Q:

Accessing Protected Attribute of Template Class

The following code doesn't work because the t member function can't access the attribute of its argument object.
How to declare template method t of template class A as a friend function of A?
For the code without template, there is no need to declare friend.
Code:
template <typename T>
class A{
    protected:
        T a;
    public:
        A(int i){
            a = i;
        }
        template <typename T1>
        void t(const A<T1> & Bb){
            a = Bb.a;
        }
};
int main(void){
    A<int> Aa(5);
    A<float> Bb(0);
    Aa.t(Bb);
}

Compiler Error (icc test.cpp):
test.cpp(11): error #308: member "A<T>::a [with T=float]" (declared at line 4) is inaccessible
              a = Bb.a;
                     ^
          detected during instantiation of "void A<T>::t(const A<T1> &) [with T=int, T1=float]" at line 17

Code without template:
class A{
    protected:
        int a;
    public:
        A(int i){
            a = i;
        }
        void t(const A & Bb){
            a = Bb.a;
        }
};
int main(void){
    A Aa(5);
    A Bb(0);
    Aa.t(Bb);
}

A:

You can make all template instantiations friends of one another.
template <typename T>
class A {
   protected:

      // This makes A<int> friend of A<float> and A<float> friend of
      // A<int>
      template <typename T1> friend class A;

      T a;
   public:
      A(int i){
         a = i;
      }
      template <typename T1>
         void t(const A<T1> & Bb){
            a = Bb.a;
         }
};
int main(void){
   A<int> Aa(5);
   A<float> Bb(0);
   Aa.t(Bb);
}

