Quick int-like type with invariant? C++

Question

Is there a quick way to define a type that has an int member and an invariant in the constructor...

e.g.

class Age {
public:
    Age(int age) {
        Expects(0 <= age && age <= 125);
        age_ = age;
    }
private:
    int age_;
}

...and then have the type as usable as an int?

Age a1 {32};
Age a2 {40};
cout << "Combined ages: " << a1 + a1;
Age a3 = a1 * 2;
// ...etc. etc.

(Or is the only way to override all operators?)

Thanks, John.

Answer 1

You can add operator int() returning the value and it will be useable as an integer.