Question

I’m trying to understand whay i get an error on this code:
(the error is under g++ unix compiler. VS is compiling OK)

template<class T> class A {
public:
    T t;
public:
    A(const T& t1) : t(t1) {}
    virtual void Print() const { cout<<*this<<endl;}
    friend ostream& operator<<(ostream& out, const A<T>& a) {
            out<<"I'm "<<typeid(a).name()<<endl;
            out<<"I hold "<<typeid(a.t).name()<<endl;
            out<<"The inner value is: "<<a.t<<endl;
            return out;
    }
};

template<class T> class B : public A<T> {
public:
    B(const T& t1) : A<T>(t1) {}
    const T& get() const { return t; }
};

int main() {
    A<int> a(9);
    a.Print();
    B<A<int> > b(a); 
    b.Print();
    (b.get()).Print();  
    return 0;
}

This code is giving the following error:

main.cpp: In member function ‘const T& B::get() const’:
main.cpp:23: error: ‘t’ was not declared in this scope

It did compiled when i changed the code of B to this:

template<class T> class B : public A<T> {
public:
    B(const T& t1) : A<T>(t1) {}
    const T& get() const { return A<T>::t; }
};

I just cant understand what is the problem with the first code…
It doesn’t make sense that i really need to write “A::” every time…

Answer 1

You can also use this->t to access the base class template member.

In B::get(), the name t is not dependent on the template parameter T, so it is not a dependent name. Base class A<T> is obviously dependent on the template parameter T and is thus a dependent base class. Nondependent names are not looked up in dependent base classes. A detailed description of why this is the case can be found in the C++ FAQ Lite.