Template partial specialization

[oswww]

I want to write a template that combines two type of resources:

class someClasses
{
    typedef someType ElementType;
}

template<class T1,class T2>
class combo
{
    ...
}

And I want to specify my T1 and T2 has the same ElementType, how can I write my combo class to partial specialize the general case so the ElementType check is at compile time?

[JLBorges]

template< class T1,class T2 >
class combo
{
    ...
};

template< class T > class combo<T,T> // partial specialization; types are the same
{
    ...
};

[oswww]

I don't mean that, T1 T2 can be different, but their ElementType should match

[JLBorges]

This is one way:

#include <type_traits>

template< typename T1, typename T2,
          bool = std::is_same< typename T1::ElementType,
                               typename T2::ElementType>::value >
class combo
{
    //...
};

// partial specialization; ElementTypes are the same
template< typename T1, typename T2 > class combo< T1, T2, true >
{
    // ...
};

[oswww]

Thanks a lot!
And now I think of writing a private constructor in the general combo class to ban the mismatch ElementType code from compiling, but the error is somewhat hard to understand to the user of my code.
I Actually just want the partial specialization version to be generated by the compiler, is there any better way?

[LB]

If you want clearer error messages, you can use static_assert in the class scope. I modified JLBorges' example:

#include <type_traits>

template< typename T1, typename T2,
          bool same = std::is_same< typename T1::ElementType,
                                    typename T2::ElementType>::value >
class combo
{
    static_assert(same, "T1::ElementType must be the same as T2::ElementType");
};

// partial specialization; ElementTypes are the same
template< typename T1, typename T2 > class combo< T1, T2, true >
{
    //actual class definition...
};

Though I don't think this is what you want (you are confusing me - how would you expect the compiler to generate either class automatically?)

[JLBorges]

This would suffice:

#include <type_traits>

template< typename T1, typename T2,
          bool = std::is_same< typename T1::ElementType,
                               typename T2::ElementType>::value >
class combo ; // not defined

template< typename T1, typename T2 > class combo< T1, T2, true >
{
    // ...
};

This would give a more cogent error diagnostic: (EDIT: this is the same as LB's post; hadn't seen it at that time)

#include <type_traits>

template< typename T1, typename T2,
          bool SAME_NESTED_TYPE = std::is_same< typename T1::ElementType,
                                                typename T2::ElementType>::value >
class combo
{
    static_assert( SAME_NESTED_TYPE,
                   "the nested type 'ElementType' must be the same for both types" ) ;
    //...
};

template< typename T1, typename T2 > class combo< T1, T2, true >
{
    // ...
};

[oswww]

Thanks, static_assert is what I want. Now my code looks like this:

#include <type_traits>

template<class T1,class T2>
class combo
{
    static_assert(std::is_same< typename T1::ElementType,
                                typename T2::ElementType>::value >, 
                    "T1::ElementType must be the same as T2::ElementType");
....
}

[LB]

Ah, I was slightly confused but now think I understand. Though I still do not see why you would want the compiler to 'generate' the design of the specialization unless you planned to allow the case where the nested types were different.

[JLBorges]

> Now my code looks like this

Yes! We don't need that specialization at all.