Does std::cout have a return value?

Question

I am curious if std::cout has a return value, because when I do this:

cout << cout << "";

some hexa code is printed. What's the meaning of this printed value?

Protectorate

Though the question itself isn't (quite) an exact duplicate, most of the answers to a previous question apply here as well. – Reposit

Answer 1

Because the operands of cout << cout are user-defined types, the expression is effectively a function call. The compiler must find the best operator<< that matches the operands, which in this case are both of type std::ostream.

There are many candidate operator overloads from which to choose, but I'll just describe the one that ends up getting selected, following the usual overload resolution process.

std::ostream has a conversion operator that allows conversion to void*. This is used to enable testing the state of the stream as a boolean condition (i.e., it allows if (cout) to work).

The right-hand operand expression cout is implicitly converted to void const* using this conversion operator, then the operator<< overload that takes an ostream& and a void const* is called to write this pointer value.

Note that the actual value resulting from the ostream to void* conversion is unspecified. The specification only mandates that if the stream is in a bad state, a null pointer is returned, otherwise a non-null pointer is returned.

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The operator<< overloads for stream insertion do have a return value: they return the stream that was provided as an operand. This is what allows chaining of insertion operations (and for input streams, extraction operations using >>).

Mcgregor

Answer 2

cout does not have a return value. cout is an object of type ostream. operator << has a return value, it returns a reference to cout.

See http://www.cplusplus.com/reference/iostream/ostream/operator%3C%3C/ for reference.

The only signature that matches is:

ostream& operator<< (ostream& ( *pf )(ostream&));

so it returns the pointer to the operator<< member.

the one in James' answer. :)

Eby

I can't see any way that this code semantically evaluates to "print the address of an operator<< function." It is far more likely that it is evaluated as (cout << cout) << "" and the inner call is displaying whatever void* that cout is converted to. – Joiner

Answer 3

I believe that would be the address of the ostream object that "" got printed to

Eastereasterday

@Luchian beat me to it with a reference – Eastereasterday