Session Aims and Instructions

Multiple inheritance
[Stroustrup 15.2]

• Slides
• Lab exercise

  A simple solution for exercise 1.a is the following:

  #include <iostream>
  #include <cassert>
  
  using namespace std;
  
  class A {
    public:
    int x; /* The state of A. Without members (fields), an object has no state. */
    A( int i = 0 ) : x(i) {
      /* for 1.c - uncomment & add similar constructors to B, C, and D, with different default argument values. */
      /* cout << "A(" << i << ")\n"; */
    }
  };
  class B : public /* virtual */ A { /* uncomment virtual for 1.b */
  };
  class C : /* virtual */ public A { /* uncomment virtual for 1.b */
  };
  class D : public B, public C {
  };
  
  int main() {
    D d;
    /** There are two ways to access the x of a particular object. */
    /** (a) By qualification: */
    d.B::x = 1; /* set the x from the B side of D to 1 */
    d.C::x = 2; /* set the x from the C side of D to 2 */
      cout << "d.B::x and d.C::x are" << (( &(d.B::x) == &(d.C::x) )? "" : " *not*") << " the same!\n";
    /** (b) By restricting the object's interface to a particular base interface: */
    B &b = d;   /* refer to the B part of the D object */
    C &c = d;   /* refer to the C part of the D object */
      cout << "b.x and c.x are" << (( &(b.x) == &(c.x) )? "" : " *not*") << " the same!\n";
  
      assert( &(d.B::x) == &(b.x) );
      assert( &(d.C::x) == &(c.x) );
  
    cout << "The B side x is " << d.B::x << " or " << b.x << endl;
    cout << "The C side x is " << d.C::x << " or " << c.x << endl;
    return 0;
  }
  

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