How to serialize boost::rational

Question

I cannot serialize boost::rational<int>. I searched for a boost/serialize/rational.h header but it does not exist.

 /usr/include/boost/serialization/access.hpp:118:9: error: ‘class boost::rational<int>’ has no member named ‘serialize’

Is there a way to achieve it?

score 2 · Accepted Answer · answered Mar 21 '14 at 16:37

Just serialize the numerator and denominator.

Here's the legwork, in semi-generic form (supports archives with named nodes, like XML serialization, too): Live On Coliru

#include <boost/archive/xml_iarchive.hpp>
#include <boost/archive/xml_oarchive.hpp>
#include <boost/serialization/serialization.hpp>
#include <boost/rational.hpp>
#include <iostream>

namespace boost { namespace serialization {

    template <typename Archive, typename T>
        void save(Archive& ar, ::boost::rational<T> const& r, unsigned /*version*/)
        {
            int n = r.numerator(), d = r.denominator();
            ar & boost::serialization::make_nvp("numerator", n);
            ar & boost::serialization::make_nvp("denominator", d);
        }

    template <typename Archive, typename T>
        void load(Archive& ar, ::boost::rational<T>& r, unsigned /*version*/)
        {
            int n, d;
            ar & boost::serialization::make_nvp("numerator", n);
            ar & boost::serialization::make_nvp("denominator", d);

            r = ::boost::rational<T>(n, d);
        }

} }

BOOST_SERIALIZATION_SPLIT_FREE(boost::rational<int>);

using namespace boost;
#include <iostream>
#include <sstream>

int main()
{
    rational<int> number(2, 3), other;

    std::stringstream ss;
    {
        archive::xml_oarchive oa(ss);
        oa << serialization::make_nvp("rational", number);
    }

    std::cout << "Serialized: '" << ss.str() << "'\n";

    {
        archive::xml_iarchive ia(ss);
        ia >> serialization::make_nvp("rational", other);
    }

    std::cout << "Deserialized: " << other;
}

Prints

Serialized: '<?xml version="1.0" encoding="UTF-8" standalone="yes" ?>
<!DOCTYPE boost_serialization>
<boost_serialization signature="serialization::archive" version="10">
<rational class_id="0" tracking_level="0" version="0">
    <numerator>2</numerator>
    <denominator>3</denominator>
</rational>
</boost_serialization>

'
Deserialized: 2/3

Thank you for answering. I was missing the obvious. Nice code fashion. — jlandercy, Mar 21 '14 at 19:03

score -1 · Answer 2 · answered Oct 18 '20 at 04:22

Use the provided input/output functions: https://www.boost.org/doc/libs/1_64_0/libs/rational/rational.html#Input%20and%20Output

Serialize to a std::string:

template <typename I>
std::string serialize(boost::rational<I>& rational) const {
  std::stringstream rational_ss;

  rational_ss << rational;
  return rational_ss.str();
}

This serializes the rational to a string that looks like "3/5" e.g.

To deserialize, use boost::rational's >>.

How to serialize boost::rational

2 Answers2

Linked