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下面的代码在使用 VC++ 的 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（适用于 Android 和 iOS），模板似乎存在问题。

“应该”匹配的模板的特定错误

..\boost/multiprecision/detail/default_ops.hpp:3745:18：注意：候选模板被忽略：无法将“表达式”与“数字”匹配 UNARY_OP_FUNCTOR(ceil, number_kind_floating_point) ^ ..\boost/multiprecision/detail/default_ops .hpp:3745:18：注意：候选模板已忽略：无法将 1 与 0 匹配 ..\boost/multiprecision/detail/default_ops.hpp:3745:18：注意：候选模板已忽略：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, void>] ..\boost/multiprecision/detail/default_ops.hpp:3745:18：注意：候选模板被忽略：无法匹配 1 和 0

VC++ 处理所有这一切都“很好”。

等待@Sehe 出现；]

4

1 回答 1

2

就像许多人已经建议的那样，表达式模板是罪魁祸首：

T 推导出为

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: 构造T(10)。相反，要么强制类型：

std::cout << floorBI<0>(BigFloat(A / (B * C))) << "\n";

或者更聪明地推断类型。

演示

住在科利鲁

#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

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于 2022-02-09T19:19:59.707 回答