Partially specializing member-function implementat

Question

I'm currently refactoring some code the explicitly specializes a member function of a class template with two template parameters.

template <class S, class T>
class Foo
{
  void bar();
};

template <class S, class T>
void Foo<S, T>::bar()
{ /* Generic stuff */ }

template <>
void Foo<SomeType, SomeType>::bar()
{ /* Some special function */ }

Now I added some more template parameters, so the class now looks like this:

template <class S, class EXTRA0, class T, class EXTRA1>
class Foo
{
  void bar();
};

These two extra parameters just add typedefs to my class, so the run-time functionality doesn't really change. Is there any way I can keep the (now partially) specialized implementation of bar? I can't seem to figure out the syntax for that and I have a hunch that it might not be possible.

Edit: I'm looking for something like:

template <class EXTRA0, class EXTRA1>
void foo<SomeType, EXTRA0, Sometype, EXTRA1>::bar()
{
   /* specialized implementation */
}

which does not seem to compile..

Answer 1

You can achieve that by using a specialize functor instead a function :

#include <iostream>

typedef int SomeType;

template <class A, class B>
class BarFunctor {
public:
    void operator()() {
        std::cout << "generic" << std::endl;
    }
};

template <>
class BarFunctor<SomeType, SomeType> {
public:
    void operator()() {
        std::cout << "special" << std::endl;
    }
};

template <class S, class T, class EXTRA0, class EXTRA1>
class Foo {
public:
    void helloWorld() {
        std::cout << "hello world !" << std::endl;
    }

    void bar() {
        return _bar();
    }

private:
    BarFunctor<S, T> _bar;
};

int main() {

    Foo<char, char, char, char> gen;
    Foo<SomeType, SomeType, char, char> spe;
    gen.helloWorld();
    spe.helloWorld();
    gen.bar();
    spe.bar();
    return 0;
}

Answer 2

I do not think that what you want is that easily possible. What about something like this:

template <class S, class EXTRA0, class T, class EXTRA1>
class FooBase
{
    void bar();
};

template <class S, class EXTRA0, class T, class EXTRA1>
void FooBase<S, EXTRA0, T, EXTRA1>::bar()
{ /* Generic stuff */ }

template <class S, class EXTRA0, class T, class EXTRA1>
class Foo
    : public FooBase <S, EXTRA0, T, EXTRA1>
{ };

template <class EXTRA0, class EXTRA1>
class Foo<int, EXTRA0, int, EXTRA1>
    : public FooBase <int, EXTRA0, int, EXTRA1>
{
    void bar ();
};

template <class EXTRA0, class EXTRA1>
void Foo<int, EXTRA0, int, EXTRA1>::bar()
{ /* Some special function */ }

Answer 3

You can make Base class , where you can define all your members except bar() and then create derivative classes(one for general purpose, one for SomeType):

template <class S, class T>
class FooBase
{
       // All other members 
};

template <class S, class EXTRA0, class T, class EXTRA1>
class Foo:public FooBase<S,T>
{
public:
      void bar()
      {

      }
};

struct SomeType {};

template <class EXTRA0, class EXTRA1>
class Foo<SomeType,EXTRA0,SomeType,EXTRA1>:public FooBase<SomeType,SomeType>
{
public:
    void bar()
    {

    }
};

int main()
{
    Foo<SomeType,int,SomeType,int> b;
    b.bar();
}

Answer 4

You are correct, it is not possible.

What you can do is create a helper member class template inside the new Foo, and place the specialized function inside it as a non-template member function. Specialize the helper class instead of the function.

Another alternative is to turn the specialization into a non-template overload.