如何使用boost序列化與shared_ptr成員的對象

Question

有抽象I1和派生C1。

有摘要I2和派生C2。

I1有shared_ptr<I2>。我怎樣才能使他們序列化使用boost serializaton？我試圖做到這一點，但我的應用程序得到例外。

#include <sstream> 
#include <boost/shared_ptr.hpp> 
#include <boost/archive/binary_oarchive.hpp> 
#include <boost/archive/binary_iarchive.hpp> 
#include <boost/serialization/shared_ptr.hpp> 

struct I1 
{ 
    I1() {} 
    virtual ~I1() = 0 {} 

    template<class Archive> 
    void serialize(Archive &ar, const unsigned int version) 
    { 
    } 
}; 

struct C1 : I1 
{ 
    virtual ~C1() {} 

    template<class Archive> 
    void serialize(Archive &ar, const unsigned int version) 
    { 
     ar & boost::serialization::base_object<I1>(*this); 
    } 
}; 

struct I2 
{ 
    virtual ~I2() = 0 {} 

    template<class Archive> 
    void serialize(Archive &ar, const unsigned int version) 
    { 
     ar & p; 
    } 

    boost::shared_ptr<I1> p; 
}; 

struct C2 : I2 
{ 
    C2() { p = boost::shared_ptr<I1>(new C1); } 
    virtual ~C2() { } 

    template<class Archive> 
    void serialize(Archive &ar, const unsigned int version) 
    { 
     ar & boost::serialization::base_object<I2>(*this); 
    } 
}; 

int main() 
{ 
    C2 c2; 

    std::string s; 
    std::stringstream ss(s); 

    boost::archive::binary_oarchive oa(ss); 
    oa.register_type<I1>(); 
    oa.register_type<C1>(); 
    oa.register_type<I2>(); 
    oa.register_type<C2>(); 

    oa << c2; 

    boost::archive::binary_iarchive ia(ss); 
    //ia.register_type<I1>(); // cannot instantiate abstract class 
    ia.register_type<C1>(); 
    //ia.register_type<I2>(); // cannot instantiate abstract class 
    ia.register_type<C2>(); 

    ia >> c2; 
} 

Answer 1

升壓序列化文檔說here約BOOST_CLASS_EXPORT有趣的事情：

... ... BOOST_CLASS_EXPORT

因此，需要導出是通過使用通過指針或對其基類的引用操作的派生類來使用的。

您的p指針確實如此。添加這些宏代碼也擺脫從你的主醜明確register_type()的電話，這也是不錯的:)

所以，這段代碼看起來編譯和工作在VS2014：

#include <sstream> 
#include <boost/shared_ptr.hpp> 
#include <boost/archive/binary_oarchive.hpp> 
#include <boost/archive/binary_iarchive.hpp> 
#include <boost/serialization/shared_ptr.hpp> 
#include <boost/serialization/export.hpp> 

struct I1 
{ 
    I1() {} 
    virtual ~I1() = 0 {} 

    template<class Archive> 
    void serialize(Archive &ar, const unsigned int version) 
    { 
    } 
}; 

BOOST_CLASS_EXPORT(I1) 

struct C1 : I1 
{ 
    virtual ~C1() {} 

    template<class Archive> 
    void serialize(Archive &ar, const unsigned int version) 
    { 
     ar & boost::serialization::base_object<I1>(*this); 
    } 
}; 

BOOST_CLASS_EXPORT(C1) 

struct I2 
{ 
    virtual ~I2() = 0 {} 

    template<class Archive> 
    void serialize(Archive &ar, const unsigned int version) 
    { 
     ar & p; 
    } 

    boost::shared_ptr<I1> p; 
}; 

BOOST_CLASS_EXPORT(I2) 

struct C2 : I2 
{ 
    C2() { p = boost::shared_ptr<I1>(new C1); } 
    virtual ~C2() { } 

    template<class Archive> 
    void serialize(Archive &ar, const unsigned int version) 
    { 
     ar & boost::serialization::base_object<I2>(*this); 
    } 
}; 

