Re: [Boost-users] help with boost/numeric/operators.hpp

Subject: Re: [Boost-users] help with boost/numeric/operators.hpp
From: Ryan Gallagher (ryan.gallagher_at_[hidden])
Date: 2012-01-03 15:38:56

Robert Ramey <ramey <at> rrsd.com> writes:
> template<class T>
> struct my_type : boost::numeric::operators<T> {
>     T & operator+=(T & rhs);
>     // automatically generate operator+, etc
>    // ...
> };
> 
> so that one can use
>     my_type<int> x, y, z;
>     z = x + y;
>     // etc..

I'm not familiar with boost::numeric::operators<T> but I have used 
Boost.Operators to do similar.  In this case I'd write:

template<typename T>
struct my_type : boost::addable<my_type<T> >
{
   my_type& operator+=(my_type<T> const& rhs);
   ...
};

> There is a simple example like this in the documentation.  So far so good.
> 
> Now given:
> 
> int & operator+=(int  & lhs, my_type<int>);
> 
> I want to use operators<T, U> to generate
> 
> int operator+(int  & lhs, my_type<int>);

If you want that exact return value ("int" not "my_type<int>") then I suppose
you've defined the proper operator+= overload for it and just need that 
visible and to modify my_type as follows:

template<typename T>
struct my_type : boost::addable<my_type<T>,
                 boost::addable<int, my_type<T> > >
{
   ...
};

HTH,

-Ryan