Let's consider the perfectly-forwarding version:
When called with an rvalue, it will return final_action<F>(static_cast<F&&>(f))
.
When called with an lvalue, it will return final_action<F&>(f)
.
Let's now consider the const F&
overload:
- When called both an lvalue or rvalue, it will return
final_action<F>(f)
.
As you can see, there is an important difference:
live example on wandbox
I am not sure why it was deemed necessary to have the const F&
overload.
This is the implementation of final_action
:
template <class F>
class final_action
{
public:
explicit final_action(F f) noexcept : f_(std::move(f)), invoke_(true) {}
final_action(final_action&& other) noexcept
: f_(std::move(other.f_)), invoke_(other.invoke_)
{
other.invoke_ = false;
}
final_action(const final_action&) = delete;
final_action& operator=(const final_action&) = delete;
~final_action() noexcept
{
if (invoke_) f_();
}
private:
F f_;
bool invoke_;
};
Unless I am missing something, instanting final_action<F&>
doesn't really make sense, as f_(std::move(f))
will not compile.
live example on wandbox
So I think this should have just been:
template <class F>
inline final_action<F> finally(F&& f) noexcept
{
return final_action<std::decay_t<F>>(std::forward<F>(f));
}
Ultimately, I think that the implementation of finally
in GSL incorrect/unoptimal (i.e. redundant, has code repetition).