[boost] Re: boost::dereferenceable<>::operator->() problem

From: Pavel Kuznetsov (pavel_at_[hidden])
Date: 2004-08-23 22:04:18

David,

>> Ok, I got your point. But maybe then it might be a good idea to add
>> some kind of compile-time assert to
>> boost::dereferenceable<>::operator->()?

> I'd be happy to evaluate your patch containing an assert.

Here it is. First, we should add:

   #include <boost/type_traits/detail/yes_no_type.hpp>
   #include <boost/static_assert.hpp>

to the top of <boost/operators.hpp>. Then dereferenceable class
template definition (around 300 line of the same file):

   template <class T, class P, class B = ::boost::detail::empty_base>
   struct dereferenceable : B
   {
     P operator->() const
     {
       return &*static_cast<const T&>(*this);
     }
   };

should be changed as follows:

namespace detail {

template<class T>
struct has_lvalue_const_dereference_
{
private:
   template<class U>
   static ::boost::type_traits::yes_type discriminator_(U&(T::*)()const);

   template<class V>
   static ::boost::type_traits::no_type discriminator_(V(T::*)()const);

   static ::boost::type_traits::no_type discriminator_(...);

public:
   BOOST_STATIC_CONSTANT(bool, value =
        sizeof( discriminator_( &T::operator* ) ) ==  
sizeof( ::boost::type_traits::yes_type ) );
};

struct empty_base { };

} // namespace detail

template <class T, class P, class B = ::boost::detail::empty_base>
struct dereferenceable : B
{
   P operator->() const
   {
     BOOST_STATIC_ASSERT( ::boost::detail::has_lvalue_const_dereference_<T>::value  
);
     return &*static_cast<const T&>(*this);
   }
};

That's it, I think.

-- 
Pavel Kuznetsov
MetaCommunications Engineering
http://www.meta-comm.com/engineering