Question

For my homework assignment, I have to create a linked list and it must be a template class. The assignment calls for one template class called List and another nested template class called Link.

The methods should be defined outside of the class definition.

I created a simple version of the assignment first and it’s working, but I feel there is an error in my design.

#include <iostream>
using namespace std;

template <class T1>
class A
{
public:
  template <class T2>
  class B;

  T1    _a;
};

template <class T1>
template <class T2>
class A<T1>::B
{
public:
  T2 _b;
};

int main()
{
  A<int> a;
  a._a = 5;

  A<int>::B<int> b;
  b._b = 10;

  cout << a._a << " " << b._b << "\n";


  return 0;
}

My main question is, since the nested class will be of the same type, as you can see from the statement A<int>::B<int> b;, is there a way to have class B automatically be of the same type as A? Does my question even make sense? In other words, can I do A<int>::B b; and b automatically the same type as A? There will never be a situation where I would have A<double>::B<int>; for example.

Answer 1

Yes, just make B not a template. T1 in B‘s definition still refers to the template argument of A.

template <class T1>
class A
{
public:
  class B;

  T1    _a;
}

Now, instead of A<int>::B<int>, it’s simply A<int>::B.

List<int>::Link<double> wouldn’t make sense anyway, so there’s no point in it being a template.