Question

I thought that in

cout << "Hello world" 

cout object has an operator overloading so we can pass strings into cout objects member function.

But in some example code I saw a class which has an operator overloading defined in it.

class GenericPlayer : public Hand
{
    ..
    friend ostream& operator <<(ostream& os, const GenericPlayer& aGenericPlayer);
    ..
};

...
cout << aGenericPlayer << endl;
...

Even if it is not, what if both cout and aGenericPlayer overload operator<< ?

Answer 1

Even if it is not, what if both cout and aGenericPlayer overload operator<< ?

std::cout is an std::ostream object, so any std::ostream& operator<<(std::ostream, SomeType) will work with std::cout. But the point is that the second parameter of the operator is different, so the overloads are different. The first “string” one is something like

std::ostream& operator<<(std::ostream&, const char*);

and the second

std::ostream& operator <<(std::ostream& os, const GenericPlayer& aGenericPlayer); 

So, they are different operator overloads and there is no ambiguity.