Convert user-defined type to basic type

The compiler cannot convert a user-defined type (class object) to a basic type automatically, because it does not know the meaning of the conversion. We tell it how by writing a conversion function (casting operator) as a member of the class.

Syntax:

operator type_name() {
    // code to compute and return a value of type_name
}

Rules:

• It must be a member function of the source class.
• It has no return type written in the declaration (the type name after operator is the return type).
• It takes no arguments.
• It is called automatically when an object is assigned to a variable of that basic type, or explicitly with a cast.

Example - convert a Distance object (feet, inches) to a float giving the total in feet:

#include <iostream>
using namespace std;

class Distance {
    int feet;
    float inch;
public:
    Distance(int f, float i) { feet = f; inch = i; }
    operator float() {                 // Distance -> float
        return feet + inch / 12.0;
    }
};

int main() {
    Distance d(5, 6);
    float ft = d;                      // calls d.operator float()
    cout << "Distance in feet = " << ft << endl;   // 5.5
    cout << (float)d << endl;          // explicit cast, also 5.5
    return 0;
}

Output: Distance in feet = 5.5 then 5.5.