c++ template specialization of template container

Question

what i would do is like...

template < template <typename ... > typename __Container, typename ... __Args >
ostream& operator<<(ostream& os, __Container<__Args ...> const& rhs){
  stringstream ss;
  int i = 0;
  for (const auto& it : rhs) 
    ss << "idx[" << i << "] " << it << "\n";
  return os << ss.str();
}

/// 
/// ... and something template specialization code
/// 

int main (){
  std::vector<int> vec_data = { ... };
  std::cout << vec_data << std::endl; 

  std::deque<int> dq_data = { ... };
  std::cout << dq_data << std::endl; 

  std::map<std::string, double> map_data = { {}, {}, {}, ... };
  std::cout << map_data << std::endl; 


  return 0;
}

in this example, deque and vector are not a problem, but when i tried to struggle to specialize std::map i dumped it. is there any possible way to do it?

Answer 1

Function templates can't be partially specialized, but can be overloaded:

template <typename ...P>
std::ostream &operator<<(std::ostream &os, const std::map<P...> &map)

But it's not the best idea, because you'd also have to specialize for std::multimap and std::unordered_[multi]map , and any non-standard containers you encounter in the future.

Instead you could create a function to print the single element, and overload it for std::pair (which is the element type of maps). Then call it from your operator.

There are more problems with the code:

• Operators should only be overloaded where ADL can find them. Otherwise you won't be able to call them from some places (from any namespace that defines any operator<< ; or from a template function defined above your operator).

  Since the only suitable namespace here would be std , and adding declarations there is not allowed, you shouldn't create such operator in the first place. Write a function instead.

• Identifiers containing __ or starting with _[AZ] are reserved, don't use them.

• The operator<< should accept any basic_ostream<...> to support wide streams.

• stringstream should be ostringstream , or removed altogether (print directly to the ostream).