Question

我正在使用boost序列化，主要是侵入版本。对于模板Matrix类，我希望non-intrusive version能够在Visual Studio上使用以下代码：

namespace boost 
{
    namespace serialization 
    {

        template<class Archive, int R, int C, class ElementType>
        void serialize(Archive & ar, Matrix<R, C, ElementType> & m, const unsigned int version)
        {
            ar & ...
        }

    } 
}

int R，int C是行和列，ElementType是double或float。

但是，这不适用于带有错误

的GCC 4.3.2

error: 'class Matrix<1u, 3u, double>' has no member named 'serialize'

如果我使用像

这样的特殊表格

namespace boost 
{
    namespace serialization 
    {

        template<class Archive>
        void serialize(Archive & ar, Matrix<3,1,double> & m, const unsigned int version)
        {
            ar & ...
        }

    } 
}

它在GCC上编译，但当然只针对一组特殊的模板参数。

我可以做些什么来使其适用于所有R，C和ElementType的两个编译器？

编辑：这些是导致错误的行：

/[myfolder]/lib/BOOST/1_44_0/include/boost/serialization/access.hpp: In static member function 'static void boost::serialization::access::serialize(Archive&, T&, unsigned int) [with Archive = boost::archive::binary_iarchive, T = Matrix<3u, 1u, double>]':
/[myfolder]/lib/BOOST/1_44_0/include/boost/serialization/serialization.hpp:70: instantiated from 'void boost::serialization::serialize(Archive&, T&, unsigned int) [with Archive = boost::archive::binary_iarchive, T = Matrix<3u, 1u, double>]'
/[myfolder]/lib/BOOST/1_44_0/include/boost/serialization/serialization.hpp:129: instantiated from 'void boost::serialization::serialize_adl(Archive&, T&, unsigned int) [with Archive = boost::archive::binary_iarchive, T = Matrix<3u, 1u, double>]'
/[myfolder]/lib/BOOST/1_44_0/include/boost/archive/detail/iserializer.hpp:182: instantiated from 'void boost::archive::detail::iserializer<Archive, T>::load_object_data(boost::archive::detail::basic_iarchive&, void*, unsigned int) const [with Archive = boost::archive::binary_iarchive, T = Matrix<3u, 1u, double>]'

Answer 1

它看起来像是签名/未签名的不匹配给我。您的模板函数使用int声明，但错误表示它尝试与模板匹配的对象具有参数1u和3u。当您实例化您尝试序列化的对象时，您是否使用维度的无符号值？尝试更改序列化模板函数以使用unsigned或使用int实例化Matrix。

模板类的非侵入式序列化方法

1 个答案: