可以将 boost multiprecision int 乘以浮点数吗?这不支持吗?
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
可以将 boost multiprecision int 乘以浮点数吗?这不支持吗?
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
不,不支持,因为它是有损的。
您可以显式进行转换:
#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";
}
印刷
x: 12345678
y: 0.26
Result: 3.20988e+06
z1: 3209876
z2: 3209876
一个常见的陷阱是惰性求值的表达式模板。使用临时对象时,它们是一个陷阱:
auto result = x.convert_to<bigfloat>() * bigfloat("0.26");
result
此后使用的是Undefined Behavior,因为临时对象已被销毁。分配到bigfloat
强制评估。