const Point& and const Point are same

if the Prototype is

Code:

---

Point operator-=(const Point);

---

andFunction declearation is

Code:

---

Point Point::operator-=(const Point pt){
        this->x -= pt.x;
        this->y -= pt.y;
        return *this;
}

---

then I can Use it in this way

Code:

---

*pt2 -= *pt1;

---

-----------------
But when

Code:

---

Point operator-=(const Point&);

---

andFunction declearation is

Code:

---

Point Point::operator-=(const Point& pt){
        this->x -= pt.x;
        this->y -= pt.y;
        return *this;
}

---

then ALSO I can use it in this way

Code:

---

*pt2 -= *pt1;

---

WHY ??
Isnt there any difference between that Point& ??

The difference is that in the former there is an additional copy constructor call.

Quote:

---

Originally Posted by laserlight

The difference is that in the former there is an additional copy constructor call.

---

Is there anything bad on using with & or using with & ??
Is there anything good while Calling copy constructor.

Quote:

---

Is there anything bad on using with & or using with & ??

---

In this case, no. You should do it, in fact. I would return a reference too:

Code:

---

Point& Point::operator-=(const Point& pt){
    x -= pt.x;
    y -= pt.y;
    return *this;
}

---

Quote:

---

Is there anything good while Calling copy constructor.

---

Yes, for initialising an object to be a copy of an existing object. In this case it is unneeded, so having such a call is bad.

K thanks

Quote:

---

Originally Posted by laserlight

The difference is that in the former there is an additional copy constructor call.

---

In principle, yes, as a temporary would be created to be passed to the function (and subsequently destroyed). However, the compiler is allowed to eliminate temporaries if the only means of detecting their existance is by tracking constructor and destructor calls.

The statement above is more accurately written as "The difference is that in the former an additional temporary object is potentially introduced, which would yield an additional call of the copy constructor and destructor."

Sorry; I'm in the mood for nitpicking.