Thread: Variable Reference Question

Ok, I have this operator overloaded like this:

Code:

Vector3 operator + (Vector3& add);

The code works fine but my question is, if I'm returning a non reference the overloaded operator will create a new Vector3 class and than return it. Now when it returns it, does it create a replica of the temp Vector3 class and than returns that? Here is the code for the overloaded operator:

Code:

Vector3 Vector3::operator + (Vector3 &add)
{
	return Vector3(x + add.x, y + add.y, z + add.z);
}

Also if I derefernece something like this: *this will it create a whole new variable of this? So if I have an operator that returns Vector& and I return the dereferenced of this, will it create a whole new this and than return a reference to it?

if I'm returning a non reference the overloaded operator will create a new Vector3 class and than return it. Now when it returns it, does it create a replica of the temp Vector3 class and than returns that?

Your code creates a temporary vector3 object, and then returns a copy of it. However, if I remember correctly this extra copy will probably be optimised away.

Also if I derefernece something like this: *this will it create a whole new variable of this?

No, though if you "return *this;" in a function that returns an object rather than a reference to an object, it will make a copy of the current object and return that.

So if I have an operator that returns Vector& and I return the dereferenced of this, will it create a whole new this and than return a reference to it?

No, it will return a reference to the current object.

Incidentally, you might want to be more const-correct by passing a const reference instead, and make operator+ a const member function. Another way to do it would be to have an operator+= member function, then implement operator+ as a free function in terms of operator+=.