c++ boost::any 定义我自己的 print ，

Question

我正在努力寻找如何使用 boost::any 创建一个可以首先使用模板打印任何类型的打印函数。

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

我需要首先定义 print()。 我希望对any有真正的运算符<<，这个想法很简单：将类的实例附加到每个any对象 printer 具有合适的 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 any 不是（也不应该是）类模板。

拜托，我真的需要今晚或明天的帮助，我明天有截止日期，但我希望今晚就可以完成。

Answer 1

有一个非常简单的方法可以做到这一点，如“超越 C++ 标准库：简介”中所述Boost”：

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;
  }
};

然后使用 any_out 而不是 boost::any。

Answer 2

我这样做，我认为这是干净和安全的：

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;
}
}

对于您想要支持的任何类型，只需确保已为其类型和运算符调用 AnyWriter::register() 即可;<为它而存在。

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

Pawel Zubrycki 的回答非常好（提到 Björn Karlsson 的书）。

但代码在以下几行中存在一些错误：

// ...
o << *boost::any_cast<T>(a);    // should be:   o << *boost::any_cast<T>(&a);
// ...
a.streamer_->print(o, a);       // should be:   a.streamer_->print(o, a.o_);

这是 Pawel Zubrycki 答案的正确版本，该版本有效（部分......）

#ifndef ANY_OUT_H
#define ANY_OUT_H

#include <iostream>
#include <boost/any.hpp>

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

template <class T>
struct streamer_impl: streamer{
  void print(std::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);
};

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

#endif

此测试代码有效：

{
   any_out a = 5;
   std::cout << a << std::endl;
}

但是！！！

以下内容< strong>不起作用：

int main()
{
  char str[] = "mystring";
  any_out a = str;
  std::cout << a << std::endl;

  a = "myconststring";
  std::cout << a << std::endl;
}

这里什么也没打印。

为什么？？

那么在下面的构造函数中，类型就会混乱

any_out(const T& value)

那么，如果我们将 Streamer 实例化为从

new streamer_impl<T>()

构造函数中删除引用，

  any_out(const T value)

，即...是一种解决方案。

另一个解决方案是保留引用并调整 streamer_impl 的模板实例化。参见下文

以下推荐解决方案带来的是：

#ifndef ANY_OUT_H
#define ANY_OUT_H

#include <iostream>
#include <boost/any.hpp>

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

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

class any_out {
  boost::any o_;
  streamer *streamer_;
  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)
    : o_(value),
#if 1
      streamer_(new streamer_impl<typename std::decay<decltype(value)>::type>)
#else
      streamer_((o_.type() == typeid(const char *))
                ? static_cast<streamer *>(new streamer_impl<const char *>)
                : static_cast<streamer *>(new streamer_impl<T>))
#endif
  {
  }

  // template<class T> any_out(const T value)
  //   : o_(value),
  //     streamer_(new streamer_impl<T>)
  // {
  // }

  any_out(const any_out& a)
    : o_(a.o_), streamer_(a.streamer_ ? a.streamer_->clone() : 0) {}

  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);
};

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

#endif

上面带来一些麻烦的测试代码现在可以很好地工作（使用“推荐解决方案”）：

int main()
{
  char str[] = "mystring";
  any_out a = str;
  std::cout << a << std::endl;

  a = "myconststring";
  std::cout << a << std::endl;
}

Answer 4

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，其中一些看起来不错，而另一些（对我来说）则不然。不过，总的来说，这似乎是一项很难以一般方式完成的任务，因为（如该线程中提到的），某些类型永远不会是 ostream'able，但可以包含在 boost::any 中代码>对象。