can I specialize operator<< ?

Question

Hi,

I want to specialize operator<< but this code is not compiling;

template<>

std::ostream& operator<< < my_type >( std::ostream& strm, my_type obj);

Answer 1

Why not just overload?

// no template <>
std::ostream& operator<<( std::ostream& strm, my_type obj);

You only specialize when there exists a template to specialize.

_{Your parameter should probably be const my_type&, to avoid a needless copy.}

Answer 2

To specialize a template, first you have to have a template declared.

In the case of a free operator<< you don't need a template; you can just overload it for your my_type class:

std::ostream& operator<<( std::ostream& strm, my_type obj );

If your object isn't trivial in size, you may want to consider passing via a const reference so that you don't copy it every time you stream it:

std::ostream& operator<<( std::ostream& strm, const my_type& obj );

(Technically you can explicitly specialize an operator<<, but I don't think that this is what you want or need. In order to be able to use a template operator<< with the usual << syntax you need to make the template specialization deducible from one of the parameter types.

E.g.

// template op <<
template< class T >
std::ostream& operator<<( std::ostream&, const MyTemplClass<T>& );

// specialization of above
template<>
std::ostream& operator<< <int>( std::ostream&, const MyTemplClass<int>& );

)