Boost multiprecision cpp_int multiplied by a float

Question

It is possible to multiply a boost multiprecision int by a floating point number? Is this is not supported?

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

Answer 1

No that's not supported because it is lossy.

You can do the conversion explicitly:

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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";
}

Prints

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

CAVEAT

A common pitfall are lazily-evaluated expression templates. They're a trap when using temporaries:

auto result = x.convert_to<bigfloat>() * bigfloat("0.26");

Using result thereafter is Undefined Behaviour, because the temporaries have been destroyed. Assigning to bigfloat forces the evaluation.