Overloading assignment operator for type deduction

Question

Here's the ideone code: http://ideone.com/Qp8Eqg

My question is, is it possible to force a conversion based on the lvalue alone? For example,

[Seconds] s = 2_h + 60_s;
cout <<s.getSeconds()<<endl;

Obviously, I would have to write something like 2_h.toSeconds(), but that would be too verbose and doesn't achieve the idea.

Answer 1

To allow this (which is more likely your question than what you wrote, correct me if I'm wrong):

Seconds s = 2_h;

the following would work: Add operator Seconds() const to class Hours:

class Hours {
    unsigned long long _hours;
public:
    Hours(unsigned long long hours) : _hours(hours) { }

    operator Seconds() const;

    unsigned long long getHours() const {
        return this->_hours;
    }
};

and define it after class Seconds:

Hours::operator Seconds() const { return this->_hours * 3600; }