Specialize a template class with a member function pointer

Question

I'm working with VS 2008 (sorry, can't update). I need template specialization for my list class. The argument I want to specialize is a member function pointer.

In short, how can I make the following code work (I can' use variadic templates):

// two or three args
template <typename T1, typename T2, void (T1::* FUNC)(const T2&) = 0> struct list;
// specialized code for 2 args.
template <typename T1, typename T2> struct list<T1, T2> { };

The error message is:

C2754: 'specialization' : a partial specialization cannot have a dependent non-type template parameter

Maybe there is no solution for my problem. But that also an answer.

Thanks for your help in advance.

This might be one of those instances where you would use a delegate. msdn.microsoft.com/en-us/library/213x8e7w%28v=vs.90%29.aspx — Tebc
why do you need that specialization? the primary template has a default argument, meaning you can omit it. what would you like to achieve? — Andy Prowl
If the member function pointer is null I have to implement a complete different list class. And would like to separate these two implementation to have clean code (no if ... else). — zussel
@Tebc I don't want to mix it with c# because the code needs to run under linux as well and should be native c++. Sorry. — zussel

Andy Prowl Andy Prowl · Accepted Answer · 2013-01-16T13:57:57

If I do understand correctly what you are trying to achieve, it cannot be done. Let's consider an even simpler case of illegal specialization:

template <typename T1, typename T2, void (T1::* FUNC)(const T2&)> struct list;
template <typename T1, typename T2> struct list<T1, T2, 0> { };

Here you always have to supply three arguments to instantiate your template, and you try to specialize the primary template for the case where the last argument is a null pointer.

Per § 14.5.5/8 of the C++ Standard:

"The type of a template parameter corresponding to a specialized non-type argument shall not be dependent on a parameter of the specialization. [ Example:

template <class T, T t> struct C {};
template <class T> struct C<T, 1>; // error

template< int X, int (*array_ptr)[X] > class A {};
int array[5];
template< int X > class A<X,&array> { }; // error

—end example ]"

You cannot specialize a non-type argument whose type is dependent on other types in the template parameter list. Thus, the above specialization is illegal.

Your original design is just a particular case of this specialization, where the last (non-type) parameter has a default argument value, which you omit in the specialization (omitting it is legitimate per se). The reason why this is illegal is not related with the presence of a default argument, but rather with the fact that you are trying to specialize a non-type argument that has a dependent type.

Specialize a template class with a member function pointer

1 Answers