Uniqueness of array of references

[John_Nagle]

Rust Playground

Can't have a HashSet of Rc<RefCell>. I understand why. But I need to at least be able to eliminate duplicates from an array of such things. Any good ideas? Safe Rust only, please.

I know that you can't compare references easily. Why is that? Religious issue from the functional crowd?

[steffahn]

You can put it in a wrapper struct and implement Hash and Eq in an appropriate manner.

Some relevant answers from a previous thread:

HashMap keyed by Rc<RefCell<..>>

Note that if you want to force this to work, you could always make a wrapper type and implement Hash for it manually: use std::{ cell::RefCell, hash::{Hash, Hasher}, ops::Deref, rc::Rc, }; #[derive(Debug, PartialEq, Eq)] pub struct HashingRcRefCell<T>(pub Rc<RefCell<T>>); impl<T> From<Rc<RefCell<T>>> for HashingRcRefCell<T> { fn from(val: Rc<RefCell<T>>) -> Self { HashingRcRefCell(val) } } impl<T> Deref for HashingRcRefCell<T> { type Target = RefCell<T>; …

HashMap keyed by Rc<RefCell<..>>

If you are willing to hash by ADDRESS of Rc, rather than the value inside the RefCell, you can use the same technique as: Instead of using Rc<RefCell> do pub struct MyWrapper(Rc<RefCell<T>>); then, using Hash based on address (not value) of Rc you can have "hashing on MyWrapper == hashing on address the Rc points to"

[quinedot]

If you just need a one-time elimination:

    let mut set = vec![one, two, three];
    let mut seen = HashSet::new();
    set.retain(|item| seen.insert(item.as_ptr()));

[Michael-F-Bryan]

@John_Nagle

I know that you can't compare references easily. Why is that? Religious issue from the functional crowd?

Probably because this impl means trying to compare two &T's will result in dereferencing and comparing the underlying T's. So comparing references with == means you'll always get value equality and not reference equality.

I think that has a big flow-on effect, and contributes to why Rust cares so much more about whether two things are equivalent (value equality) rather than being the same object (reference equality).