返回传递给函数的不可变引用后面的可变引用 [英] Returning a mutable reference that is behind an immutable reference, passed to the function

问题描述

如何返回在不可变引用后面的可变引用(作为参数传递给该函数)，得到处理?

How is returning a mutable reference that is behind an immutable reference, passed as an argument to the function, handled?

struct Foo { i: i32 }

struct Bar<'b> {
    f: &'b mut Foo
}

impl<'a: 'b, 'b> Bar<'b> {
    fn func(&'a self) -> &'b mut Foo {
         self.f
    }
}

fn main() {
    let mut foo = Foo { i: 1 };

    let bar = Bar { f: &mut foo };
    bar.func(); 
}

出现以下错误:

error[E0389]: cannot borrow data mutably in a `&` reference
 --> src/main.rs:9:14
  |
8 |         fn func(&'a self) -> &'b mut Foo {
  |                 -------- use `&'a mut self` here to make mutable
9 |              self.f
  |              ^^^^^^ assignment into an immutable reference

我(某种程度上)了解编译器在此处试图阻止的事情.但是我对错误消息assignment into an immutable reference感到困惑.

I (sort of) understand what the compiler is trying to prevent here. But I am confused with the error message assignment into an immutable reference . What exactly is being assigned into self (or inside ofself.f?) ?

我编写了以下代码来模拟此问题，并得到了相同的错误消息，与上面的代码不同，我能够理解.这是代码:

I wrote the following code to simulate this problem and got the same error message, which unlike the above one, I am able to understand. Here's the code:

fn main() {
    let mut foo = Foo { i: 1 };

    let bar = Bar { f: &mut foo };
    let pbar = &bar;

    pbar.f.i = 2; // assignment into an immutable reference
}

在第一个示例中，它是否尝试将可变引用f从self中移出(因为&mut不是Copy类型)，将其视为不可变引用self内部的一种突变，因此错误消息assignment into an immutable reference?

In the first example, is it trying to move the mutable reference f out of self (since &mut is not a Copy type), treating it as a mutation inside the immutable reference self, hence the error message assignment into an immutable reference?

推荐答案

您不能从不可变的引用创建可变引用.这意味着您需要将&self更改为&mut self:

You can't create a mutable reference from an immutable one. This means that you need to change &self into &mut self:

impl<'a: 'b, 'b> Bar<'b> {
    fn func(&'a mut self) -> &'b mut Foo {
         self.f
    }
}

现在您的变量需要可变，以便您可以对该方法进行可变引用:

And now the your variable needs to be mutable so that you can take a mutable reference to it for the method:

let mut bar = Bar { f: &mut foo };
bar.func(); 

到底是什么被分配给self(或在self.x内部?)?

该错误消息可能有点偏离.您的代码中没有分配，但是您返回的是 mutable 引用.可变引用唯一可以让您在这里做的事情就是分配self.f或self.f.i.

The error message might be a little off. There is no assignment in your code, but you are returning a mutable reference. The only extra thing that a mutable reference would let you do here is to assign self.f or self.f.i.

绝对可以改善此错误消息，但确实包含使&'a self可变以解决问题的提示.

Definitely this error message can be improved, but it does include a hint to make the &'a self mutable to fix the problem.

现在，您的原始问题:

如何返回在不可变引用后面的可变引用(作为参数传递给该函数)，得到处理?

How is returning a mutable reference that is behind an immutable reference, passed as an argument to the function, handled?

Rust为内部可变性提供了多种容器类型，例如Cell和RefCell.这些类型负责确保编译器的正确性，并使其成为运行时检查.将RefCell应用于代码的一种方法可能是这样的:

Rust provides a variety of container types for interior mutability, such as Cell and RefCell. These types take the responsibility for ensuring correctness away from the compiler and make it a runtime check. One way of applying a RefCell to your code might be like this:

use std::cell::RefCell;
use std::ops::DerefMut;

struct Foo { i: i32 }

struct Bar<'b> {
    // store the data in a RefCell for interior mutability
    f: &'b RefCell<Foo>
}

impl<'a: 'b, 'b> Bar<'b> {
    // Return a RefMut smart pointer instead of mutable ref, but hide the implementation,
    // just exposing it as something that can be mutably dereferenced as a Foo
    fn func(&'a self) -> impl DerefMut<Target = Foo> + 'b {
         self.f.borrow_mut()
    }
}

fn main() {
    let foo = RefCell::new(Foo { i: 1 });
    let bar = Bar { f: &foo };

    let mut f = bar.func(); 
    f.i = 3;
}

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