Returning a Reference to a const Object

The usual reason for using a const reference is efficiency, but there are restrictions on when this choice can be used. If a function returns an object that is passed to it, either by object invocation or as a method argument, you can increase the efficiency of the method by having it return a reference. For example, suppose you wanted to write a function Max() that returned the larger of two Vector objects, where Vector is the class developed in Chapter 11. The function would be used in this manner:

Vector force1(50,60);
Vector force2(10,70);
Vector max;
max = Max(force1, force2);

Either of the following two implementations would work:

// version 1
Vector Max(const Vector & v1, const Vector & v2)
{
    if (v1.magval() > v2.magval())
        return v1;
    else
        return v2;
}

// version 2
const Vector & Max(const Vector & v1, const Vector & v2)
{
    if (v1.magval() > v2.magval())
        return v1;
    else
        return v2;
}

There are three important points here. First, recall that returning an object invokes the copy constructor, whereas returning a reference doesn’t. Thus Version 2 does less work and is more efficient. Second, the reference should be to an object that exists when the calling function is executing. In this example, the reference is to either force1 or force2, and both are objects defined in the calling function, so this requirement is met. Third, both v1 and v2 are declared as being const references, so the return type has to be const to match.