Question

I have a problem with nested templates and their template specialization. Given the following classes:

A small template class

template<class U>
class T {
public:
    T(){}
    virtual ~T (){}

};

And some kind of nested template

template<typename T, template<typename> class U>
class A {
public:
    void foo() 
    { 
        std::cerr << "A generic foo"; 
    }
};

And a small main.cpp

int main(int argc, const char *argv[])
{
    A<int,T> *a = new A<int,T>;
    a->foo();

    //This wont work:
    A<double,T*> *b = new A<double,T*>;
    b->foo();

    return 0;
}

Now I need a specialization if U is a pointer:

    A<double,T*> *b = new A<double,T*>;
    b->foo();

How to achieve this? I tried something like:

template<typename T, template<typename> class U>
class A< T, U* >
{
public:
void foo()
{
    std::cerr << "A specialized foo";
}
};

But it just resolves in

A.h:18:16: Error: Templateargument 2 is invalid

Answer 1

What you’re tying to do is not possible, because T* has no meaning. Neither is it a proper type, nor does it match a template, which requires additional parameters. If U were to represent T*, what would U<int> be? You probably mean T<int>* but that doesn’t match your declaration, so there is no way to plug that type into A.

Since you asked for a way to get around this, from the top of my head, something like this.

Accept a third template argument to A, which I would call Expander and set it by default to this:

template <typename T> struct Expander {
  typedef T type;
};

Then, when invoking A you could say

A<int,T> normal;
A<int,T,PtrExpander> pointer;

with

template <typename T> struct PtrExpander {
  typedef T* type;
};

and A would be:

template<typename T, template<typename> class U, template <typename> class E = Expander> class A {
  typedef typename E<U<Your_Args_to_U> >::type;