In order to use cout as such : std::cout << myObject, why do I have to pass an ostream object? I thought that was an implicit parameter.
ostream &operator<<(ostream &out, const myClass &o) {
out << o.fname << " " << o.lname;
return out;
}
Thanks
You aren't adding another member function to ostream
, since that would require redefining the class. You can't add it to myClass
, since the ostream
goes first. The only thing you can do is add an overload to an independent function, which is what you're doing in the example.