Class Operator Overloading

Question

This is a very basic operator overload question. Say I had a class like this...

class xy { public: double x, y; XY(double X, double Y) { x = X; y = Y;} XY operator+(const XY & add) const { return XY(this->x + add.x, this->y + add.y); } XY & operator+=(const XY & add) const {?} } }

And I want operator+= do to what its supposed to do (you know, add to the current value of x and y). Wouldn't the code be the same for operator+ and operator +=?

I like this tutorial: cs.caltech.edu/courses/cs11/material/cpp/donnie/cpp-ops.html — leftaroundabout
– leftaroundabout, Commented May 21, 2011 at 19:00
I recommend the boost operators library for this sort of operator overloading. It allows you to define the minimal set of operations (+= in this case), and then automatically fills out the others (like operator +). — Alan Stokes
– Alan Stokes, Commented May 21, 2011 at 19:21

Šimon Tóth · Accepted Answer · 2011-05-21 17:58:08Z

5

How could it be the same? They do different things.

If you don't care about optimizations you can use += in your + operator implementation:

XY operator + (const XY& right) const { XY left(*this); left += right; return left; }

answered May 21, 2011 at 17:58

Šimon Tóth

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3 Comments

leftaroundabout Over a year ago

As long as you declare these as inline functions (like they automatically are when defined inside the class) it should not even slow down the performance.

Šimon Tóth Over a year ago

@leftaroundabout It depends on your copy constructor implementation. It might be faster to call the default constructor and construct the values manually.

leftaroundabout Over a year ago

Right. I was referring to this specific class with just two double variables, where default constructor + manual value assignment should not be faster than copy constructor. But of course this does not hold for all classes.

eduffy · Accepted Answer · 2011-05-21 17:56:45Z

Yep, do the add operation (stick to the += operator), and return a reference to itself. Oh, and this can't be a const method.

XY & operator+=(const XY & add) { this->x += add.x; this->y += add.y; return *this; }

haha that "const" was screwing me up when I tried to change anything pointed to by "this"

NPE · Accepted Answer · 2011-05-21 17:57:29Z

No. Conventionally, operator+ stores the result in a new object and returns it by value, whereas operator+= adds the right-hand side to *this and returns *this by reference.

The two operators are related -- and can often be implemented in terms of one another -- but they have different semantics and therefore can't have identical implementations.

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