How does operator<< overloading work?

Question

Given a class:

struct employee {
    string name;
    string ID;
    string phone;
    string department;
};

How does the following function work?

ostream &operator<<(ostream &s, employee &o)
{
 s << o.name << endl;
 s << "Emp#: " << o.ID << endl;
 s << "Dept: " << o.department << endl;
 s << "Phone: " << o.phone << endl;

 return s;
}

cout << e; produces formatted output for a given employee e.

Example output:

Alex Johnson
Emp#: 5719
Dept: Repair
Phone: 555-0174

I can't understand how the ostream function works. How does it get the parameter "ostream &s"?How does it overload the "<<" operator and how does the << operator work? How can it be used to write all of the information about an employee? Can someone please answer these questions in the layman's terms?

Answer 1

This is called overload resolution. You've written cout << *itr. Compiler takes it as operator<<(cout, *itr);, where cout is an instance of ostream and *itr is an instance of employee. You've defined function void operator<<(ostream&, employee&); which match most closely to your call. So the call gets translated with cout for s and *itr for o

Answer 2

Given an employee e;. the following code: cout << e;

will call your overloaded function and pass references to cout and e.

ostream &operator<<(ostream &s, const employee &o)
{
    // print the name of the employee e to cout 
    // (for our example parameters)
    s << o.name << endl; 

    // ...

    // return the stream itself, so multiple << can be chained 
    return s;
}

Sidenote: the reference to the employee should be const, since we do not change it, as pointed out by πάντα ῥεῖ