Re: template specialization

From:
"Alf P. Steinbach" <alfps@start.no>
Newsgroups:
comp.lang.c++
Date:
Mon, 15 Sep 2008 06:13:52 +0200
Message-ID:
<uM-dnVcaxKycfFDVnZ2dnUVZ_q3inZ2d@posted.comnet>
* Konstantin:

mqrk wrote:

The simplest (and probably the best) solution is like this:


The problem still remains:

template <typename T> struct Container< T, std::set >
{
    std::set<T> data;
};

template <typename T> struct Container< T, std::list >
{
    std::list<T> data;
};

These are specializations, and one needs to write a primary template.
However, std::set template has three parameters: set<Key, Compare,
Alloc>, and std::list has only two: list<Type, Alloc>

Thus, if the primary template is

template < typename, template < typename, typename, typename > class >
    class Container;

then the second specialization yields an error ("error C3201: the
template parameter list for class template 'std::list' does not match
the template parameter list for template parameter ...")

and if the primary template is

template < typename T, template<typename, typename> class >
    class Container;

then the "set" template yields an error, and "list" works fine.


<code>
#include <set>
#include <list>

template< typename T > class AccessPolicy;

template< typename T > class AccessPolicy< std::set<T> >
{
public:
     typedef std::set<T> Cont;
     static void add( Cont& cont, T const& id )
     {
         cont.insert( id );
     }
};

template< typename T > class AccessPolicy< std::list<T> >
{
public:
     typedef std::list<T> Cont;
     static void add( Cont& cont, T const& id )
     {
         cont.push_back( id );
     }
};

template<
     typename CType,
     template< typename T > class Access = AccessPolicy
     >
class Container
{
private:
     CType data;
public:
     typedef typename CType::value_type Id;

     void add( Id const& id )
     {
         Access<CType>::add( data, id );
     }
};

int main()
{
     using namespace std;
     Container< std::list<double> > a;
     Container< std::set<double> > b;

     a.add( 0 );
     b.add( 0 );
}
</code>

Cheers & hth.,

- Alf

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