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/*
Copyright (c) 2007, 2017, Oracle and/or its affiliates. All rights reserved.
This program is free software; you can redistribute it and/or modify
it under the terms of the GNU General Public License, version 2.0,
as published by the Free Software Foundation.
This program is also distributed with certain software (including
but not limited to OpenSSL) that is licensed under separate terms,
as designated in a particular file or component or in included license
documentation. The authors of MySQL hereby grant you an additional
permission to link the program and your derivative works with the
separately licensed software that they have included with MySQL.
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, version 2.0, 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 02110-1301 USA */
#ifndef MY_BIT_INCLUDED
#define MY_BIT_INCLUDED
/**
@file include/my_bit.h
Some useful bit functions.
*/
#include <sys/types.h>
#include "my_config.h"
#include "my_inttypes.h"
#include "my_macros.h"
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)];
}
/**
Determine if a single bit is set among some bits.
@tparam IntType an integer type
@param bits the bits to examine
@retval true if bits equals to a power of 2
@retval false otherwise
*/
template <typename IntType>
constexpr bool is_single_bit(IntType bits) {
/*
Proof of correctness:
(1) is_single_bit(0)==false is left as an exercise to the reader.
(2) is_single_bit(1)==true is left as an exercise to the reader.
(3) is_single_bit(1<<(N+1))==true because the most significant set bit
in (bits - 1) would be 1<<N, and obviously (bits & (bits - 1)) == 0.
(4) In all other cases, is_single_bit(bits)==false.
In these cases, we must have multiple bits set, that is,
bits==m|1<<N such that N>=0 and m!=0 and the least significant
bit that is set in m is greater than 1<<N. In this case,
(bits-1)==m|((1<<N)-1), and (bits&(bits-1))==m, which we defined to be
nonzero. So, m==0 will not hold.
Note: The above proof (3),(4) is applicable also to the case where
IntType is signed using two's complement arithmetics, and the most
significant bit is set, or in other words, bits<0.
*/
return bits != 0 && (bits & (bits - 1)) == 0;
}
#endif /* MY_BIT_INCLUDED */
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