C ++：使用类型名作为基础的模板类中的函数调用编译时错误

Question

I don't know and unable to find the concept name to form my question simple (And also the title too). So i posted the entire code. Below is the code, with two base classes. And one derived class created using class template. Derive class inherits the two base class using typename T. Each base class are having unique function called foo() & boo().

My questions are:
1) In this condition, is it possible to use both the function call in the derived class (in callMethods())? i am getting error and no idea to solve it.
2) Or do i need to change the class design if the existing design is wrong?
3) Is it a good practice to design the class like this if the design is correct?

Error Message:

error: 'boo' is not a member of 'mybase_1'
         T::boo();     
error: 'foo' is not a member of 'mybase_2'
         T::foo();      
                ^                ^

Test code:

enum class myenum : int {one, two};

class mybase_1{
protected:
    void foo(){
        qDebug() << "foo called\n";
    }
};

class mybase_2{
protected:
    void boo(){
        qDebug() << "boo called\n";
    }
};

template <typename T>
class myderived : public T{
public:
    myderived(myenum _enm);
    void callMethods();
private:
    myenum enm;
};

template <typename T> myderived<T>::myderived(myenum _enm):enm{_enm}{
    qDebug() << "derived constructor\n";
}
template <typename T> void myderived<T>::callMethods(){
    switch(enm){
    case myenum::one:
        T::foo();
        break;
    case myenum::two:
        T::boo();
        break;
    }
}

int main(int argc, char *argv[])
{
    QCoreApplication a(argc, argv);

    myderived<mybase_1> tmp1(myenum::one);
    tmp1.callMethods();
    myderived<mybase_2> tmp2(myenum::two);
    tmp2.callMethods();

    return a.exec();
}

Answer 1

1) In this condition, is it possible to use both the function call in the derived class?

In your implementation, only if the derived or the base class implements both functions.

i am getting error and no idea to solve it.

The type is known at compile time, so the simplest solution is to specialize the template:

template<>
void myderived<mybase_1>::callMethods(){
    foo();
}
template<>
void myderived<mybase_2>::callMethods(){
    boo();
}

If you don't need the enum for anything else, then you can throw it away along with the switch statement.

Answer 2

1) Just to rectify your errors— Your switch case is

switch(enm){
case myenum::one:
    T::foo();
    break;
case myenum::two:
    T::boo();
    break;
}

which assumes that T has both foo() and boo(). Define these two functions in both the base classes(it doesn't matter if they are empty.)

What I guess you are trying to do is call a function of base from derived type, which is inheritance has been created I guess.

enum class myenum : int {one, two};

class mybase_1{
protected:
    void foo(){
        qDebug() << "foo in mybase_1 is called \n";
    }
};

class mybase_2{
protected:
    void foo(){
        qDebug() << "foo in mybase_2 is called\n";
    }
};

template <typename T>
class myderived : public T{
public:
    myderived(myenum _enm);
    void callMethods();
private:
    myenum enm;
};

template <typename T> myderived<T>::myderived(myenum _enm):enm{_enm}{
    qDebug() << "derived constructor\n";
}
template <typename T> void myderived<T>::callMethods(){
    foo(); // the foo called depends on the parameter supplied to template, which determines from which class your derived class has been derived.
}

int main(int argc, char *argv[])
{
    QCoreApplication a(argc, argv);

    myderived<mybase_1> tmp1(myenum::one);
    tmp1.callMethods();
    myderived<mybase_2> tmp2(myenum::two);
    tmp2.callMethods();

    return a.exec();
}