Question

我很难找到如何使用boost::any创建一个可以先使用模板打印任何类型的打印功能。

template <typename T>
struct printer {
    void print(ostream& os, const boost::any& a);
};

我需要先定义print()。我希望拥有真正的operator <<，这个想法很简单：将每个任何对象附加到一个类的实例 printer<T>使用合适的T并在any的值类型更改时更改此对象。第一个技术问题是打印机对象依赖于T，而任何不是（也不应该是）类模板。

我真的需要一只手为今晚或明天我有一个明天的截止日期，但我希望今晚继续努力。

Answer 1

“Beyond the C++ Standard Library: An Introduction to Boost”中描述了很简单的方法：

struct streamer {
  virtual void print(ostream &o, const boost::any &a) const =0;
  virtual streamer * clone() const = 0;
  virtual ~streamer() {}
};

template <class T>
struct streamer_impl: streamer{
  void print(ostream &o, const boost::any &a) const { o << *boost::any_cast<T>(a); }
  streamer *clone() const { return new streamer_impl<T>(); }
};

class any_out {
  streamer *streamer_;
  boost::any o_;
  void swap(any_out & r){
    std::swap(streamer_, r.streamer_);
    std::swap(o_, r.o_);
  }
public:
  any_out(): streamer_(0) {}
  template<class T> any_out(const T& value)
    : streamer_(new streamer_impl<T>()), o_(value) {}
  any_out(const any_out& a)
    : streamer_(a.streamer_ ? a.streamer_->clone() : 0), o_(a.o_) {}

  template <class T>
  any_out & operator=(const T& r) { 
    any_out(r).swap(*this);
    return *this;
  }
  ~any_out() { delete streamer_; }

  friend std::ostream &operator<<(std::ostream& o, const any_out & a) {
    if(a.streamer_)
      a.streamer_->print(o, a);
    return o;
  }
};

然后您使用any_out代替boost::any。

Answer 2

我是这样做的，我认为这是干净安全的：

any_extension.hpp：

namespace cpputil
{

struct AnyWriter
{
    /// Register a type with the AnyWriter.
    /// @pre T must have an ostream << operator available somewhere
    template<class T> static bool registerType()
    {
        return registeredTypes().emplace(std::type_index(typeid(T)),
                                         std::bind(&AnyWriter::write<T>,
                                                   std::placeholders::_1,
                                                   std::placeholders::_2)).second;
    }

    /// Write any registred object to a stream
    /// @pre Underlying type must have been registered with a call to AnyWriter::registerType<T>
    /// @param os is reference to a std::ostream
    /// @param anyObject is a reference to a boost::any
    static void writeAny(std::ostream& os, const boost::any& anyObject);
private:

    // A function object that converts an any to a type and streams it to an ostream
    using WriteFunction = std::function<void (std::ostream&, const boost::any&)>;

    // a map of typeinfo to WriteFunction
    using RegisteredTypes = std::unordered_map<std::type_index, WriteFunction >;

    // retrieve the WriteFunction map in a safe way
    static RegisteredTypes& registeredTypes();

    // Convert an any to a type, and write it to a stream
    template<class T> static void write(std::ostream& os, const boost::any& anyObject) {
        try {
            const T& typedObject = boost::any_cast<const T&>(anyObject);
            os << typedObject;
        }
        catch(boost::bad_any_cast& e) {
            os << "<exception in conversion: " << e.what() << ">";
        }
    }

};
}

namespace std {
    ostream& operator<<(ostream& os, const ::boost::any& anyObject);
}

any_extension.cpp：

#include "any_extension.h"
#include <string>

namespace cpputil {

namespace AnyWriterRegistration {
    const bool stringRegistered = AnyWriter::registerType<std::string>();
    const bool intRegistered = AnyWriter::registerType<int>();
    const bool doubleRegistered = AnyWriter::registerType<double>();
}



AnyWriter::RegisteredTypes& AnyWriter::registeredTypes()
{
    static RegisteredTypes _registrationMap;
    return _registrationMap;
}

void AnyWriter::writeAny(std::ostream &os, const boost::any &anyObject)
{
    auto registered = registeredTypes();
    auto iFind = registered.find(anyObject.type());
    if(iFind == registered.end()) {
        os << "<unregistered type: " << anyObject.type().name() << ">";
    }
    else {
        iFind->second(os, anyObject);
    }
}

}

namespace std {
ostream& operator<<(ostream& os, const ::boost::any& anyObject)
{
    if(anyObject.empty()) {
        os << "<empty>";
    }
    else {
        cpputil::AnyWriter::writeAny(os, anyObject);
    }
    return os;
}
}

对于您支持的任何类型，只需确保已为其类型调用AnyWriter :: register（）并且运算符＆lt;＆lt;为它存在。

例如：

any_test.cpp：

struct chicken {};
std::operator<<(std::ostream& os, const chicken& aChicken) {
    os << "cluck!";
    return os;
}

namespace {
    const bool chickenRegistered = AnyWriter::register<Chicken>();
}

void chickenTest() {
    boost::any animal = chicken();
    std::cout << animal << std::endl;
}

输出：咕！

Answer 3

在Boost邮件列表中查看此主题： http://lists.boost.org/Archives/boost/2005/01/79232.php

它有一些想法，其中一些似乎没有问题，其中一些没有（对我而言）。但总的来说，这似乎是一项难以完成的任务，因为（如该线程所述），某些类型永远不会被ostream，但可能包含在boost::any对象中。

c ++ boost :: any来定义我自己的打印，

3 个答案: