Question

I have an overloaded << operator from my reckful class implemented as follows:

ostream& operator << (ostream& os, const reckful& p)
{
    os << p.PrintStuff();

    return os;
}

PrintStuff() just being a member function of reckful that returns a string.

The way I understand things, if I were to write something like cout << reckobject << endl; in main(), cout << reckobject would take precedence and my overloaded << (using cout as the left operand and reckobject as the right operand) would return the ostream object os, leaving the expression os << endl; to be evaluated which would output the string and then end the line. So, the first << is the one I declared and the second is the standard << right?

However, my main question is… what is the sequence of events, which are the left and right operands, and which << operators are which when I run a statement like this:

cout << "reckful object = " << reckobject << "!" << endl;

Why does this work if there isn’t an ostream object and a reckful object on either side of one << ?

Thanks.

Answer 1

If you notice standard way to implement << it returns the ostream itself. This is the critical piece

So something like

 cout << "reckful object = " << reckobject << "!" << endl;

will be called once for

cout << "reckful object = " 

This function call will return a ostream with which the second call will be made

namely

   cout  << reckobject;

so on an so forth.

You can test is out by implementing your << as

void operator << (ostream& os, const reckful& p)
{  
    os << 1;
}

in which case you can do

std::cout << p;

but not

std::cout << p << std::endl;

The operators make it harder to understand but consider this Point class

struct Point
{
    Point& setX( int x) { X = x; return *this;}
    Point& setY( int y) { Y = y; return *this;}

    int X;
    int Y;
};

The way setX and setY are defined, allows

    Point p;
    p.setX( 2 ).setY( 4 );

This is the same mechanism << is using to chain function calls.