_my_bits_nbits[(uchar) (v >> 8)] +
_my_bits_nbits[(uchar) (v >> 16)] +
_my_bits_nbits[(uchar) (v >> 24)]);
}
/*
Next highest power of two
SYNOPSIS
my_round_up_to_next_power()
v Value to check
RETURN
Next or equal power of 2
Note: 0 will return 0
NOTES
Algorithm by Sean Anderson, according to:
http://graphics.stanford.edu/~seander/bithacks.html
(Orignal code public domain)
Comments shows how this works with 01100000000000000000000000001011
*/
static inline uint32 my_round_up_to_next_power(uint32 v)
{
v--; /* 01100000000000000000000000001010 */
v|= v >> 1; /* 01110000000000000000000000001111 */
v|= v >> 2; /* 01111100000000000000000000001111 */
v|= v >> 4; /* 01111111110000000000000000001111 */
v|= v >> 8; /* 01111111111111111100000000001111 */
v|= v >> 16; /* 01111111111111111111111111111111 */
return v+1; /* 10000000000000000000000000000000 */
}
static inline uint32 my_clear_highest_bit(uint32 v)
{
uint32 w=v >> 1;
w|= w >> 1;
w|= w >> 2;
w|= w >> 4;
w|= w >> 8;
w|= w >> 16;
return v & w;
}
static inline uint32 my_reverse_bits(uint32 key)
{
return
(_my_bits_reverse_table[ key & 255] << 24) |
(_my_bits_reverse_table[(key>> 8) & 255] << 16) |
(_my_bits_reverse_table[(key>>16) & 255] << 8) |
_my_bits_reverse_table[(key>>24) ];
}
C_MODE_END
#endif /* MY_BIT_INCLUDED */
|
