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

Robert Ramey 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& 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 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 > { ... }; HTH, -Ryan