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e4715aa82a
Previously, integral `exp2()` would produce the incorrect result for exponents above 31.
71 lines
1.4 KiB
C++
71 lines
1.4 KiB
C++
/*
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* Copyright (c) 2022, Leon Albrecht <leon2002.la@gmail.com>
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*
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* SPDX-License-Identifier: BSD-2-Clause
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*/
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#pragma once
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#include <AK/BuiltinWrappers.h>
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#include <AK/Concepts.h>
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#include <AK/Types.h>
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namespace AK {
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template<Integral T>
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constexpr T exp2(T exponent)
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{
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return static_cast<T>(1) << exponent;
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}
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template<Integral T>
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constexpr T log2(T x)
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{
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return x ? (8 * sizeof(T) - 1) - count_leading_zeroes(static_cast<MakeUnsigned<T>>(x)) : 0;
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}
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template<Integral T>
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constexpr T ceil_log2(T x)
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{
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if (!x)
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return 0;
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T log = AK::log2(x);
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log += (x & ((((T)1) << (log - 1)) - 1)) != 0;
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return log;
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}
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template<Integral I>
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constexpr I pow(I base, I exponent)
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{
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// https://en.wikipedia.org/wiki/Exponentiation_by_squaring
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if (exponent < 0)
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return 0;
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if (exponent == 0)
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return 1;
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I res = 1;
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while (exponent > 0) {
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if (exponent & 1)
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res *= base;
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base *= base;
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exponent /= 2u;
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}
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return res;
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}
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template<auto base, Unsigned U = decltype(base)>
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constexpr bool is_power_of(U x)
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{
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if constexpr (base == 2)
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return is_power_of_two(x);
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// FIXME: I am naive! A log2-based approach (pow<U>(base, (log2(x) / log2(base))) == x) does not work due to rounding errors.
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for (U exponent = 0; exponent <= log2(x) / log2(base) + 1; ++exponent) {
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if (pow<U>(base, exponent) == x)
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return true;
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}
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return false;
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}
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}
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