Is it possible to cout object in C++ like this obj << cout << endl

Question

We had a class where professor asked us to overload ostream to print object this way (saying we have object t)

cout << t << endl;

Then we were asked to cout the same object this way

t << cout << endl;

How does this work and why?

ostream& operator<<(ostream& o, T& t)
{
   return o << t.member;
}

// This is usual way and "normal" that I know about but won't work on both ways

Expected output is the same, but second way is confusing. Why would anyone want to use it?

It's unclear to me which of these questions you want an answer to. — Drew Dormann, Jun 05 '19 at 18:31
The professor isn't saying, "You should always write `t << cout << endl`", but the professor is instead testing your understanding of operators and operator overloading. — Justin, Jun 05 '19 at 18:31
`ostream& operator<<(ostream& o, T& t)` should be `ostream& operator<<(ostream& o, const T& t)` — πάντα ῥεῖ, Jun 05 '19 at 18:55

score 4 · Answer 1 · answered Jun 05 '19 at 18:31

As any good book or tutorial should tell you, for any operator X the expression a X b will (if a suitable overload is found) be equal to operatorX(a, b).

Or if a (in a X b) have overloaded the operator as a member function, then it's equal to a.operatorX(b).

If we now take cout << t, that will call either operator<<(cout, t) or cout.operator<<(t) depending on the type of t.

As should be easy to guess, reversing the order to t << cout would then be operator<<(t, cout) or t.operator<<(cout).

Is it possible to cout object in C++ like this obj << cout << endl

1 Answers1