Question

Is it possible to create a type (let’s say degrees) and define specific operators for it? Such as: =, +, *, -, /, +=, *=, -=, /=.

I’m wondering this because I need to use degrees for one of my programs and I don’t want to use a float object because using degrees a; a.value = 120; a.value = b.value + a.value; is redundant over a simple degrees a = 120; a = b+a;.

Now why don’t I just use:

typedef float degrees;

? Well, because I need one more thing. When I write

degrees a;
a = 120;
a += 300;

a should be equal to 60 (420-360) because I don’t really need a = 6150 when I can have a = 30 with the same effect. So I’d overload those operators to keep the value between 0 and 360.

Is it possible? And, if so, how?

Answer 1

The solution to your problem doesn’t need Boost or any other libraries. You can achieve what you want by using C++ classes, and overloading both the mathematical operators you want (+, -, *, /, etc) and the assignment operators you want (=, +=, -=, etc) and the comparison operators you want (<, >, <=, >=, etc)… or really any operators you want!

For example:

#include <cmath>

class Degrees {
public:
    // *** constructor/other methods here ***
    Degrees& operator=(float rhs) {
        value = rhs;
        normalize();
        return *this;
    }
    Degrees& operator+=(const Degrees &rhs) {
        value += rhs.value;
        normalize();
        return *this;
    }
    Degrees operator+(const Degrees &rhs) {
        return Degrees(value + rhs.value);
    }

private:
    float value;
    void normalize() {
        value = std::fmod(value, 360);
    }
};

Then you can do things like this:

Degrees a, b; // suppose constructor initializes value = 0 in all of them
a = 10;
b = 20;
a += b; // now, a = 30.
Degrees c = a + b; // now, c = 50.

I’ve given you an example for overloading assignment and plus operators, but you can try this same thing with any other kind, and it should work.