返回RefCell中Vec的迭代器 [英] Returning iterator of a Vec in a RefCell

问题描述

鉴于以下 struct 和 impl ：

use std::slice::Iter;
use std::cell::RefCell;

struct Foo {
    bar: RefCell<Vec<u32>>,
}

impl Foo {
    pub fn iter(&self) -> Iter<u32> {
        self.bar.borrow().iter()
    }
}

fn main() {}

我收到有关终身问题的错误消息：

I get an error message about a lifetime issue:

error: borrowed value does not live long enough
  --> src/main.rs:9:9
   |
9  |         self.bar.borrow().iter()
   |         ^^^^^^^^^^^^^^^^^ does not live long enough
10 |     }
   |     - temporary value only lives until here
   |
note: borrowed value must be valid for the anonymous lifetime #1 defined on the body at 8:36...
  --> src/main.rs:8:37
   |
8  |       pub fn iter(&self) -> Iter<u32> {
   |  _____________________________________^ starting here...
9  | |         self.bar.borrow().iter()
10 | |     }
   | |_____^ ...ending here

我如何返回并使用 bar s iterator？

How am I able to return and use bars iterator?

推荐答案

你不能这样做，因为它可以让你绕过运行时检查对于唯一性违规行为。

You cannot do this because it would allow you to circumvent runtime checks for uniqueness violations.

RefCell 为您提供了一种将可变性排他性检查推迟到运行时的方法，以换取允许它通过共享引用保存在内部的数据的变异。这是使用RAII警卫完成的：您可以使用对 RefCell 的共享引用获取一个保护对象，然后访问 RefCell 使用这个保护对象：

RefCell provides you a way to "defer" mutability exclusiveness checks to runtime, in exchange allowing mutation of the data it holds inside through shared references. This is done using RAII guards: you can obtain a guard object using a shared reference to RefCell, and then access the data inside RefCell using this guard object:

&'a RefCell<T>        -> Ref<'a, T> (with borrow) or RefMut<'a, T> (with borrow_mut)
&'b Ref<'a, T>        -> &'b T
&'b mut RefMut<'a, T> -> &'b mut T

这里的关键点是'b 与'a 不同，它允许人们获得& mut T 引用，而无需& mut 对 RefCell 的引用。但是，这些参考文献将与警卫相关联，并且不能比守卫更长寿。这是故意完成的： Ref 和 RefMut 析构函数切换其 RefCell 强制进行可变性检查并强制 borrow（）和 borrow_mut（）如果这些检查失败则会发生混乱。

The key point here is that 'b is different from 'a, which allows one to obtain &mut T references without having a &mut reference to the RefCell. However, these references will be linked to the guard instead and can't live longer than the guard. This is done intentionally: Ref and RefMut destructors toggle various flags inside their RefCell to force mutability checks and to force borrow() and borrow_mut() panic if these checks fail.

你能做的最简单的事情就是在 Ref 周围返回一个包装器，这个引用将实现 IntoIterator ：

The simplest thing you can do is to return a wrapper around Ref, a reference to which would implement IntoIterator:

use std::cell::Ref;

struct VecRefWrapper<'a, T: 'a> {
    r: Ref<'a, Vec<T>>
}

impl<'a, 'b: 'a, T: 'a> IntoIterator for &'b VecRefWrapper<'a, T> {
    type IntoIter = Iter<'a, T>;
    type Item = &'a T;

    fn into_iter(self) -> Iter<'a, T> {
        self.r.iter()
    }
}

（试试在操场上）

您无法直接为 VecRefWrapper 实施 IntoIterator ，因为内部 Ref 将由 into_iter（）消耗，给你基本上与你现在相同的情况。

You can't implement IntoIterator for VecRefWrapper directly because then the internal Ref will be consumed by into_iter(), giving you essentially the same situation you're in now.

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