AlkantarClanX12
Your IP : 18.191.27.78
/* Copyright (c) 2007, 2011, Oracle and/or its affiliates.
Copyright (c) 2009, 2020, MariaDB Corporation.
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
/*
Some useful bit functions
*/
C_MODE_START
extern const uchar _my_bits_reverse_table[256];
/*
my_bit_log2_xxx()
In the given value, find the highest bit set,
which is the smallest X that satisfies the condition: (2^X >= value).
Can be used as a reverse operation for (1<<X), to find X.
Examples:
- returns 0 for (1<<0)
- returns 1 for (1<<1)
- returns 2 for (1<<2)
- returns 2 for 3, which has (1<<2) as the highest bit set.
Note, the behaviour of log2(0) is not defined.
Let's return 0 for the input 0, for the code simplicity.
See the 000x branch. It covers both (1<<0) and 0.
*/
static inline CONSTEXPR uint my_bit_log2_hex_digit(uint8 value)
{
return value & 0x0C ? /*1100*/ (value & 0x08 ? /*1000*/ 3 : /*0100*/ 2) :
/*0010*/ (value & 0x02 ? /*0010*/ 1 : /*000x*/ 0);
}
static inline CONSTEXPR uint my_bit_log2_uint8(uint8 value)
{
return value & 0xF0 ? my_bit_log2_hex_digit((uint8) (value >> 4)) + 4:
my_bit_log2_hex_digit(value);
}
static inline CONSTEXPR uint my_bit_log2_uint16(uint16 value)
{
return value & 0xFF00 ? my_bit_log2_uint8((uint8) (value >> 8)) + 8 :
my_bit_log2_uint8((uint8) value);
}
static inline CONSTEXPR uint my_bit_log2_uint32(uint32 value)
{
return value & 0xFFFF0000UL ?
my_bit_log2_uint16((uint16) (value >> 16)) + 16 :
my_bit_log2_uint16((uint16) value);
}
static inline CONSTEXPR uint my_bit_log2_uint64(ulonglong value)
{
return value & 0xFFFFFFFF00000000ULL ?
my_bit_log2_uint32((uint32) (value >> 32)) + 32 :
my_bit_log2_uint32((uint32) value);
}
static inline CONSTEXPR uint my_bit_log2_size_t(size_t value)
{
#ifdef __cplusplus
static_assert(sizeof(size_t) <= sizeof(ulonglong),
"size_t <= ulonglong is an assumption that needs to be fixed "
"for this architecture. Please create an issue on "
"https://jira.mariadb.org");
#endif
return my_bit_log2_uint64((ulonglong) value);
}
/*
Count bits in 32bit integer
Algorithm by Sean Anderson, according to:
http://graphics.stanford.edu/~seander/bithacks.html
under "Counting bits set, in parallel"
(Original code public domain).
*/
static inline uint my_count_bits_uint32(uint32 v)
{
v = v - ((v >> 1) & 0x55555555);
v = (v & 0x33333333) + ((v >> 2) & 0x33333333);
return (((v + (v >> 4)) & 0xF0F0F0F) * 0x1010101) >> 24;
}
static inline uint my_count_bits(ulonglong x)
{
return my_count_bits_uint32((uint32)x) + my_count_bits_uint32((uint32)(x >> 32));
}
/*
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
(Original 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
((uint32)_my_bits_reverse_table[ key & 255] << 24) |
((uint32)_my_bits_reverse_table[(key>> 8) & 255] << 16) |
((uint32)_my_bits_reverse_table[(key>>16) & 255] << 8) |
(uint32)_my_bits_reverse_table[(key>>24) ];
}
/*
a number with the n lowest bits set
an overflow-safe version of (1 << n) - 1
*/
static inline uint64 my_set_bits(int n)
{
return (((1ULL << (n - 1)) - 1) << 1) | 1;
}
/* Create a mask of the significant bits for the last byte (1,3,7,..255) */
static inline uchar last_byte_mask(uint bits)
{
/* Get the number of used bits-1 (0..7) in the last byte */
unsigned int const used = (bits - 1U) & 7U;
/* Return bitmask for the significant bits */
return (uchar) ((2U << used) - 1);
}
#ifdef _MSC_VER
#include <intrin.h>
#endif
/*
Find the position of the first(least significant) bit set in
the argument. Returns 64 if the argument was 0.
*/
static inline uint my_find_first_bit(ulonglong n)
{
if(!n)
return 64;
#if defined(__GNUC__)
return __builtin_ctzll(n);
#elif defined(_MSC_VER)
#if defined(_M_IX86)
unsigned long bit;
if( _BitScanForward(&bit, (uint)n))
return bit;
_BitScanForward(&bit, (uint)(n>>32));
return bit + 32;
#else
unsigned long bit;
_BitScanForward64(&bit, n);
return bit;
#endif
#else
/* Generic case */
uint shift= 0;
static const uchar last_bit[16] = { 32, 0, 1, 0,
2, 0, 1, 0,
3, 0, 1, 0,
2, 0, 1, 0};
uint bit;
while ((bit = last_bit[(n >> shift) & 0xF]) == 32)
shift+= 4;
return shift+bit;
#endif
}
C_MODE_END
#endif /* MY_BIT_INCLUDED */