c++functionreturnnoreturn

How can a [[noreturn]] function have a return type?


I encountered this "beautiful" example of "highly readable" and "elegant" code, but I'm having troubles understanding it:

struct S {

    [[noreturn]] virtual inline auto f(const unsigned long int *const)
                         –> void const noexcept;
};

Here's what I understand (please correct me if I'm wrong):

Here are my main concerns:

I couldn't get this code to run on VS2013 Preview, so I guess these features weren't implemented yet.

I'm very curious about this, so I'll appreciate it if someone can explain! Cheers


Solution

  • The [[noreturn]] is an attribute which has whatever semantics it has. It doesn't change, however, how the function is declared: all normal functions in C++ (i.e., all functions except the constructors, destructors, and conversion operators) have a declared return type. Adding any attribute doesn't change this rule.

    The purpose of the [[noreturn]] attribute is probably to indicate that the function never returns in a normal way. Given that the function is also declared to be noexcept it basically means that the corresponding function also can't thrown an exception. One example of a function with similar behavior is exit() which terminates the program. I could imagine that functions implementing some sort of application-loop could also qualify. In any case, the [[noreturn]] tells the system that the corresponding function will never normally return, i.e., falling off the function ("after }") will probably result in undefined behavior.