could not able to understand how comma operator works

Question

In this below program I have overloaded commaoperator. But, why comma operator is not taking considering the first element/object.

class Point {
  int x, y;
public:
  Point() {}
  Point(int px, int py) 
  {x = px;y = py;}
  void show() {
    cout << x << " ";
    cout << y << "\n";
  }
  Point operator+(Point op2);
  Point operator,(Point op2);
};

// overload comma for Point
Point Point::operator,(Point op2)
{
  Point temp;
  temp.x = op2.x;
  temp.y = op2.y;
  cout << op2.x << " " << op2.y << "\n";
  return temp;
}

// Overload + for Point
Point Point::operator+(Point op2)
{
  Point temp;
  temp.x = op2.x + x;
  temp.y = op2.y + y;
  return temp;
}

int main()
{
  Point ob1(10, 20), ob2( 5, 30), ob3(1, 1);
  Point ob4;
  ob1 = (ob1, ob2+ob2, ob3);//Why control is not reaching comma operator for ob1?
  ob1 = (ob3, ob2+ob2, ob1);//Why control is not reaching comma operator for ob3?
  ob4 = (ob3+ob2, ob1+ob3);//Why control is not reaching comma operator for ob3+ob2?
  system("pause");
  return 0;
}

I have tried to understand , operator also but could not able to find the solution.

  ob1 = (ob1, ob2+ob2, ob3);//Why control is not reaching comma operator for ob1?
  ob1 = (ob3, ob2+ob2, ob1);//Why control is not reaching comma operator for ob3?
  ob4 = (ob3+ob2, ob1+ob3);//Why control is not reaching comma operator for ob3+ob2?

Any help appreciated.

Answer 1

Why control is not reaching comma operator for ob1?

I guess you're asking, why does this line only output two points: 10 60 for ob2+ob2, and 1 1 for ob3. This is because you only invoke the comma operator twice; each time, it outputs its right-hand argument and ignores its left-hand argument.

The line of code is equivalent to

ob1.operator,(ob2+ob2).operator,(ob3);

making it clear that it's only called twice. ob1 is evaluated, but the operator doesn't do anything with it.

Answer 2

It does reach it. But since you're only printing out the values of the argument of operator, and never of this, you don't get to see it printed out.