Re: [Boost-users] [test] BOOST_CHECK_EQUAL type to stream

From: John C. Femiani (john.femiani_at_[hidden])
Date: 2008-07-17 20:54:02

Steven Watanabe wrote:
> AMDG
>
> John C. Femiani wrote:
>> What aspect wouldn't work as I expected?  Do you mean that overloads 
>> of repr that aren't in namespace boost could occlude ::boost::repr? 
>> In that case its a lot like ::boost::begin I think.
>
> You do realize that boost::begin is implemented using
> range_begin which should be defined in the user's namespace, right?
> http://www.boost.org/doc/libs/1_35_0/libs/range/doc/boost_range.html#method2 
>
>
Yes, but I thought it worked with std::vector because of the overload of 
range_begin that is in the boost namespace. A similar strategy would 
_not_ seem to fix your example though.

>
> It fails for exactly the same reason that overloading operator<< does.
>
> In Christ,
> Steven 
Aaargh., the overload came too late :(
Will this approach using template specialization work?

//in repr.hpp
namespace boost {
    template<class T>  
    struct repr {
        typedef T const& type;
        T const& make(T const& arg){
              return arg;  
         }
    };
  
    template<class T>  
    typename repr<T>::type make_repr(T const& arg){
        return ::boost::repr<T>().make(arg);
    }
}

//In some code
namespace boost {
   template<class T>
   void print(const T& t) {
       std::cout << make_repr(t);
   }
}

//This is how you make a proxy
namespace boost {
   template<class T>
   struct repr<std::vector<T> >  {
        typedef  std::string type;
        std::string make(std::vector<T> const& arg){
             return "<vector>";
        }
   };
}

--John