问题
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?
回答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.
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 >>).
回答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. :)
回答3:
I believe that would be the address of the ostream object that "" got printed to
来源：https://stackoverflow.com/questions/10987156/does-stdcout-have-a-return-value