Template specialization to use default type if class member typedef does not exist

Question

I\'m trying to write code that uses a member typedef of a template argument, but want to supply a default type if the template argument does not have that typedef. A simpli

Answer 1

First step: stop using "Type" and use the mpl standard "type".

BOOST_MPL_HAS_XXX_DEF(Type)

template < typename T >
struct get_type { typedef typename T::Type type; };

template < typename T >
struct calculate_type : boost::mpl::if_
<
  has_Type<T>
, get_type<T>
, boost::mpl::identity<default_type>
>::type {}

typedef calculate_type<A>::type whatever;

If you used "type" instead of "Type" in your metafunctions you wouldn't require the fetcher "get_type" to convert it and could just return T in that case.

Answer 2

To answer your addition - your specialization argument passes the member typedef and expects it to yield void as type. There is nothing magic about this - it just uses a default argument. Let's see how it works. If you say Get_Type<Foo>::type, the compiler uses the default argument of Enable, which is void, and the type name becomes Get_Type<Foo, void>::type. Now, the compiler checks whether any partial specialization matches.

Your partial specialization's argument list <T, typename T::Type> is deduced from the original argument list <Foo, void>. This will deduce T to Foo and afterwards substitutes that Foo into the second argument of the specialization, yielding a final result of <Foo, NonDefaultType> for your partial specialization. That doesn't, however, match the original argument list <Foo, void> at all!

You need a way to yield the void type, as in the following:

template<typename T>
struct tovoid { typedef void type; };

template<typename T, typename Enable = void> struct Get_Type { 
    typedef DefaultType Type; 
};
template<typename T> 
struct Get_Type< T, typename tovoid<typename T::Type>::type > {
    typedef typename T::Type  Type; 
};

Now this will work like you expect. Using MPL, you can use always instead of tovoid

typename apply< always<void>, typename T::type >::type

Answer 3

You can do that by utilizing SFINAE:

template<class T> struct has_type {
    template<class U> static char (&test(typename U::Type const*))[1];
    template<class U> static char (&test(...))[2];
    static const bool value = (sizeof(test<T>(0)) == 1);
};

template<class T, bool has = has_type<T>::value> struct Get_Type {
    typedef DefaultType Type;
};

template<class T> struct Get_Type<T, true> { 
    typedef typename T::Type Type;
};