Thread: Polymorphism and Composition

Hi all, the code below doesn't do what I would expect and I'm having some problems understanding why. I was expecting to see "Bar method" as output however it gives "Foo method", also if I use a pointer to Foo in Testing class it works as expected.

Code:

#include <iostream>

class Foo
{
  public:
    Foo() {}
    virtual ~Foo() {}

    virtual void method() {std::cout << "Foo method\n";}
};

class Bar: public Foo
{
  public:
    Bar() {}
   ~Bar() {}

    virtual void method() { std::cout << "Bar method\n";}
};

class Testing
{
  public:
    Testing() {}
    ~Testing() {}

    void setFoo(Foo &f) {f_ = f;}

    void start() {f_.method();}

  private:
    Foo f_;
};


int
main()
{
  Foo f0;
  Bar b0;

  Testing t0;

  t0.setFoo(b0);
  t0.start();
  return 0;
}

Thanks!

Originally Posted by sugarfree

Hi all, the code below doesn't do what I would expect and I'm having some problems understanding why. I was expecting to see "Bar method" as output however it gives "Foo method", also if I use a pointer to Foo in Testing class it works as expected.

Code:

#include <iostream>

class Foo
{
  public:
    Foo() {}
    virtual ~Foo() {}

    virtual void method() {std::cout << "Foo method\n";}
};

class Bar: public Foo
{
  public:
    Bar() {}
   ~Bar() {}

    virtual void method() { std::cout << "Bar method\n";}
};

class Testing
{
  public:
    Testing() {}
    ~Testing() {}

    void setFoo(Foo &f) {f_ = f;}

    void start() {f_.method();}

  private:
    Foo f_;
};


int
main()
{
  Foo f0;
  Bar b0;

  Testing t0;

  t0.setFoo(b0);
  t0.start();
  return 0;
}

Thanks!

Because "f_" is a "Foo", not a "Bar". Change that to a "Foo&" and everything works just fine.

Because f_ is a Foo, you are copying the "Foo" subset of Bar (if such an instance is passed) into the f_ member. Hence it is a Foo, and Foo::method is called.
Can be solved with a reference as Sebastiani pointed out.