Why output of second program differs from first program?

Question

When returning a reference to the object on which the function is invoked, the returned reference can be used to chain function calls on a single object.

Here, I am applying the same concept. But I am getting different output if I initialize objects differently.

First example:

#include<iostream>
using namespace std;

class Test
{
private:
  int x;
  int y;
public:
  Test(int x = 0, int y = 0) { this->x = x; this->y = y; }
  Test &setX(int a) { x = a; return *this; }
  Test &setY(int b) { y = b; return *this; }
  void print() { cout << "x = " << x << " y = " << y << endl; }
};

int main()
{
  Test obj1(5, 5);

  // Chained function calls.  All calls modify the same object
  // as the same object is returned by reference
  obj1.setX(10).setY(20);

  obj1.print();
  return 0;
}

Output is 10 and 20, which is correct.

However, output is not correct for the second example:

#include<iostream>
using namespace std;

class Test
{
private:
  int x;
  int y;
public:
  Test (int x = 0, int y = 0) { this->x = x; this->y = y; }
  Test setX(int a) { x = a; return *this; }
  Test setY(int b) { y = b; return *this; }
  void print() { cout << "x = " << x << " y = " << y << endl; }
};

int main()
{
  Test obj1;
  obj1.setX(10).setY(20);
  obj1.print();
  return 0;
}

Output is 10 and 0.

Why? I think both are the same, the output of the second program should be 10 and 20 too. What is the reason it's different?

You tagged this C# but your code appears to be C++, which is it? — Scott Chamberlain
– Scott Chamberlain, Commented Nov 28, 2013 at 20:00

Sergey Kalinichenko · Accepted Answer · 2013-11-28 20:02:52Z

5

The difference is that the second version of your program returns by value. This means that the second call (i.e. setY) is performed on a copy of obj1, not on the obj1 itself. That's why only X ends up being set, but not Y.

In your first program, on the other hand, the setters return their results as a reference. This means that no copy is being made, so setY is called on the same object as setX, not on its copy.

answered Nov 28, 2013 at 20:02

Sergey Kalinichenko

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2 Comments

vivek jain Over a year ago

then Why its setting x only why not y also

Sergey Kalinichenko Over a year ago

@vivekjain Because the initial call (i.e. setX) is done on the object itself, before the setter gets to make a copy (implicitly, through returning by value).

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