Question

I was doing this as a personal exercise and wanted to make sure I got this right and understand it correctly. I have a coordinate class with members row and column. I wanted to overload the + and += operator. Here is my code:

Coordinate& Coordinate :: operator+= (const Coordinate& rhs){
    this->m_Row += rhs.m_Row;
    this->m_Column += rhs.m_Column;

    return *this;
}

Coordinate& operator+ (const Coordinate& lhs, const Coordinate& rhs) {
    return Coordinate(lhs) += rhs;
}

where

friend Coordinate& operator + (const Coordinate& lhs, const Coordinate& rhs);

is a friend function defined in Coordinate class.

Are there any pitfalls to this code ?

Here is my understanding of how these work:

operator += 

Add rhs m_Row and m_Column to this members. Return a reference to the object pointed by this pointer and thereby avoid creating another object due to copy constructor.

operator +

Create a local object of lhs (since lhs is a constant and we don’t want to modify its contents) using the copy constructor (lets call it localObj). Invoke the += member operator on localObj which performs the addition. Return a reference to this localObj so that we don’t create another object due to copy constructor, otherwise.

Now, the very last statement concerns me, since I am returning a reference to a local object. As soon as the function (operator +) goes out of scope, localObj will be destroyed and returned reference will point to an object which has been destroyed. Am I correct in understanding this ?

If so, how should I fix it ??

EDIT:
After all the answers and what I learnt: here is what my Coordinate class looks like now: http://rextester.com/MJJI7394

Answer 1

You are right to be worried, you are returning a reference to a temporary here:

Coordinate& operator+ (const Coordinate& lhs, const Coordinate& rhs) {
    return Coordinate(lhs) += rhs;
}

You need to return a Coordinate by value, for example like this:

Coordinate operator+ (Coordinate lhs, const Coordinate& rhs) {
    return lhs += rhs;
}

In the example above, we make a copy of the first parameter instead of taking a reference and then copying in the body of the function. Then we return the result of += on that, by value.

With this setup, there is no need to declare operator+ as a friend.

See this SO link for more information, and thanks to @Blastfurnace for pointing it out.