Why don't I have to define the same members when I do total specialization of a class template in C++?

Question

I'm very surprised to find that the following compiles:

#include <iostream>

using namespace std;

template<typename T>
class SomeCls {
public:
  void UseT(T t) {
    cout << "UseT" << endl;
  }
};

template<>
class SomeCls<int> {
  // No UseT? WTF?!??!?!
};

int main(int argc, char * argv[]) {
  SomeCls<double> d;
  SomeCls<int> i;

  d.UseT(3.14);
  // Uncommenting the next line makes this program uncompilable.
  // i.UseT(100);

  return 0;
}

Why is this allowed? It just seems wrong that class SomeCls<int> doesn't need to have a void UseT(T t) method. I'm sure I'm missing the point of specialization here (I'm not a C++ expert). Can someone please enlighten me?

Answer 1

Because SomeCls<double> is a completely different type than SomeCls<int> or any other SomeCls<T>. They are not related in any way, so they can have whatever members they want. Just be sure not to call i.UseT(), this is where the compiler would start to complain, of course.