Question

I have a doubt about how C++ casts types when it has to do math.

The code below as it is ( i.e. with only the cast to int without casting to double ) works and builds without problem. If I define ENABLE_DOUBLE_CAST doesn’t build and complains about operator*. Do you know why?

The doubts I have are 2:

1. why without the operator cast to double, it is used the one to int in a multiplication between double. Is it because of the implicit cast?
2. why WITH the cast to double enable ( adding more clarity to the syntax ) that is not taken into consideration?

Thanks

AFG

class CA{
    int _m;
public:
    CA( int a, int b ){
     _m=a*b;
    }
    operator int (){
        std::cout<< "(CA int cast)" ;
        return _m;
    }
#ifdef ENABLE_DOUBLE_CAST
    operator double(){
        std::cout << "(CA double cat)";
        return 2.3 * _m;
    }
#endif
};

int main( int argc, const char** argv){
    CA obj_2( 10,20 );                  

    double total_1 = 100.0 * obj_2;   
    double total_2 = obj_2 * 100.0;
    return 0;
}

Answer 1

why without the operator cast to double, it is used the one to int in a multiplication between double. Is it because of the implicit cast?

Yes that is true.
It uses the operator int () to convert obj_2 to an int and then uses the inbuilt operator*(double, int)

why WITH the cast to double enabled ( adding more clarity to the syntax ) that is not taken into consideration?

When you provide both operator double() and operator int () for your class, it creates an ambiguity, whether to convert obj_2 to an int or double because there are two inbuilt operator overloads which it can use, namely:

operator*(double, int)
operator*(double, double)

Suggestion:
You should overload the operator* for your class if you want to avoid this problem of implicit conversions and ambiguity arising over it.

operator*(double,CA) and
operator*(CA,int)