An Peter's answer suggested, there is is no such abstraction in the stdlib. You can define a small enum to handle both cases:
use std::rc::{Rc, Weak};
enum MaybeStrong<T> {
Strong(Rc<T>),
Weak(Weak<T>),
}
impl<T> MaybeStrong<T> {
fn get(&self) -> Option<Rc<T>> {
match self {
MaybeStrong::Strong(t) => Some(Rc::clone(t)),
MaybeStrong::Weak(w) => w.upgrade(),
}
}
}
struct Foo<T> {
bar: MaybeStrong<T>
}
impl<T> Foo<T> {
fn from_weak(inner: Weak<T>) -> Self {
Self { bar: MaybeStrong::Weak(inner) }
}
fn from_strong(inner: Rc<T>) -> Self {
Self { bar: MaybeStrong::Strong(inner) }
}
fn say(&self) where T: std::fmt::Debug {
println!("{:?}", self.bar.get())
}
}
fn main() {
let inner = Rc::new("foo!");
Foo::from_weak(Rc::downgrade(&inner)).say();
Foo::from_strong(inner).say();
}
self.bar() will always return a Some if it was created from a strong pointer and return None in case it's a Weak and it's dangling. Notice that due to the fact that get() needs to create an owned Rc first, the method can't return a &T (including Option<&T>) because that &T could be dangling. This also means that all users of bar() will own one strong count on the inner value while processing, making it safe to use in any case.
Rc<Child>) while the child points to its owner:Weak<Parent>. Being unable to distinguish this fundamental difference can be confusing. - Boiethios