[FIXED] how slicing of derived classes occurs?

Issue

I'm having trouble understanding how slicing occurs? for example, in this piece of code:

class A {
public:
  virtual h() {
    cout << "type A" << endl;
  }
};

class B : public A {
public:
  virtual h() override {
    cout << "type B" << endl;
  }
};

void f(A a) {
  a.h();
}

void g(A& a) {
  a.h();
}

int main() {
  A a1 = B(); // a1 is A and doesn't recognize any B properties
  A *a = new B();
  f(*a);
  g(*a);
}

I noticed that:

1. variable a1 doens't know it's a B, but variable a does know. I refer this is happening because in variable a1 the assignment to B is by value, in contrst to variable a, where I create a pointer to B.

2. same thing happens when I pass variable a to different functions - when I pass by value, it thinks it's an A, but when i pass by reference, it thinks it's B.

I would be happy if anyone could give me more extensive and deeper explanation. Thank you in advance!

Solution

In this statement

A a1 = B();

there is used the default copy constructor of the class A because the object a1 has the type A.

This constructor looks like

constexpr A( const A & );

So only subobject of the type A of the object of the class B is copied to the object a1. The table of pointers to virtual functions of the object a1 will contain pointer to its own virtual function.

In this declaration

A *a = new B();

the pointer a points to an object of the type B that contains the table of pointers to virtual functions that in turn contains a pointer to the virtual function of the class B.

So in this call

  g(*a);

the reference to the object of the dynamic type B is passed (the object itself was not changed when it was passed to the function by reference).

So within the function

void g(A& a) {
  a.h();
}.

there will be access to the table of virtual function pointers that contains a pointer to the virtual function of the class B. Here a new object of the class A is not created because the original object is passed by reference.

In this call

  f(*a);

the argument of the function is passed by value. That is there is a copy initialization of the parameter of the function declared like

void f(A a) {
  a.h();
}

So it is in fact the same case as in the declaration

A a1 = B();

considered above.

Answered By - Vlad from Moscow

Answer Checked By - Marilyn (PHPFixing Volunteer)