c++ boost::any to define my own print ,

Question

Am struggling a lot to find how to do to use boost::any to create a print function that can print any type using template first.

template 

        

Answer 1

Very nice answer by Pawel Zubrycki (mentioning Björn Karlsson's book).

But the code has a few errors in the following lines:

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

Here's a corrected version of Pawel Zubrycki's answer that works (partially...)

#ifndef ANY_OUT_H
#define ANY_OUT_H

#include 
#include 

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

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

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

  template 
  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

This test-code works:

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

However!!!!

The following does not work:

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

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

Here nothing gets printed.

Why??

Well the type is messed up, in the following constructor

any_out(const T& value)

if we then instantiate streamer as

new streamer_impl()

Removing the reference from the constructor, i.e.

  any_out(const T value)

... is one solution.

Another solution is to leave the reference and tweak the template instantiation of streamer_impl. See below

Which brings as to the following recommened solution is:

#ifndef ANY_OUT_H
#define ANY_OUT_H

#include 
#include 

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

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

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 any_out(const T& value)
    : o_(value),
#if 1
      streamer_(new streamer_impl::type>)
#else
      streamer_((o_.type() == typeid(const char *))
                ? static_cast(new streamer_impl)
                : static_cast(new streamer_impl))
#endif
  {
  }

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

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

  template 
  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

The test-code that gave some trouble above, now works nicely (with the "recommeded solution"):

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

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