Reputation: 29193
I have this type:
struct immobile {
// other stuff omitted
immobile(immobile&) = delete;
immobile(immobile&&) = delete;
};
immobile mk_immobile();
// e.g. this compiles
// mk_immobile() is a prvalue and i is its result object
immobile i(mk_immobile());
I also have this class template:
template<typename T>
struct container {
std::variant<T, other_stuff> var;
template<typename... Args>
container(Args&&... args)
: var(std::in_place_index<0>, std::forward<Args>(args)...) {}
};
I want to construct a container
around the object produced by mk_immobile()
, with the immobile
object used to initialize one of the variants of var
.
container<immobile> c(mk_immobile());
However, this does not work. For one, std::variant
's constructor wants std::is_constructible_v<immobile, immobile>
, which doesn't hold. Worse, even this simplified version fails:
template<typename T>
struct demonstration {
T t;
template<typename... Args>
demonstration(Args&&... args) : t(std::forward<Args>(args)...) {}
};
demonstration<immobile> d(mk_immobile());
Which seems to imply that std::forward
does not, in fact, perfectly forward—prvalues do not forward as prvalues. (This makes sense to me; I don't think doing that would be possible.) I can make demonstration
work by changing it to this:
template<typename T>
struct demonstration {
T t;
template<typename F>
demonstration(F&& f) : t(std::forward<F>(f)()) {}
};
demonstration<immobile> d([] { return mk_immobile(); });
But I do not see a way to change container
in a similar manner. How do I change container
so that it can construct a std::variant
(or other tagged union) out of a prvalue? I can change container
but cannot change immobile
.
Upvotes: 7
Views: 370
Reputation: 21130
You abuse casts
template<typename F>
struct initializer
{
F f;
template<typename T>
operator T()
{
return f();
}
};
template<typename F>
initializer(F&&) -> initializer<F>;
And use as
container<immobile> c{initializer{[]{
return mk_immobile();
}}};
Upvotes: 6