Re: [boost] [Units] A first step to ease the usage of Units' quantities inside of functors created with the Lambda library

From: Steven Watanabe (watanabesj_at_[hidden])
Date: 2008-06-06 12:01:52

AMDG

Torsten Maehne wrote:
> I have tried to add specializations to lambda.hpp like the following:
>
>     /// Partial specialization of return type trait for action
>     /// boost::lambda::functor<T> * unit<Dim, System>.
>     template<typename System, typename Dim, typename T>
>     struct plain_return_type_2<arithmetic_action<multiply_action>,
>                                boost::lambda::lambda_functor<T>,
>                                boost::units::unit<Dim, System> > {
>         typedef typename boost::units::multiply_typeof_helper<
>             boost::lambda::lambda_functor<T>,
>             boost::units::unit<Dim, System> >::type type;
>     };
>
> , to disambiguate the overloads. However the specialization seems to be
> not picked up by the compiler -- it still uses either:
>
> boost::lambda::functor<T> * Y
>
> or
>
> Y * unit<Dim, System>
>
> when uncommenting the check for bl::_1 * bu::meters in test_lambda.cpp.
> (Note: I didn't add the specialization to the attached corrected
> lambda.hpp.) Do you have an idea, what I'm doing wrong here?
>   

Yes.  The correct solution is to add overloads for the operators, rather 
than specializations
of plain_return_type_2.

template<typename System, typename Dim, typename T>
... operator*(const boost::lambda::lambda_functor<T>&, const 
boost::units::unit<Dim, System>&);

You'll have to peek at the internals of lambda to figure out what the 
return type should be.

In Christ,
Steven Watanabe