If I want to specialise just one method in a template, how do I do it?

Question

Say I have a templated class like

template  struct Node
{
    // general method split
    void split()
    {
        // ... actual code her         


        

Answer 1

Your case is a fortunate example because you are fully-specializing the template. In other words there is only one template argument, and for the function you wish to specialize, you are providing a fully-specialized instantiation of that function definition. Unfortunately for partial class template specialization, it is a requirement that you re-implement the class.

For example, this works:

template<typename T, typename U>
struct Node
{
    void function() { cout << "Non-specialized version" << endl; }
};

template<>
void Node<int, char>::function() { cout << "Specialized version" << endl; }

The partially specialized version though will not work:

template<typename T, typename U>
struct Node
{
    void function() { cout << "Non-specialized version" << endl; }
};

template<typename U>
void Node<int, U>::function() { cout << "Specialized version" << endl; }

What you may want to-do if you find yourself in the second scenario, in order to avoid the needless duplication of code, is pack all the common elements into a common base-class, and then place all the elements that will be varying with the partial specialization in a derived class. That way you do not need to reduplicate all the common code in the base-class, you would only need to re-write the definitions for the specialized derived class.

Answer 2

Awkward method that I usually use: Declare a "basic" template implementation. Then declare your template implementation, the default implementation would just inherit the "basic" one. The customize would inherit the "basic", plus override specific methods. Like this:

template <typename T> struct NodeBase
{
    // methods
};

template <typename T> struct Node
    :public NodeBase<T>
{
    // nothing is changed
};

template <> struct Node<SpecificType>
    :public NodeBase<SpecificType>
{
    // re-define methods you want
};

Note that the "method overriding" has nothing to do with virtual functions or etc. It's just a declaring a function with the same name and parameters as in the base - the compiler will automatically use it.

Answer 3

You can provide a specialization for only that function outside the class declaration.

template <typename T> struct Node
{
    // general method split
    void split()
    {
        // implementation here or somewhere else in header
    }
};

// prototype of function declared in cpp void splitIntNode( Node & node );

template <>
void Node<int>::split()
{
     splitIntNode( this ); // which can be implemented
}

int main(int argc, char* argv[])
{
   Node <char> x;
   x.split(); //will call original method
   Node <int> k;
   k.split(); //will call the method for the int version
}

If splitIntNode needs access to private members, you can just pass those members into the function rather than the whole Node.

Answer 4

Just define some

template<>
void Node<Triangle*>::split()
{
}

after the primary template, but before the first time you ever instantiate it.