Overloading problem Again

This is a discussion on Overloading problem Again within the C++ Programming forums, part of the General Programming Boards category; I am overloading += operator in this way Code: Point Point::operator+=(const Point& pt){ int tmp_x =this->x + pt.x; int tmp_y ...

  1. #1
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    Overloading problem Again

    I am overloading += operator in this way
    Code:
    Point Point::operator+=(const Point& pt){
    	int tmp_x =this->x + pt.x;
    	int tmp_y =this->y + pt.y;
    	return Point(tmp_x, tmp_y);
    }
    and this is my Copy Constructor
    Code:
    Point::Point(const Point& pt){
    	*this = pt;
    }
    Why teh Operator Is not getting Overloaded ??
    I know making x += pt.x or this->x += pr.x Works. I wanna Know Why it doesnt Work ??

  2. #2
    ZuK
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    Think for a moment what operator += does for an int variable.
    it assignes a new value to that variable.
    Does your operator change the object ( this )?
    Kurt

  3. #3
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    Yes
    As if you wanna Do x+=y;
    It will chnage the x just like that the Object needs to be changed here.

  4. #4
    Cat without Hat CornedBee's Avatar
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    Because the semantics are off. The semantics you implemented are those of +, not +=. += modifies the left object.
    This is the canonical form for overloading mathematical operators:
    Code:
    class Point {
      // ...
      Point &operator += (const Point &o);
    };
    
    Point &Point::operator += (const Point &o)
    {
      x += o.x;
      y += o.y;
      return *this;
    }
    
    Point operator +(const Point &lhs, const Point &rhs)
    {
      Point tmp(lhs);
      tmp += rhs;
      return tmp;
    }
    Note that the overload of + is exactly the same for every class that needs it. It's neither a member nor a friend. All it needs is a += in the class.

    As for +=, it returns a Point& for the same reason that = does: chaining.
    All the buzzt!
    CornedBee

    "There is not now, nor has there ever been, nor will there ever be, any programming language in which it is the least bit difficult to write bad code."
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  5. #5
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    Quote Originally Posted by CornedBee
    Code:
    Point &Point::operator += (const Point &o)
    {
      x += o.x;
      y += o.y;
      return *this;
    }
    here you are Using x += o.x;
    So you are updating the x and y DIRECTLY then is there any need of returning that this ??.
    cause The Item Thats on the left side of += Is (this) here and in x += o.x; teh First x is of the Left ones So Is there any need of return as you are Updating it Directly ??
    -----------------------------------
    I didnt Understand
    Because the semantics are off. The semantics you implemented are those of +, not +=. += modifies the left object.
    This is the canonical form for overloading mathematical operators:
    Properly. can you explain more ??
    Last edited by noobcpp; 07-01-2007 at 07:24 AM.

  6. #6
    Cat without Hat CornedBee's Avatar
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    Your original code:
    Code:
    Point Point::operator+=(const Point& pt){
    	int tmp_x =this->x + pt.x;
    	int tmp_y =this->y + pt.y;
    	return Point(tmp_x, tmp_y);
    }
    Doesn't modify this. It just creates a new point whose components are the sum of the components of this and pt. Those are the semantics of +. The semantics of += demand that this is modified. That's what my implementation does.

    The object is then returned for chaining, as I said.
    All the buzzt!
    CornedBee

    "There is not now, nor has there ever been, nor will there ever be, any programming language in which it is the least bit difficult to write bad code."
    - Flon's Law

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    Sorry I am not so good in English.
    What does semantics maen ??

  8. #8
    Cat without Hat CornedBee's Avatar
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    Sorry for that. In linguistic, the semantics of a sentence are its meaning. In programming languages, therefore, the semantics of a code snippet are what it does. (As opposed to the syntax, which just says what it looks like.)

    Syntax and semantics are important concepts when studying programming languages.
    All the buzzt!
    CornedBee

    "There is not now, nor has there ever been, nor will there ever be, any programming language in which it is the least bit difficult to write bad code."
    - Flon's Law

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    So as you are Modifying `this` Directly in your Code. Why do you need to return that ?? Isnt void Ok for that ??

  10. #10
    ZuK
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    It has been mentioned. You want to return a reference to the object to be able to do chaining.
    like this
    Code:
    sometype a;
    sometype b;
    b = a += 2;
    for this to work operator += has to return an object of sometype
    if you want to be able to do
    Code:
    (a+=2) = 3;
    then operator += has to return a reference

    If you don't want anything like this you return void.
    Kurt

  11. #11
    Cat without Hat CornedBee's Avatar
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    But you do want it, because it is consistent with the behaviour of built-in types. Such consistency is important. If overloaded operators are inconsistent with the built-in operators (unless the problem domain is completely different) the chaos ensues.
    All the buzzt!
    CornedBee

    "There is not now, nor has there ever been, nor will there ever be, any programming language in which it is the least bit difficult to write bad code."
    - Flon's Law

  12. #12
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    OK
    I've Understood it Properly
    Thanks
    a OP b OP c
    here Up to b OP C is OK
    When a cant Use the OP with <blank> Void
    Thats Why it needs Pointer or refference or Object.
    Am I right ??
    //I Meant Operator with OP

  13. #13
    Algorithm Dissector iMalc's Avatar
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    Quote Originally Posted by ZuK View Post
    It has been mentioned. You want to return a reference to the object to be able to do chaining.
    like this
    Code:
    sometype a;
    sometype b;
    b = a += 2;
    for this to work operator += has to return an object of sometype
    if you want to be able to do
    Code:
    (a+=2) = 3;
    then operator += has to return a reference

    If you don't want anything like this you return void.
    Kurt
    Code like that last example is why they should have made it standard to return a const reference.
    My homepage
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  14. #14
    ZuK
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    Quote Originally Posted by iMalc View Post
    Code like that last example is why they should have made it standard to return a const reference.
    I just said that you can do that.
    Kurt

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