C++ partial method specialization

Question

Is there a partial specialization for template class method?

 template <class A, class B>
 class C
 {
  void foo();
 }

it doesn't work to specialize it like this:

template <class A> void C<A, CObject>::foo() {};

Any help?

Answer 1

If you are already have specialized class you could give different implementation of foo in specialized class:

template<typename A, typename B>
class C
{
public:
    void foo() { cout << "default" << endl; };
};

template<typename A>
class C<A, CObject>
{
public:
  void foo() { cout << "CObject" << endl; };
};

---

To specialize member function in Visual C++ 2008 you could make it template too:

template<typename A, typename B>
class C
{
  template<typename T>
  void foo();

  template<>
  void foo<CObject>();
};

The solution above seems to will be available only in future C++ Standard (according to draft n2914 14.6.5.3/2).

Answer 2

I think there is a misunderstanding there.

There are two kinds of templates:

• the template classes
• the template methods

In your example, you have a template class, which of course contains some methods. In this case, you will have to specialize the class.

template <class A>
class C<A,CObject>
{
  void foo() { ... } // specialized code
};

The problem in your example is relatively simple: you define the method foo for the specialization C but this specialization has never been declared beforehand.

The problem here is that you have to fully specialize your C class (and thus copying a lot of data). There are a number of workarounds.

• Inheritance (Composition ?): do all the common work in a base class, then have the C class inherits and specialize as appropriate
• Friend: instead of having the 'foo' method being a member of C, define it as a friend free functions and specialize only this method
• Delegation: have your 'foo' method call another method 'bar', which is a free function, and specialize 'bar' appropriately

Which in code gives:

// 1- Inheritance
template <class A, class B>
class CBase
{
  // Everything that does not require specialization
};

template <class A, class B>
class C: public CBase<A,B>
         // depending on your need, consider using another inheritance
         // or even better, composition
{
  void foo(); // generic
};

template <class A>
class C<A,CObject> : public CBase<A,CObject>
{
  void foo(); // specialized
};

// 2- Friend
// note the change in signature:
// - now you need to pass the attributes to be changed
// - the last parameter helps differentiating the overload
//   as there is no specialization for functions
template <class A, class B> void foo(Arg1&, Arg2&, const B&);
template <class A> void foo(Arg1&, Arg2&, const CObject&);

template <class A, class B>
class C
{
  friend template <class, class> foo;
};

// 3- Delegation
// same signature as foo in (2)
template <class A, class B> void bar(Arg1&, Arg2&, const B&);
template <class A> void bar(Arg1&, Arg2&, const CObject&);

template <class A, class B>
class C
{
  void foo() { bar(member1, member2, B()); }
};

Hope it clarifies, and helps!

Answer 3

No, there is no partial function template specialization in C++0x to be added.

As correctly mentioned above, with regards to function templates basically 2 things were done:

• default template arguments were made available;
• variadic templates were introduced.

So as before, workarounds should be used to "emulate" partial function templates specialization.

Answer 4

Since the class is the template, you need to specialize that:

template <class A>
class C<A, CObject> 
{
   void foo() { ... }
}

Answer 5

If I remember correctly, you cannot make partial template specialization for functions. Not sure whether it is included in C++0X

Update: (Awaiting confirmation) As noted in the comments, partial template specialization of functions is possible in C++0X.