Understanding (simple?) C++ Partial Template Specialization

This is a very often found problem, and there is a surprisingly simple solution. I will show it in an artificial example, because it's more clearer than to use your code, and you will have to understand it to adapt it to your code

template<typename A, typename B>
struct TwoTypes { };

template<typename A, typename B>
struct X {
  /* forwards ... */
  void f() { fImpl(TwoTypes<A, B>()); }

  /* special overload for <A, int> */
  template<typename A1>
  void fImpl(TwoTypes<A1, int>) {
    /* ... */
  }

  /* generic */
  template<typename A1, typename B1>
  void fImpl(TwoTypes<A1, B1>) {
    /* ... */
  }
};

Explicitly specializing functions is never (almost never?) the right way. In my work as a programmer, I've never explicitly specialized a function template. Overloading and partial ordering is superior.

Partial specialization of a function template, whether it is member function template or stand-alone function template, is not allowed by the Standard:

template<typename T, typename U> void f() {} //okay  - primary template
template<typename T> void f<T,int>() {}      //error - partial specialization
template<> void f<unsigned char,int>() {}    //okay  - full specialization

But you can partially specialize the class template itself. You can do something like this:

template <class A>
class Thing<A,int>  //partial specialization of the class template
{
    //..
    int doSomething();
};

template <class A>
int Thing<A,int>::doSomething()  { /* do whatever you want to do here */ }

Note that when you partially specialize a class template, then the template parameter-list of member function (in its definition outside the class), must match the template parameter list of the class template partial specialization. That means, for the above partial specialization of the class template, you cannot define this:

template <class A>
int Thing<A,double>::doSomething(); //error

Its not allowed, because the template parameter-list in function definition didn't match the template parameter-list of the class template partial specialization. §14.5.4.3/1 from the Standard (2003) says,

The template parameter list of a member of a class template partial specialization shall match the template parameter list of the class template partial specialization.[...]

For more on this, read my answer here:

C++ - Overload templated class method with a partial specilization of that method

---

So what is the solution? Would you partially specialize your class along with all the repetitive work?

A simple solution would be work delegation, instead of partially specializing the class template. Write a stand-alone function template and specialize this as:

template <class B>
B doTheActualSomething(B & b) { return b;  }

template <>
int doTheActualSomething<int>(int & b) { return b + 1; }

And then call this function template from doSomething() member function as:

template <class A, class B>
B Thing<A,B>::doSomething() { return doTheActualSomething<B>(b_); }

---

Since in your particular case, doTheActualSomething needs to know the value of only one member, namely b_, the above solution is fine, as you can pass the value to the function as argument whose type is the template type argument B, and specialization for int is possible being it full-specialization.

But imagine if it needs to access multiple members, type of each depends on the template type argument-list, then defining a stand-alone function template wouldn't solve the problem, because now there will be more than one type argument to the function template, and you cannot partially specialize the function for just, say, one type (as its not allowed).

So in this case you can define a class template instead, which defines a static non-template member function doTheActualSomething. Here is how:

template<typename A, typename B>
struct Worker
{
   B doTheActualSomething(Thing<A,B> *thing)
   {
      return thing->b_;
   }
};

//partial specialization of the class template itself, for B = int
template<typename A>
struct Worker<A,int>
{
   int doTheActualSomething(Thing<A,int> *thing)
   {
      return thing->b_ + 1;
   }
};

Notice that you can use thing pointer to access any member of the class. Of course, if it needs to access private members, then you've to make struct Worker a friend of Thing class template, as:

//forward class template declaration
template<typename T, typename U> struct Worker

template <class A, class B>
class Thing
{
    template<typename T, typename U>  friend struct Worker; //make it friend
   //...
};

Now delegate the work to the friend as:

template <class A, class B>
B Thing<A,B>::doSomething()
{
    return Worker<A,B>::doTheActualSomething(this); //delegate work
}

Two points to be noted here:

• In this solution, doTheActualSomething is not a member function template. Its not enclosing class which is template. Hence we can partially specialize the class template anytime, to get the desired effect of the partial member function template specialization.
• Since we pass this pointer as argument to the function, we can access any member of the class Thing<A,B>, even private members, as Worker<T,U> is also a friend.

Complete online demo : http://www.ideone.com/uEQ4S

---

Now there is still a chance of improvement. Now all instantiations of Worker class template are friends of all instantiation of Thing class template. So we can restrict this many-to-many friendship as:

template <class A, class B>
class Thing
{
    friend struct Worker<A,B>; //make it friend
   //...
};

Now only one instantiation of Worker class template is a friend of one instantiation of Thing class template. That is one-to-one friendship. That is, Worker<A,B> is a friend of Thing<A,B>. Worker<A,B> is NOT a friend of Thing<A,C>.

This change requires us to write the code in somewhat different order. See the complete demo, with all the ordering of class and function definitions and all:

http://www.ideone.com/6a1Ih