C++, copy constructor

Posted on 2013-01-17
Last Modified: 2013-01-18
Hello experts, please have a look at the following code:

#include <iostream>

using namespace std;

class Line
      int getLength( void );
      Line( int len );             // simple constructor, headers only
      Line( const Line &obj);      // copy constructor, headers only
      ~Line();                      // destructor

      int *ptr;

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

    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);//****** why does it call copy constructor here? ******

   return 0;

I do not understand why calling 'display' with 'line' as argument makes a call to constructor here...

Any help?

Thank you

Question by:panJames
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LVL 22

Accepted Solution

ambience earned 500 total points
ID: 38788153

void display(Line obj)

does not take Line by reference. It takes it by value so a copy happens
LVL 22

Expert Comment

ID: 38788160
void display(Line& obj)

should not invoke the constructor.

Looking at the purpose of display its better to have it like

void display(const Line& obj)

but that wont compile unless you also change

int getLength( void ) const;

int Line::getLength( void ) const
    return *ptr;
LVL 22

Expert Comment

ID: 38788176
As a general rule, always consider passing objects as const references unless there is a compelling reason not to do so. Passing it by reference ensures there is no overhead of copying and const ensure there are no side-affects because you can only access const methods and data on the object.

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