c++ - cout a class object using conversion operator -

#include <iostream> #include <cmath>  using namespace std;  class complex { private:     double real;     double imag;  public:     // default constructor     complex(double r = 0.0, double = 0.0) : real(r), imag(i)     {}      // magnitude : usual function style     double mag()     {         return getmag();     }      // magnitude : conversion operator     operator int ()     {         return getmag();     }  private:     // class helper magnitude     double getmag()     {         return sqrt(real * real + imag * imag);     } };  int main() {     // complex object     complex com(3.0, 4.0);      // print magnitude     cout << com.mag() << endl;     // same can done     cout << com << endl; } 

i don't understand how compiler resolving call conversion operator cout << com << endl;.

i can have more 1 conversion operator in same class. how resolution done in case?

you have declared conversion operator int. since operator not explicit, compiler considers when finding best overload of ostream::operator<<. keep in mind c++ compiler attempts automatically convert types find matching call, including conversion constructors, conversion operators , implicit type conversions.

if not want behavior, starting c++11 can mark operator explicit:

explicit operator int() {      return getmag(); } 

regarding second part of question, if there multiple conversions equally good, compile error invoked. if added, say, operator double, there exists ostream::operator(double), call ambiguous , need cast com desired type.