overloading the assignment operator when the class contains pointer to another object

Question

I have the below code for class A

class A
{

int *ptr;

    public:
       A();
       A(const A &);
       ~A();
       A&  operator = (const A &);
       void display();
};

void A::display()
{

cout<<" ptr ="<<*ptr<<endl;
}

A::A()
{
    cout<<"A's constructor called"<<endl;
    ptr = new int;
    *ptr = 100;
}

A::A(const A &src)
{
    cout<<"copy constructor called"<<endl;
    ptr = new int;
    *ptr = *src.ptr;
}

A::~A()
{
    delete ptr;
    cout<<" destructor called"<<endl;
}

A&  A::operator = (const A &src)
{
    cout<<"A::assignmnet operator called"<<endl;
  if(&src != this)
  {

   delete ptr ;
   ptr = new int;
   *ptr = *src.ptr;

  }

    return *this;

}

Now there is another class B which contains a pointer to class A as a member variable

class B
{
    A *a;
    public:
       B()
       {
           cout<<" B's constructor called"<<endl;
           a = new A();
       }

    B& operator = (const B &);
    ~B()
    {
        cout<<"B:destructor called"<<endl;
        delete a;
    }

    void display()
    {
     cout<<"inside B's display"<<endl;
     a->display();   
    }

};

Now the assignment operator for class B can be written as 1.

B& B::operator=( const B & src)
{

 cout<<"B's assignment operator called"<<endl;
    if(this != &src)
    {
         *a = *src.a;
    }

    return *this;

 }

Or as 2.

 B& B::operator=( const B & src)
{

 cout<<"B's assignment operator called"<<endl;
    if(this != &src)
    {

        delete a;
        a = new A();
        *a = *src.a;
    }

    return *this;

 }

Are these two scenario correct.

Where is the difference between the scenarios?

nvoigt
– nvoigt

2016-11-11 10:04:33 +00:00
Commented Nov 11, 2016 at 10:04 — nvoigt
– nvoigt, Commented Nov 11, 2016 at 10:04
And why is a a pointer when it's owned by the class?

nvoigt
– nvoigt

2016-11-11 10:05:27 +00:00
Commented Nov 11, 2016 at 10:05 — nvoigt
– nvoigt, Commented Nov 11, 2016 at 10:05

Franck · Accepted Answer · 2016-11-11 10:37:45Z

In the code you show, a should not be a pointer but simply declared as A a;.

From your code B has the ownership of A. The reason is twofolds:

you explicitly create A objects in B and not derived types of A
B has the ownership of a: constructors, destructor and assignment.

If you need some kind of polymorphism on A (use derived types of A), you can use smart pointers with std::unique_ptr.

Your first scenario reasons as if the class B contains A a. Your second scenario reasons as if the class B contains std::unique_ptr<A> a. Both scenarios are correct. But I think the first one is more consistent (and more efficient) w.r.t. the current whole code.

If you need polymorphism, you should use the second with std::unique_ptr. Note that in this second scenario, you could replace

a = new A();
*a = *src.a;

by

a = new A(*src.a);

if the copy constructor is consistent w.r.t. the assignment operator.

With A a the code would be

class B
{
    A a;
    public:
       B() {} // calls A()
       B& operator = (const B & src) // or = default;
         { a = src.a; // this != &src is managed inside a
           return *this;
         }
      ~B() {} // no more usefull
      void display()
      {
        cout<<"inside B's display"<<endl;
        a.display();   
      }
};

With unique_ptr the code would be

class B
{
    std::unique_ptr<A> a;
    public:
       B() : a(new A()) {}
       B& operator = (const B & src)
         { if (this != &src)
              a.reset(src.a.get() ? new A(*src.a) : nullptr);
           return *this;
         }
      ~B() {} // no more usefull
      void display()
      {
        cout<<"inside B's display"<<endl;
        a->display();   
      }
};

plus 1, but please state the advantages with regard to the assignment operator (and maybe also a link or example for the assignment with a unique_ptr)

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