inheritance

[gizzmo]

Hi all, I would like to do something as follows, though I don't know if it is even possible. Here is a simple code

#include <iostream>

template<typename T>
class A
{
public:
  typedef T atype;
  A() {std::cout << "constructor A\n";}
  ~A() {std::cout << "destructor A\n";}
  T m() const { return 1; }
  void print() const
  {
  std::cout << m() << std::endl;
  }
};

template<typename T>
class B : public A<T>
{
public:
  B(): A<T>::A() {std::cout << "constructor B\n";}
  ~B() {std::cout << "destructor B\n";}
  T m() const { return 2; }
};

int main()
{
A<double> a;
B<double> b;
a.print();
b.print();
return 0;
}

The output that the above peace of code gives is

constructor A
constructor A
constructor B
1
1
destructor B
destructor A
destructor A

but I would like to see

constructor A
constructor A
constructor B
1
2
destructor B
destructor A
destructor A

Basically, I want the print method which is inherited in the B class to evoke m() method which is declared in B class and not the one from the A class. I guess I have to use virtual functions, but I don't know which methods should be made virtual.

[bluecoder]

#include <iostream>

template<typename T>
class A
{
public:
  typedef T atype;
  A() {std::cout << "constructor A\n";}
  ~A() {std::cout << "destructor A\n";}
  T m() const { return 1; }
  void print() const
  {
  std::cout << m() << std::endl;
  }
};

template<typename T>
class B : public A<T>
{
public:
  B(): A<T>::A() {std::cout << "constructor B\n";}
  ~B() {std::cout << "destructor B\n";}
  T m() const { return 2; }
void print() const
  {
  std::cout << m() << std::endl;  //adding print function  in the derived class 
  }
};

int main()
{
A<double> a;
B<double> b;
a.print();
b.print();
return 0;
}

[gizzmo]

That is precisely what I would like to avoid. The code I have posted is just a huge simplification of my real code. The print function is much, much longer in my real code and I have couple of classes that are derived from A. So even though, this is the simplest solution I would like to avoid it, because than I would have to duplicate the print() method in all my derived classes.

[Disch]

make 'm' virtual:

template<typename T>
class A
{
public:
  typedef T atype;
  A() {std::cout << "constructor A\n";}
  ~A() {std::cout << "destructor A\n";}
  virtual T m() const { return 1; }  // <- make it virtual
  void print() const
  {
  std::cout << m() << std::endl;
  }
};