Re: [Boost-users] Template instantiation problem

From: François Duranleau (duranlef_at_[hidden])
Date: 2006-05-10 22:58:01

On Wed, 10 May 2006, Rush Manbert wrote:

> I have a template class that is designed to contain built in types or
> std::string types. I always need to be able to cast an instance of the
> class as a std::string, and I also need to be able to cast it as the
> parameter type with which it was instantiated.
>
> In my class definition, I have defined two cast operators, like so:
> (Extraneous stuff omitted)
[...]
> I thought that if I could conditionally define the cast as T operator
> (or the cast as std::string) I would be okay, but that means testing
> against the value of T with the preprocessor. I don't see how to do
> that, but I know that some of you folks are really good at bending the
> preprocessor to your will.
>
> Can anyone show me a way out of this predicament?

No need for the preprocessor. It can all be done with templates. I could 
imagine something like this:

template < typename Derived ,
            typename T >
struct convert_operator
{
     operator const T& ()
     {
         // ... your code, you can access the subclass with:
         // static_cast< Derived& >( * this )
         // Add a const before Derived if the operator is const
     }
} ;

template < typename Derived >
struct convert_operator< Derived , ::std::string >
{
    // nothing
} ;

template < typename T >
class MyDataObject
     : public MyDataObjectBase
     , public convert_operator< MyDadaObject< T > , T >
{
     // Cast as std::string
     operator const std::string& ()
     {
         // ...some code here
     }

     // ...
} ;

-- 
François Duranleau
LIGUM, Université de Montréal
"As the caterpillar chooses the fairest leaves to lay her eggs on, so the
  priest lays his curse on the fairest joys."
                               - William Blake, _Marriage of Heaven & Hell_