Question

Consider the following example:

template <typename T>
class A {
    public:
        void f() {
            cout << "A::f()\n";
        }
};

template<>
class A<int> {
};

template<typename T>
class B: public A<T> {
    public:
        void g() {
            cout << "B::g()\n";
            A<T>::f();
        }
};

int main() {
    B<int> b;     // (1)
    b.g();        // (2)

    return 0;
}

Obviously the call to A::f() inside B::g() will fail for int template type. My question is at what point does the call fail? At (1) or (2)? I thought it should be (1) because at that point the compiler creates a new class with the template type int and compiles it. That compilation should fail in f() correct?

Answer 1

It fails at 2). Member function of templates are instantiated when called.

More precisely: When a class template is instantiated, the declaration of its member functions are instantiated, but not their definition. The definition is instantiated when the function is used.