CODEWITHSUNDEEP
c++pointerscopy-constructor
khan shah
khan shah

Reputation: 1251

Why doing assignment does not make objects to point same location

I am trying to understand the concept of copy constructor. With copy constructor I get desired result. But with out copy constructor I get the same result. The code is given here:

#include<iostream>
using namespace std;

class Point
{
private:
    int x, y;
public:
    Point(int x1, int y1) { x = x1; y = y1; }

    // Copy constructor
    // Point(const Point &p2) {x = p2.x; y = p2.y; }
    void set()
    {
        x=50;
        y = 100;
    }
    int getX()            {  return x; }
    int getY()            {  return y; }
};

int main()
{
    Point p1(10, 15); // Normal constructor is called here
    Point p2 = p1 ;
    // p2 = p1;
    cout << "p1.x = " << p1.getX() << ", p1.y = " << p1.getY();
    cout << "\np2.x = " << p2.getX() << ", p2.y = " << p2.getY();
    p2.set();
    cout << "\np1.x = " << p1.getX() << ", p1.y = " << p1.getY();
    cout << "\np2.x = " << p2.getX() << ", p2.y = " << p2.getY();
     // Copy constructor is called here

    // Let us access values assigned by constructors
    return 0;
}

The out put is : 

p1.x = 10, p1.y = 15
p2.x = 10, p2.y = 15
p1.x = 10, p1.y = 15
p2.x = 50, p2.y = 100

Shouldn't it be :

p1.x = 10, p1.y = 15
p2.x = 10, p2.y = 15
p1.x = 50, p1.y = 100
p2.x = 50, p2.y = 100

Edit 1 : What if I initialize objects like this:

Point p1(10, 15); // Normal constructor is called here
    Point p2(11,10);
    p2 = p1;

Will it call copy constructor? If not, why in this case the result is the same?

Upvotes: 1

Views: 78

Answers (3)

Devang Jayachandran
Devang Jayachandran

Reputation: 101

A copy constructor is called usually in cases of declaration and initialization of object to another object at same time, or while passing to functions. What the first statement means is:

Point p1(10,15);
Point p2(p1);

Is an example of usage of copy constructor while,

Point p1(10,15);
Point p2;
p2=p1;

is just plain assignment. And No. Your output is right in the first case. A copy constructor just copies the values of one object into another, it doesn't make the new one a reference object. Also, the main use of copy constructor is usually in functions. As a footnote, to understand constructors and when they are called, you could use different cout statements in them. Then you'll know when a particular constructor is called. P.S. I'm a beginner, so please tell me if I've got anything wrong.

Upvotes: 0

AndyG
AndyG

Reputation: 41110

C++ makes a distinction between objects and references to objects. If you want

p2 and p1 to point to the same Point, you should do this:

Point& p2 = p1;

Because you never defined a copy constructor for your Point class, the compiler gave you one. It will effectively look like this:

Point(const Point& other) : x(other.x), y(other.y)
{}

And your code, Point p2 = p1 will call this copy constructor.

This causes the integer values to be copied, not point to the same location in memory.

Upvotes: 2

Sergey Kalinichenko
Sergey Kalinichenko

Reputation: 726779

But without copy constructor I get the same result.

This is because you get a default copy constructor and a default assignment operator inserted for you by the compiler. It does precisely the same thing as your hand-written copy constructor does, so there is no need to code one manually.

Shouldn't it [the optput] be ...

No, it should not. In your code p1 and p2 are different unrelated objects, even though the initial state of p2 comes from p1.

You need to use pointers or references to get the behavior that you want:

// Point p2 = p1;
Point& p2(p1);

Now p2 behaves as an "alias" for p1, so there is really one object.

demo

Upvotes: 2

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