Why friend overloaded operator is preferred to conversion operator in this case
- by skydoor
Hi I have a code like this, I think both the friend overloaded operator and conversion operator have the similar function. However, why does the friend overloaded operator is called in this case? What's the rules?
Thanks so much!
class A{
double i;
public:
A(int i):i(i) {}
operator double () const { cout<<"conversion…