[英]Boost multiprecision rounding towards 0
下面的代码使用 VC++ 在 Windows 上编译得很好:
#include <boost/multiprecision/cpp_int.hpp>
#include <boost/multiprecision/cpp_dec_float.hpp>
#include <iostream>
namespace mp = boost::multiprecision;
using BigFloat = mp::cpp_dec_float_50;
using BigInt = mp::uint256_t;
template <int decimals = 0, typename T> T floorBI(T const& v)
{
static const T scale = pow(T(10), decimals);
if (v.is_zero())
return v;
// ceil/floor is found via ADL and uses expression templates for
// optimization
if (v < 0)
return ceil(v * scale) / scale;
else
// floor is found via ADL and uses expression templates for optimization
return floor(v * scale) / scale;
}
int main()
{
BigFloat A = 3;
BigFloat B = 2;
static_cast<BigInt>(floorBI<0>(static_cast<BigFloat>(A) / static_cast<BigFloat>(B)));
return 0;
}
,但是对于 GCC(对于 Android 和 iOS),模板似乎存在问题。
“应该”匹配的模板的特定错误读取
..\boost/multiprecision/detail/default_ops.hpp:3745:18: 注意:候选模板被忽略:无法将“expression”与“number”匹配.hpp:3745:18: 注意:候选模板被忽略:无法匹配 1 与 0..\boost/multiprecision/detail/default_ops.hpp:3745:18: 注意:候选模板被忽略:要求 'boost::multiprecision:: number_category<boost::multiprecision::backends::cpp_int_backend<256, 256, boost::multiprecision::unsigned_magnitude, boost::multiprecision::unchecked, void> >::value == number_kind_floating_point' 不满足 [with Backend = boost::multiprecision::backends::cpp_int_backend<256, 256, boost::multiprecision::unsigned_magnitude, boost::multiprecision::unchecked, 无效>]..\boost/multiprecision/detail/default_ops.hpp:3745:18 : 注意:候选模板被忽略:无法将 1 与 0 匹配
VC++ 可以“很好地”处理所有这些。
等待@Sehe 出现;]
就像许多人已经建议的那样,表达式模板是罪魁祸首:
T推导为
boost::multiprecision::detail::expression<
boost::multiprecision::detail::divides,
boost::multiprecision::number<
boost::multiprecision::backends::cpp_dec_float<50>>,
boost::multiprecision::detail::expression<
boost::multiprecision::detail::multiply_immediates,
boost::multiprecision::number<
boost::multiprecision::backends::cpp_dec_float<50>>,
boost::multiprecision::number<
boost::multiprecision::backends::cpp_dec_float<50>>,
void, void>,
void, void>
不能从 int: T(10)
构造。 相反,要么强制类型:
std::cout << floorBI<0>(BigFloat(A / (B * C))) << "\n";
或者更聪明地推导类型。
#include <boost/multiprecision/cpp_dec_float.hpp>
#include <boost/multiprecision/cpp_int.hpp>
#include <boost/multiprecision/number.hpp>
#include <iostream>
namespace bmp = boost::multiprecision;
using BigFloat = bmp::cpp_dec_float_50;
template <int decimals = 0, typename T>
std::enable_if_t<not bmp::is_number_expression<T>::value, T>
floorBI(T const& v)
{
static const T scale = pow(T(10), decimals);
if (v.is_zero())
return v;
// ceil/floor is found via ADL and uses expression templates for
// optimization
if (v < 0)
return ceil(v * scale) / scale;
else
// floor is found via ADL and uses expression templates for optimization
return floor(v * scale) / scale;
}
template <int decimals = 0, typename Expr>
auto floorBI(Expr const& expr,
std::enable_if_t<bmp::is_number_expression<Expr>::value, void>* enable = nullptr)
{
return floorBI<decimals, typename Expr::result_type>(expr);
}
int main()
{
BigFloat A(3), B(2), C(2);
std::cout << floorBI<1>(BigFloat(A / (B * C))) << "\n";
std::cout << floorBI<1>(A / (B * C)) << "\n";
std::cout << floorBI<2>(BigFloat(A / (B * C))) << "\n";
std::cout << floorBI<2>(A / (B * C)) << "\n";
B *= -1;
std::cout << floorBI<1>(BigFloat(A / (B * C))) << "\n";
std::cout << floorBI<1>(A / (B * C)) << "\n";
std::cout << floorBI<2>(BigFloat(A / (B * C))) << "\n";
std::cout << floorBI<2>(A / (B * C)) << "\n";
}
印刷
0.7
0.7
0.75
0.75
-0.7
-0.7
-0.75
-0.75
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