C++ Boost将对象序列化为json

假设我有一个来自type

Animal

的对象，并将其序列化

在下面的示例中，要序列化的内容是

22 serialization::archive 16 0 0 4 1 5 Horse

很好。但是，如果我需要将其序列化为

json

，该怎么办呢。是否可以通过

boost

序列化

我寻找这样一条线：

{
  "legs": 4,
  "is_mammal": true,
  "name": "Horse"
}

代码：

---

不，没有这样的事

您可以编写自己的（通过实现归档概念）。但我认为这不值得努力。只需使用JSON库

下面是与您的示例一起使用的最小输出归档模型的草图：

#include <boost/serialization/serialization.hpp>
#include <boost/serialization/nvp.hpp>
#include <iostream>
#include <iomanip>
#include <sstream>

struct MyOArchive {
    std::ostream& _os;
    MyOArchive(std::ostream& os) : _os(os) {}
    using is_saving = boost::true_type;

    template <typename T>
        MyOArchive& operator<<(boost::serialization::nvp<T> const& wrap) {
            save(wrap.name(), wrap.value());
            return *this;
        }

    template <typename T>
        MyOArchive& operator<<(T const& value) {
            return operator<<(const_cast<T&>(value));
        }

    template <typename T>
        MyOArchive& operator<<(T& value) {
            save(value);
            return *this;
        }

    template <typename T> MyOArchive& operator&(T const& v) { return operator<<(v); }

    bool first_element = true;
    void start_property(char const* name) {
        if (!first_element) _os << ", ";
        first_element = false;
        _os << std::quoted(name) << ":";
    }
    template <typename T> void save(char const* name, T& b) {
        start_property(name);
        save(b);
    }
    void save(bool b) { _os << std::boolalpha << b; }
    void save(int i) { _os << i; }
    void save(std::string& s) { _os << std::quoted(s); }

    template <typename T>
    void save(T& v) {
        using boost::serialization::serialize;
        _os << "{";
        first_element = true;
        serialize(*this, v, 0u);
        _os << "}\n";
        first_element = false;
    }
};

class Animal {
  public:
    Animal() {}
    void set_leg(int l) { legs = l; };
    void set_name(std::string s) { name = s; };
    void set_ismammal(bool b) { is_mammal = b; };
    void print();

  private:
    friend class boost::serialization::access;

    template <typename Archive> void serialize(Archive &ar, unsigned) 
    {
        ar & BOOST_SERIALIZATION_NVP(legs)
           & BOOST_SERIALIZATION_NVP(is_mammal)
           & BOOST_SERIALIZATION_NVP(name);
    }

    int legs;
    bool is_mammal;
    std::string name;
};

void Animal::print() {
    std::cout << name << " with " << legs << " legs is " << (is_mammal ? "" : "not ") << "a mammal" << std::endl;
}

void save_obj(const Animal &animal, std::stringstream &stream) {
    MyOArchive oa{ stream };
    oa << animal;
}

int main() {
    std::stringstream stream;
    {
        Animal animal;
        animal.set_name("Horse");
        animal.set_leg(4);
        animal.set_ismammal(true);

        save_obj(animal, stream);
    }

    std::cout << "stream print: " << stream.str() << std::endl;
}

警告 我不推荐这种方法。事实上，上面有许多遗漏的东西——最明显的是，它只是输出

---

查看：boost:：property_tree:：write_json@FrankS101，经常使用它。但是如何将它与

存档连接起来呢

？非常感谢您花费的时间。除了创建一个boost树，然后将其序列化为json之外，我还考虑了一个类似的想法来处理更糟糕的情况。

object print: Horse with 4 legs is a mammal
stream print: 22 serialization::archive 16 0 0 4 1 5 Horse

#include <boost/serialization/serialization.hpp>
#include <boost/serialization/nvp.hpp>
#include <iostream>
#include <iomanip>
#include <sstream>

struct MyOArchive {
    std::ostream& _os;
    MyOArchive(std::ostream& os) : _os(os) {}
    using is_saving = boost::true_type;

    template <typename T>
        MyOArchive& operator<<(boost::serialization::nvp<T> const& wrap) {
            save(wrap.name(), wrap.value());
            return *this;
        }

    template <typename T>
        MyOArchive& operator<<(T const& value) {
            return operator<<(const_cast<T&>(value));
        }

    template <typename T>
        MyOArchive& operator<<(T& value) {
            save(value);
            return *this;
        }

    template <typename T> MyOArchive& operator&(T const& v) { return operator<<(v); }

    bool first_element = true;
    void start_property(char const* name) {
        if (!first_element) _os << ", ";
        first_element = false;
        _os << std::quoted(name) << ":";
    }
    template <typename T> void save(char const* name, T& b) {
        start_property(name);
        save(b);
    }
    void save(bool b) { _os << std::boolalpha << b; }
    void save(int i) { _os << i; }
    void save(std::string& s) { _os << std::quoted(s); }

    template <typename T>
    void save(T& v) {
        using boost::serialization::serialize;
        _os << "{";
        first_element = true;
        serialize(*this, v, 0u);
        _os << "}\n";
        first_element = false;
    }
};

class Animal {
  public:
    Animal() {}
    void set_leg(int l) { legs = l; };
    void set_name(std::string s) { name = s; };
    void set_ismammal(bool b) { is_mammal = b; };
    void print();

  private:
    friend class boost::serialization::access;

    template <typename Archive> void serialize(Archive &ar, unsigned) 
    {
        ar & BOOST_SERIALIZATION_NVP(legs)
           & BOOST_SERIALIZATION_NVP(is_mammal)
           & BOOST_SERIALIZATION_NVP(name);
    }

    int legs;
    bool is_mammal;
    std::string name;
};

void Animal::print() {
    std::cout << name << " with " << legs << " legs is " << (is_mammal ? "" : "not ") << "a mammal" << std::endl;
}

void save_obj(const Animal &animal, std::stringstream &stream) {
    MyOArchive oa{ stream };
    oa << animal;
}

int main() {
    std::stringstream stream;
    {
        Animal animal;
        animal.set_name("Horse");
        animal.set_leg(4);
        animal.set_ismammal(true);

        save_obj(animal, stream);
    }

    std::cout << "stream print: " << stream.str() << std::endl;
}

stream print: {"legs":4, "is_mammal":true, "name":"Horse"}