Template specialization and inheritance

Question

Suppose I have a template class with a lot of functions and I want to specialize them to change only a few of them and keep the other ones exactly as specified in the base template class.

How can I do that?

Below is what I want to achieve, but the solution isn't good, because it does not allow me to refer to the specialization for int as Base<int> – I need to use IntSpec for that.

#include <iostream>  using namespace std;  template<typename T> struct Base {   void print1() {cout << "Base::print1" << endl;};   void print2() {cout << "Base::print2" << endl;}; };  struct IntSpec : public Base<int> {   void print2() {cout << "Base<int>::print2()" << endl;}; };  int main() {   Base<double> d;   // Base<int> i;  <-- I want this kind of instantiation   IntSpec i;    d.print1();   d.print2();   i.print1();   i.print2(); } 

The output is:

Base::print1 Base::print2 Base::print1 Base<int>::print2() 

Answer 1

Nicol's solution works fine, but this is an alternative:

template<typename T> struct Base {   void print1() {cout << "Base::print1" << endl;};   void print2() {cout << "Base::print2" << endl;}; };  template<> void Base<int>::print2() {cout << "Base<int>::print2()" << endl;}; 

That way you can specialize only specific member functions and still use those that you haven't specialized(in this case, print1) without any problem. So now you'd use it just like you wanted:

Base<int> i; i.print1(); i.print2(); // calls your specialization 

Demo here.