如何在不破坏封装的情况下返回对RefCell内部内容的引用? [英] How do I return a reference to something inside a RefCell without breaking encapsulation?

问题描述

我有一个具有内部可变性的结构.

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();
}

产生错误:

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 | |     }
   | |_____^

问题是我在Foo上没有返回借用的vec的函数，因为借用的vec仅在Ref的生存期内有效，但是Ref立即超出范围.

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.

我认为Ref必须坚持因为 :

RefCell<T>使用Rust的生存期来实现动态借用"，这一过程使人们可以要求对内部值进行临时，排他，可变的访问. RefCell<T>的借用是在运行时"进行跟踪的，这与Rust的本机引用类型不同，后者在编译时是完全静态跟踪的.因为RefCell<T>借位是动态的，所以可以尝试借入已经可变借值的值；当这种情况发生时，会导致任务出现紧急情况.

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.

现在，我可以改写这样的函数来返回整个内部空间:

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

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

但是，这可能会将真正是私有实现详细信息的字段(在本示例中为MutableInterior.hide_me)暴露给Foo.

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

理想情况下，我只想公开vec本身，并可能带有实现动态借用行为的保护措施.这样，呼叫者就不必了解hide_me.

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.

推荐答案

您可以创建类似于RefCell::borrow()返回的Ref<'a,T>防护的新结构，以包装此Ref并避免其外出范围，例如:

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>,
}

然后，您可以为其实现Deref特征，以便可以像使用&Vec<i32>一样使用它:

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
    }
}

之后，更新您的get_items()方法以返回FooGuard实例:

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(),
        }
    }
}

和Deref发挥了魔力:

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;
}

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