Re: [Boost-users] How to change the default enums serialization inBoost.Serialization

4 Oct 2011


      Andrés Senac wrote:
...
Hi,
By default in Boost.Serialization, enum types are serialized as a
32-bit integer. But I need to serialize some enum types as different
width integer. I've tried to specialize the
boost::serialization::serialize method, but it seems it doesn't work
for enums.    
Here is my attempt:
#include <iostream>
#include 
#include 
enum MyEnum_t
{
    HELLO, BYE
};
namespace boost
{
namespace serialization
{
template< class Archive >
void save(Archive & ar, const MyEnum_t & t, unsigned int version)
{
    unsigned char c = (unsigned char) t;
    ar & c;
}
template< class Archive >
void load(Archive & ar, MyEnum_t & t, unsigned int version)
{
    unsigned char c;
    ar & c;
    t = (MyEnum_t) c;
}
} // namespace serialization
} // namespace boost
BOOST_SERIALIZATION_SPLIT_FREE(MyEnum_t)
int main(int argc, const char *argv[])
{
    boost::asio::streambuf buf;
    boost::archive::binary_oarchive pboa(buf);
    buf.consume(buf.size()); // Ignore headers
    MyEnum_t me = HELLO;
    pboa << me;
    std::cout << buf.size() << std::endl; // buf.size() = 4, but I
want 1 
    return 0;
}
Thanks in advance,
Andrés
Consider making a wrapper for you enum - like this;

struct enum_wrapper {
    unsigned char m_value;
    enum_wrapper(unsigned int value) :
        m_value(value)
   {
        assert(value <= 255);
  }
   template<class Archive>
   void save(Archive &ar, unsigned int version){
        ar << m_value;
    }
    void load
    ...
}

then you can use

....
    enum {
        a
        b
        ..
    } e;
    ar & enum_wrapper(e);
..

Robert Ramey