ladybird/AK/IntegralMath.h
Tim Ledbetter e4715aa82a AK: Use correct type when calculating integral exp2()
Previously, integral `exp2()` would produce the incorrect result for
exponents above 31.
2023-10-27 21:59:44 -04:00

72 lines
1.4 KiB
C++

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