best C++ solution? (fwd decl, circular ref)

[D_C]

How can I call a.Print() from a.b.Recurse()? Adding a pointer in b.Recurse(A *ptr) causes undeclared errors, and adding a forward declaration was an invalid type of forward declaration. I copied an example using header files, and the forward declaration problem still existed. I used both Cygwin and Mingw compilers.

Obviously, this is a simplified example. One solution is to nest class B within class A. However, in my actual code, class A has a vector of class B. I just tested it, that would work fine, but is there a better way to solve this?

Example code

#include <iostream.h>
#include <stdlib.h>
 
class B
{
public:
  void Recurse()
  {
    this->Print(); // want a.Print(), not b.Print()
  }
 
  void Print()
  {
    cout << "B" << endl;
  }
};
 
class A
{
  public:
  B b;
 
  void Recurse()
  {
    b.Recurse();
  }
 
  void Print()
  {
    cout << "A" << endl;
  }
};
 
int main()
{
    A a = A();
    a.Recurse();
 
    system("PAUSE");
    return 0;
}

[Banfa]

The mistake you are making is putting your class code into the class definition. If you define the class separately to where you put the class code then you will be able to successfully make a forward reference to class A without causing errors

#include <iostream.h>
#include <stdlib.h>
 
class A;
 
class B
{
public:
  void Recurse(A &a);
  void Print();
};
 
class A
{
  public:
  B b;
 
  void Recurse();
  void Print();
};
 
void B::Recurse(A &a);
{
  a.Print(); // want a.Print(), not b.Print()
}
 
void B::Print()
{
  cout << "B" << endl;
}
 
void A::Recurse()
{
  b.Recurse(*this);
}
 
void A::Print()
{
  cout << "A" << endl;
}
 
int main()
{
    A a = A();
    a.Recurse();
 
    system("PAUSE");
    return 0;
}
 

I haven't tested/compiled this code, but the principle is sound. This works because during the class definition all B needs to know is that class exists, it does not need to know what members it has. It only needs to know what members it has during the implementation of it's methods which it does once the code is re-written like this.

[D_C]

Thank you.