Explaining copy constructor example

Question

A copy constructor is used for many things such as when I need to use pointers or dynamically allocate memory for an object. But looking at this example at tutorialpoint.com:

#include <iostream>

using namespace std;

class Line
{
public:
  int getLength( void );
  Line( int len );             // simple constructor
  Line( const Line &obj);  // copy constructor
  ~Line();                     // destructor

private:
  int *ptr;
};

// Member functions definitions including constructor
Line::Line(int len)
{
cout << "Normal constructor allocating ptr" << endl;
// allocate memory for the pointer;
ptr = new int;
*ptr = len;
}

Line::Line(const Line &obj)
{
cout << "Copy constructor allocating ptr." << endl;
ptr = new int;
*ptr = *obj.ptr; // copy the value
}

Line::~Line(void)
{
cout << "Freeing memory!" << endl;
delete ptr;
}
int Line::getLength( void )
{
return *ptr;
}

void display(Line obj)
{
   cout << "Length of line : " << obj.getLength() <<endl;
}

// Main function for the program
int main( )
{
   Line line(10);

   display(line);

  return 0;
}

the result is :

Normal constructor allocating ptr
Copy constructor allocating ptr.
Length of line : 10
Freeing memory!
Freeing memory!

and when I commented out (the copy constructor) and the code inside destructor I got the same results:

Normal constructor allocating ptr
Length of line : 10

So what is the difference between using the copy constructor here or not? Also why does "Freeing Memory!" occur twice?

Answer 1

Print the address of the memory being freed.

I believe you will find the compiler generated the constructor for you, did a value copy of the contents, including the pointer, and you are double-freeing the pointer and just getting lucky that the runtime isn't complaining about it.

The compiler generated copy constructor is still being called - nothing has changed in that regard, you just aren't printing anything from it since you didn't write it.