Re: Resetable mixin / idiom
pt wrote:
On 16 Lip, 19:21, Dragan Milenkovic <dra...@plusplus.rs> wrote:
pt wrote:
template<typename T, typename... A>
void reset(T & t, A && ... a) {
t.~T();
new (&t) T(forward<A>(a)...);
}
The very first thing that comes to my mind is exception safety
(unless you require a no-throw constructor). There are a few
solutions to this problem.
Can you please explain the solution to the exception problem?
Apparently, I had solutions to another problem similar to yours
in my mind when I said "a few". However, here is one try:
T x(std::forward<A>(a)...);
t.~T()
new (&t) T(std::move(x));
.... but it would only work for a nothrow-movable T (which IMO
is not a big limitation). Others have already stated that
operator= is better suited for the job.
Another possibility is to wrap T in a class that has your "reset"
function... here is one version...
template <typename T>
class Reset {
public:
template <typename ... A>
void reset(A && ... a) {
std::unique_ptr<T> x(new T(std::forward<A>(a)...));
t.swap(x);
}
T & get() { return *t; }
const T & get() const { return *t; }
private:
std::unique_ptr<T> t;
};
I didn't consider the usefulness of this approach.
--
Dragan
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