Re: picking a base class with a tag

From:
jaybus56 <busch.juergen@gmx.net>
Newsgroups:
comp.lang.c++.moderated
Date:
Tue, 23 Jun 2009 14:04:36 CST
Message-ID:
<0ea33d34-dbdf-4f52-8477-f3850ae5b6ef@37g2000yqp.googlegroups.com>
On Jun 19, 7:22 pm, Cory Nelson <phro...@gmail.com> wrote:

I've got a class like this:

template<typename Tag>
struct BaseA;

template<typename Tag>
struct BaseB;

struct Derived : BaseA<Tag1>, BaseA<Tag2>, BaseB<Tag3> {};

static_casting to get at the bases is getting too verbose, I would
like to write a get() that returns a reference to the correct base
given a tag. ie:

Derived d;
BaseA<Tag1> &b = get<Tag1>(d);

But I'm drawing a blank.. any ideas?

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You could try something like this (works w/ CodeBlocks and GCC):

#include <iostream>

template <int name>
struct Tag
{
   static int GetName() { return name;}
};

template <class T>
struct BaseA
{
   BaseA()
   {}
   static int HasATag() { return T::GetName();}
};

template <class T>
struct BaseB
{
   BaseB()
   {}
   static int HasBTag() { return T::GetName();}
};

typedef Tag< 1> Tag1;
typedef Tag< 2> Tag2;
typedef Tag< 3> Tag3;

struct Derived : BaseA<Tag1>, BaseA<Tag2>, BaseB<Tag3> {};

template< template<class> class Cl, typename Ty, typename P>
inline Cl<Ty> TypeOfArg2TypeOfTemplateArg( P p, const Ty&)
{
   return Cl<Ty>( p);
}

template< typename Ty>
inline BaseA<Ty> Get( Derived d)
{
   return TypeOfArg2TypeOfTemplateArg<BaseA>( d, Ty());
}

int main()
{
   Derived d;

   BaseA< Tag1> ba1 = Get<Tag1>( d);
   BaseA< Tag2> ba2 = Get<Tag2>( d);
   std::cout << "ba1.HasATag " << ba1.HasATag() << std::endl;
   std::cout << "ba2.HasATag " << ba2.HasATag() << std::endl;
   return 0;
}

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