How do I return a reference to something inside a RefCell without breaking encapsulation?

问题

I have a struct that has inner mutability.

use std::cell::RefCell;

struct MutableInterior {
    hide_me: i32,
    vec: Vec<i32>,
}
struct Foo {
    //although not used in this particular snippet,
    //the motivating problem uses interior mutability
    //via RefCell.
    interior: RefCell<MutableInterior>,
}

impl Foo {
    pub fn get_items(&self) -> &Vec<i32> {
        &self.interior.borrow().vec
    }
}

fn main() {
    let f = Foo {
        interior: RefCell::new(MutableInterior {
            vec: Vec::new(),
            hide_me: 2,
        }),
    };
    let borrowed_f = &f;
    let items = borrowed_f.get_items();
}

Produces the error:

error[E0597]: borrowed value does not live long enough
  --> src/main.rs:16:10
   |
16 |         &self.interior.borrow().vec
   |          ^^^^^^^^^^^^^^^^^^^^^^ temporary value does not live long enough
17 |     }
   |     - temporary value only lives until here
   |
note: borrowed value must be valid for the anonymous lifetime #1 defined on the method body at 15:5...
  --> src/main.rs:15:5
   |
15 | /     pub fn get_items(&self) -> &Vec<i32> {
16 | |         &self.interior.borrow().vec
17 | |     }
   | |_____^

The problem is that I can\'t have a function on Foo that returns a borrowed vec, because the borrowed vec is only valid for the lifetime of the Ref, but the Ref goes out of scope immediately.

I think the Ref must stick around because:

RefCell<T> uses Rust\'s lifetimes to implement \'dynamic borrowing\', a process whereby one can claim temporary, exclusive, mutable access to the inner value. Borrows for RefCell<T>s are tracked \'at runtime\', unlike Rust\'s native reference types which are entirely tracked statically, at compile time. Because RefCell<T> borrows are dynamic it is possible to attempt to borrow a value that is already mutably borrowed; when this happens it results in task panic.

Now I could instead write a function like this that returns the entire interior:

pub fn get_mutable_interior(&self) -> std::cell::Ref<MutableInterior>;

However this potentially exposes fields (MutableInterior.hide_me in this example) that are really private implementation details to Foo.

Ideally I just want to expose the vec itself, potentially with a guard to implement the dynamic borrowing behavior. Then callers do not have to find out about hide_me.

回答1:

You can create a new struct similar to the Ref<'a,T> guard returned by RefCell::borrow(), in order to wrap this Ref and avoid having it going out of scope, like this:

use std::cell::Ref;

struct FooGuard<'a> {
    guard: Ref<'a, MutableInterior>,
}

then, you can implement the Deref trait for it, so that it can be used as if it was a &Vec<i32>:

use std::ops::Deref;

impl<'b> Deref for FooGuard<'b> {
    type Target = Vec<i32>;

    fn deref(&self) -> &Vec<i32> {
        &self.guard.vec
    }
}

after that, update your get_items() method to return a FooGuard instance:

impl Foo {
    pub fn get_items(&self) -> FooGuard {
        FooGuard {
            guard: self.interior.borrow(),
        }
    }
}

and Deref does the magic:

fn main() {
    let f = Foo {
        interior: RefCell::new(MutableInterior {
            vec: Vec::new(),
            hide_me: 2,
        }),
    };
    let borrowed_f = &f;
    let items = borrowed_f.get_items();
    let v: &Vec<i32> = &items;
}

回答2:

Instead of creating a brand new type, you can use Ref::map (since Rust 1.8). This has the same result as Levans' existing answer:

use std::cell::Ref;

impl Foo {
    pub fn get_items(&self) -> Ref<'_, Vec<i32>> {
        Ref::map(self.interior.borrow(), |mi| &mi.vec)
    }
}

You can also use new features like impl Trait to hide the Ref from the API:

use std::cell::Ref;
use std::ops::Deref;

impl Foo {
    pub fn get_items(&self) -> impl Deref<Target = Vec<i32>> + '_ {
        Ref::map(self.interior.borrow(), |mi| &mi.vec)
    }
}

回答3:

You can wrap the Vec in an Rc.

use std::cell::RefCell;
use std::rc::Rc;

struct MutableInterior {
    hide_me: i32,
    vec: Rc<Vec<i32>>,
}
struct Foo {
    interior: RefCell<MutableInterior>,
}

impl Foo {
    pub fn get_items(&self) -> Rc<Vec<i32>> {
        self.interior.borrow().vec.clone() // clones the Rc, not the Vec
    }
}

fn main() {
    let f = Foo {
        interior: RefCell::new(MutableInterior {
            vec: Rc::new(Vec::new()),
            hide_me: 2,
        }),
    };
    let borrowed_f = &f;
    let items = borrowed_f.get_items();
}

When you need to mutate the Vec, use Rc::make_mut to obtain a mutable reference to the Vec. If there are still other Rcs referring to the Vec, make_mut will dissociate the Rc from the other Rcs, clone the Vec and update itself to refer to that new Vec, then give you a mutable reference to it. This ensures that the value in the other Rcs doesn't suddenly change (because Rc by itself doesn't provide interior mutability).

来源：https://stackoverflow.com/questions/29401626/how-do-i-return-a-reference-to-something-inside-a-refcell-without-breaking-encap