Question

I’m trying to make the following bit of code to work…

#include <list>

template <typename T>
class container{
public:
    virtual T func_x(){
        T temp;
        //do stuff with list<t> test
        return temp;
    }
private:
    std::list<T> test;
};

template <typename T>
class container2 : public container<T>{
public:
    virtual T func_x(){
        T temp;
        //do different stuff with list<T> test
        return temp;
    }
};

What I want to be able to do is declare

container<T> x;
container2<T> y;

And be able to have y have access to all the public functions of x, except that it behaves differently for func_x.

The problem I have now is that func_x in class container2 cannot use;

std::list<T> test;

I have even tried making class container completely public. Still no dice. Can this be done?

Thanks!

Answer 1

Members are by default private for classes:

template <typename T>
class container2 : public container<T>{
    //************
    // no modifier
    virtual T func_x(){
        T temp;
        //do different stuff with list<T> test
        return temp;
    }
private:
    std::list<T> test;
};

means that func_x is private, since no modifier is specified.

You need to explicitly declare func_x as public, as you have for class container.

“Just because it’s public in the base class doesn’t mean it’s automatically that way for the derived”.

EDIT:

If you want base class members to be accessible in derived classes, you must declare them either protected or public. So, to answer your follow-up question, change

private:
    std::list<T> test;

to

protected:
    std::list<T> test;

Also, in the future, don’t edit the question to ask a new one. You should create a new question to deal with the new issue. It can be misleading for others who see answers that no longer apply to the new question.