Re: [Boost-users] [enable_if] enable if member function present

From: Roman Perepelitsa (roman.perepelitsa_at_[hidden])
Date: 2008-05-07 05:19:02

Bill Buklis <boostuser <at> pbjzone.com> writes:
> 
> I would like to able to provide two versions of a function where one is
> enabled for only POD types and other another if the object provides a usable
> member function (mainly used for compatible non POD types). Anything else
> should fail to compile. The first part is easy. I can do this with:
> 
> template< class T >
> typename boost::enable_if< boost::is_pod<T>, void >::type
> DoSomething( T& object )
> {
> ...
> }
> 
> But, I'm not sure how to do the second part. For example:
> 
> template< class T >
> typename boost::enable_if< has "DoSomething" member, void >::type
> DoSomething( T& object )
> {
>     Object.DoSomething();
> }

Take a look at is_call_possible (http://preview.tinyurl.com/4x86m8)
and can_be_called (http://preview.tinyurl.com/52rkj7).

> Perhaps it would be easier to check for a defined typedef inside the object?
> But, I'm not sure on that either.
> 
> Additionally, if possible: If both conditions can be satisfied then the
> second version (member function) should be preferred.
> 
> Is this possible?

Yes, it's possible.

template <class T>
struct handle_pods {
  static do(T&);
};

template <class T>
struct handle_class_with_method {
  static do(T& t) {
    t.DoSomething();
  }
};

template <class T>
void DoSomething(T& t) {
  typedef typename mpl::if_<
    has_do_something_method<T>,
    handle_class_with_method<T>,
    typename mpl::if<
      is_pod<T>,
      handle_pod<T>,
      int
    >::type
  >::type impl;
  impl::do(t);
}

HTH,
Roman Perepelitsa.