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?
I am curious if std::cout has a return value, because when I do this:
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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?
Because the operands of
cout << coutare user-defined types, the expression is effectively a function call. The compiler must find the bestoperator<<that matches the operands, which in this case are both of typestd::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::ostreamhas a conversion operator that allows conversion tovoid*. This is used to enable testing the state of the stream as a boolean condition (i.e., it allowsif (cout)to work).The right-hand operand expression
coutis implicitly converted tovoid const*using this conversion operator, then theoperator<<overload that takes anostream&and avoid const*is called to write this pointer value.Note that the actual value resulting from the
ostreamtovoid*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>>).