是否可以将boost多精度int乘以浮点数?不支持吗?

using bigint = boost::multiprecision::number<boost::multiprecision::cpp_int_backend<>>;

boost::multiprecision::bigint x(12345678);
auto result = x * 0.26   // << THIS LINE DOES NOT COMPILE

最佳答案

不,不支持,因为它有损。

您可以明确地进行转换:

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#include <boost/multiprecision/cpp_int.hpp>
#include <boost/multiprecision/cpp_dec_float.hpp>

//using bigint = boost::multiprecision::number<boost::multiprecision::cpp_int_backend<>>;
using bigint   = boost::multiprecision::cpp_int;
using bigfloat = boost::multiprecision::cpp_dec_float_50;

int main() {
    bigint x(12345678);
    bigfloat y("0.26");
    std::cout << "x: " << x << "\n";
    std::cout << "y: " << y << "\n";
    bigfloat result = x.convert_to<bigfloat>() * y;

    //bigint z = result; // lossy conversion will not compile
    bigint z1 = static_cast<bigint>(result);
    bigint z2 = result.convert_to<bigint>();

    std::cout << "Result: " << result << "\n";
    std::cout << "z1: " << z1 << "\n";
    std::cout << "z2: " << z2 << "\n";
}

版画
x: 12345678
y: 0.26
Result: 3.20988e+06
z1: 3209876
z2: 3209876

警告

常见的陷阱是延迟评估的表达式模板。使用临时对象时,它们是一个陷阱:
auto result = x.convert_to<bigfloat>() * bigfloat("0.26");

此后使用resultUndefined Behaviour,因为临时文件已被销毁。分配给bigfloat会强制执行评估。

关于c++ - boost 多精度cpp_int乘以浮点数,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/48383986/

10-11 16:02