about the const in an operator overloading definition

Question

For the following definition of

const vector3F operator*(const vector3F &v, float s);

There are two const , what are their respective usages?

Answer 1

The const-reference in the argument means that you don't change v , so you can pass constant vectors (and temporaries!) to the function. That's a Good Thing.

The constant by-value return is sort of a gimmick. It prevents you from writing things like this:

 vector3F v = get_vector();
 vector3F w = v;

 (v * 1.5) = w; // outch! Cannot assign to constant, though, so we're good.

Returning by-value as constant is problematic, though, since it interferes with C++11's rvalue references and move semantics:

 move_me(v * 1.5);  // cannot bind to `vector3F &&` :-(

Because of that, and because an abuse like the one I showed above is fairly unlikely to happen by accident, it's probably best to return by value only as non-constant.

Answer 2

The first const indicates that the return value is constant and can not be altered (which, by the way, is a bad idea for the multiplication operator):

const Vector3F v = myvector*100.0;

v.x = 0; // error: the vector is constant and can not be altered

The second const indicates that the argument "v" is constant:

const vector3F operator*(const vector3F &v, float s)
{
    v.x = 0; // error: "v" is constant
}