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Editing: my_bit.h
/* 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); } static inline uint my_bits_in_bytes(uint n) { return ((n + 7) / 8); } #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 */
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