Re: where to declare a friend operator >> (as well as >>)... in .h or .cpp file

From:
James Kanze <james.kanze@gmail.com>
Newsgroups:
comp.lang.c++
Date:
Mon, 8 Sep 2008 08:37:24 -0700 (PDT)
Message-ID:
<59496a2b-a134-435c-829a-cd40ecaf85a9@j22g2000hsf.googlegroups.com>
On Sep 8, 3:26 pm, puzzlecracker <ironsel2...@gmail.com> wrote:

I usually provide a normal,

named member function print, and derive from a template
class using the Barton and Nackman trick to provide the
operators, but


Please elaborate on this:the Barton and Nackman trick


It's pretty straight-forward, really. The base class contains
an inline definition of the operator---declared as friend,
because that's the only way you can provide an implementation
inline in a class for a non-member function. It's not really
something fundamentally necessary for operator<< and >>; it's
more useful for things like an operator+ (based on +=), etc.
Anyway, you define a base class template along the lines of:

    template< typename T >
    class Operators
    {
        friend T operator+( T const& lhs, T const& rhs )
        {
            T result( lhs ) ;
            result += rhs ;
            return result ;
        }
        // ...
        friend std::ostream& operator<<(
            std::ostream const& dest,
            T const& obj )
        {
            obj.print( dest ) ;
            return dest ;
        }
        // ...
    } ;

Then, whenever you need the operators, just derive from the
class:

    class Whatever : public Operators< Whatever >
    {
    public :
        // ...
        Whatever& operator+=( Whatever const& other ) ;
        // ...
        void print( std::ostream& dest ) const ;
        // ...
    } ;

I actually have several different template base classes for
this: ArithmeticOperators, MixedArithmeticOperators (with two
template parameters, for mixed type arithmetic),
STLIteratorOpertors (converts a "normal" iterator into STL), and
IOStreamOperators.

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James Kanze (GABI Software) email:james.kanze@gmail.com
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