Question

is there a way to make this code work as intended?

#include <iostream>
using namespace std;

template<typename T> class templated{
public:
    static void f(){
        cout<<"doing something generically"<<endl;
    }
};

template<> class templated<int>{
public:
    static void g(){
        cout<<"doing something else, but specific to int"<<endl;
        f();
    }
};

int main(){
    templated<int>::g();
}

G++ complains that f is not declared in scope. I have tried all the possible variations in calling f() (templated<int>::f(), putting a dummy declaration in templated, move the declaration outside of class definition…), all of which failed, so I’ll omit them here.

Answer 1

No, a specialization is totally separate from the base template and doesn’t “inherit” anything from it.

Perhaps you can add a free function that can be called from everywhere?