BOOST_CLASS_EXPORT(C2) 

int main() 
{ 
    C2 c2; 

    std::string s; 
    std::stringstream ss(s); 

    boost::archive::binary_oarchive oa(ss); 
    oa << c2; 

    boost::archive::binary_iarchive ia(ss); 
    ia >> c2; 
} 

這是有趣的是，來自Boost文檔的聲明顯然不適用於所有編譯器，並且互聯網上的大量代碼示例在VS2014中不起作用。

Answer 2

添加

BOOST_SERIALIZATION_ASSUME_ABSTRACT(I1) 
BOOST_SERIALIZATION_ASSUME_ABSTRACT(I2) 

按照文檔http://www.boost.org/doc/libs/1_39_0/libs/serialization/doc/traits.html#abstract

UPDATE

我剛剛驗證與VS2013RTM和Boost 1_55，它JustWorks（TM），我ve

• 除去抽象鹼基類型註冊（它們永遠不能具體從反正存檔加載）

• 在沉默文件的頂部加入

  #pragma warning(disable: 4244) 
  #include <boost/config/warning_disable.hpp> 
  

  已知健談警告

編譯代碼並運行時沒有錯誤。爲了獲得良好的作風，你應該

• 通過指針序列化C2（堆棧變量可以subtly derail object tracking）

下面是完整的代碼，我結束了：

#pragma warning(disable: 4244) 
#include <boost/config/warning_disable.hpp> 
#include <sstream> 
#include <boost/shared_ptr.hpp> 
#include <boost/archive/text_oarchive.hpp> 
#include <boost/archive/text_iarchive.hpp> 
#include <boost/serialization/shared_ptr.hpp> 

struct I1 
{ 
    I1() {} 
    virtual ~I1() = 0 {} 

    template<class Archive> 
    void serialize(Archive &ar, const unsigned int version) 
    { 
    } 
}; 

struct C1 : I1 
{ 
    virtual ~C1() {} 

    template<class Archive> 
    void serialize(Archive &ar, const unsigned int version) 
    { 
     ar & boost::serialization::base_object<I1>(*this); 
    } 
}; 

struct I2 
{ 
    virtual ~I2() = 0 {} 

    template<class Archive> 
    void serialize(Archive &ar, const unsigned int version) 
    { 
     ar & p; 
    } 

    boost::shared_ptr<I1> p; 
}; 

struct C2 : I2 
{ 
    C2() { p = boost::shared_ptr<I1>(new C1); } 
    virtual ~C2() { } 

    template<class Archive> 
    void serialize(Archive &ar, const unsigned int version) 
    { 
     ar & boost::serialization::base_object<I2>(*this); 
    } 
}; 

int main() 
{ 
    boost::shared_ptr<I2> c2(new C2); 

    std::string s; 
    std::stringstream ss(s); 

    boost::archive::text_oarchive oa(ss); 
    oa.register_type<C1>(); 
    oa.register_type<C2>(); 

    oa << c2; 
    std::cout << "Serialized form: '" << ss.str() << "'\n"; 

    boost::archive::text_iarchive ia(ss); 
    ia.register_type<C1>(); 
    ia.register_type<C2>(); 

    ia >> c2; 
} 

而這裏的輸出：

Answer 3

我認爲你不應該在輸出存檔上註冊你的純虛擬類。可以使用export。在定義你的類之後，用下面的代碼替換剩下的代碼：

#include <boost/serialization/export.hpp> 
BOOST_CLASS_EXPORT(C1) 
BOOST_CLASS_EXPORT(C2) 

int main() 
{ 
    C2 c2; 

    std::string s; 
    std::stringstream ss(s); 

    { 
     boost::archive::binary_oarchive oa(ss); 
     oa << c2; 
    } 

    boost::archive::binary_iarchive ia(ss); 
    ia >> c2; 
} 

我把輸出檔案放到一個單獨的塊中。我認爲它最有可能沒有它，但我想確保一切都被刷新（通過超出範圍）。