overloading operator << for std::tuple - possible simplications?

Question

I used an answer to the SO question "iterate over tuple" to write a method to overload <<. This method was tested and appears to work correctly with g++ 4.7 on Debian squeeze.

However this method is kind of roundabout, since it seems << cannot be explicitly instantiated (I found a post about it here). So, one is forced to define a string method and then call that. I have a similar method for vector, which is more direct. Does anyone have suggestions about how to eliminate the extra step of creating a string method, using the same approach, or otherwise? Thanks in advance.

#include <tuple>
#include <iostream>
#include <string>
#include <sstream>
#include <vector>

using std::ostream;
using std::cout;
using std::endl;
using std::vector;
using std::string;

// Print vector<T>.
template<typename T> ostream& operator <<(ostream& out, const vector<T> & vec)
{
  unsigned int i;
  out << "[";
  for(i=0; i<vec.size(); i++)
    {
      out << vec[i];
      if(i < vec.size() - 1)
    out << ", ";
    }
  out << "]";
  return out;
}

////////////////////////////////////////////////////////////////

// Print tuple.
template<std::size_t I = 0, typename... Tp>
inline typename std::enable_if<I == sizeof...(Tp), string>::type
stringval(const std::tuple<Tp...> & t)
{
  std::stringstream buffer;
  buffer << "]";
  return buffer.str();
}

template<std::size_t I = 0, typename... Tp>
inline typename std::enable_if<I < sizeof...(Tp), string>::type
stringval(const std::tuple<Tp...> & t)
{
  std::stringstream buffer;
  size_t len = sizeof...(Tp);
  if(I==0)
      buffer << "[";
  buffer << std::get<I>(t);
  if(I < len - 1)
    buffer << ", ";
  buffer << stringval<I + 1, Tp...>(t);
  return buffer.str();
}

template<typename... Tp> ostream& operator <<(ostream& out, const std::tuple<Tp...> & t)
{
  out << stringval(t);
  return out;
}

int
main()
{
  typedef std::tuple<int, float, double> T;
  std::tuple<int, float, double> t = std::make_tuple(2, 3.14159F, 2345.678);
  cout << t << endl;
}

When compiled, this gives

[2, 3.14159, 2345.68]

Answer 1

You can just pass the std::ostream& into that stringval function and use out << instead of buffer <<.

Demo:

#include <tuple>
#include <iostream>
#include <type_traits>

template <size_t n, typename... T>
typename std::enable_if<(n >= sizeof...(T))>::type
    print_tuple(std::ostream&, const std::tuple<T...>&)
{}

template <size_t n, typename... T>
typename std::enable_if<(n < sizeof...(T))>::type
    print_tuple(std::ostream& os, const std::tuple<T...>& tup)
{
    if (n != 0)
        os << ", ";
    os << std::get<n>(tup);
    print_tuple<n+1>(os, tup);
}

template <typename... T>
std::ostream& operator<<(std::ostream& os, const std::tuple<T...>& tup)
{
    os << "[";
    print_tuple<0>(os, tup);
    return os << "]";
}