How in this copy constructors is called? - c++

Question

I am reading copy constructors for the first time. Here is the link from which I am reading it: http://www.tutorialspoint.com/cplusplus/cpp_copy_constructor.htm
Now, in their first example, in the output, they have written this:
Normal constructor allocating ptr
Copy constructor allocating ptr.
Length of line : 10
Freeing memory!
Freeing memory!
I just want to know, how exactly the copy constructor is getting called here? In the main function, they have written it as:
#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;
}
So, it's just Line line(10) which will call the constructor for setting length. Why exactly will the copy constructor be called? Wont' the constructor be called when we pass it an actual object of class Line?

Answer 1

You are passing line by value to display.
When you pass an object to a function by value, a copy of the object is created on the stack. In your case, the obj parameter of the display function is created calling the copy constructor of the class Line.

Answer 2

When you passed solid object as display(Line) function parameter, copy constructor is called because in C++ arguments are passed by value when you pass solid object. So when you passed Line object into function, copy constructor is called to create object copy from existing one. To avoid copying you need to pass parameters by reference or pointer(smart or normal).

Answer 3

display is defined like:
void display(Line obj)
This is pass by value and not pass by reference. A copy must be made which is why the copy constructor is called.

Answer 4

Your function display is defined like:
void display(Line obj)
This is a call-by-value, and in the case of call-by-valuethe copy of the argument is made, not the actual object is passed. Hence when your function is called copy constructoris invoked.
If you want to avoid it, pass parameter by reference or pointer.
Change the definition of Display from void display(Line obj) to
void display(Line *obj)
and in mainchange the function call from display(line); to display(&line);