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
result
是Undefined 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/