具有多个收益的RVO运营商

Question

We're working with RVO in class to show how we can reduce the number of temporaries created.

I get the basics of it but I'm having difficulty understanding how to combine multiple values to return into one line.

For single temp optimization I was able to understand it fairly easy

const A operator + ( const A &tmp)
{
    A sum;
    sum = this->x + tmp.x;
    return sum;
}

can be reduced to:

const A operator + ( const A &tmp)
{
    return A(this->x + tmp.x);
}

However I'm uncertain how to apply this to a function with multiple values to return. For example:

Vect2D operator - ( const Vect2D &tmp ) const;

Vect2D Vect2D::operator - ( const Vect2D &tmp ) const
{
    Vect2D sum;

    sum.x = this->x - tmp.x;
    sum.y = this->y - tmp.y;

    return ( sum );
};

My process behind it would be:

Vect2D Vect2D::operator - ( const Vect2D &tmp ) const
{
    return Vect2D((this->x - tmp.x), (this->y - tmp.y));
};

which comes back with an error telling me " no argument takes the value (float, float) ".

would I have to reorganize the initial

Vect2D operator - ( const Vect2D &tmp ) const;

to take two float arguments within? or am I thinking about this in the wrong way?

Thank you,

E: Thanks to Matt and Jerry for affirming what I thought I needed to do with the double float operator.

Answer 1

The solution to my issue was not understanding that operators require the Friend function in order to have two arguments. While not entirely RVO the solution does reduce the number of temps created by having a return with both variables in it.

Thanks again to Matt and Jerry for getting me started on the process.