How to specialize only some members of a template class?

Question

Code:

template<class T>
struct A {
  void f1() {};
  void f2() {};

};

template<>
struct A<int> {
  void f2() {};
};


int main() {
  A<int> data;
  data.f1();
  data.f2();
};

ERROR:

test.cpp: In function 'int main()':
test.cpp:16: error: 'struct A<int>' has no member named 'f1'

Basically, I only want to specialize one function and use the common definition for other functions. (In actual code, I have many functions which I don't want to specialize).

How to do this? Thanks!

Answer 1

Consider moving common parts to a base class:

template <typename T>
struct ABase
{
    void f1();
};


template <typename T>
struct A : ABase<T>
{
    void f2();
}  


template <>
struct A<int> : ABase<int>
{
    void f2();
};

You can even override f1 in the derived class. If you want to do something more fancy (including being able to call f2 from f1 code in the base class), look at the CRTP.

Answer 2

Would this help:

template<typename T>
struct A
{
  void f1()
  {
    // generic implementation of f1
  }
  void f2()
  {
    // generic implementation of f2
  }
};

template<>
void A<int>::f2()                                                               
{
  // specific  implementation of f2
}

Answer 3

When we declare specializations for a template class, we must also define all its members, even those exactly equal to the generic template class, because there is no inheritance of members from the generic template to the specialization. So, in your specialization you have to implement void f1(); too.