是否可以在模式中创建可变引用的可变值? [英] Is it possible to create a mutable value of a mutable reference in a pattern?

问题描述

在进行模式匹配时，您可以使用 ref mut 指定您希望获得对所包含值的可变引用:

When pattern-matching, you can specify that you'd like to get a mutable reference to the contained value by using ref mut:

let mut score = Some(42);
if let Some(ref mut s) = score {
    &mut s;
}

但是，内部值是不可变的:

However, the inner value is not mutable:

error[E0596]: cannot borrow immutable local variable `s` as mutable
 --> src/main.rs:4:14
  |
4 |         &mut s;
  |              ^
  |              |
  |              cannot reborrow mutably
  |              try removing `&mut` here

我尝试添加另一个 mut，但这是无效的:

I tried to add in another mut, but that was not valid:

if let Some(mut ref mut s) = score {
    &mut s;
}

error: the order of `mut` and `ref` is incorrect
 --> src/main.rs:3:17
  |
3 |     if let Some(mut ref mut s) = score {
  |                 ^^^^^^^ help: try switching the order: `ref mut`

error: expected identifier, found keyword `mut`
 --> src/main.rs:3:25
  |
3 |     if let Some(mut ref mut s) = score {
  |                         ^^^ expected identifier, found keyword

error: expected one of `)`, `,`, or `@`, found `s`
 --> src/main.rs:3:29
  |
3 |     if let Some(mut ref mut s) = score {
  |                             ^ expected one of `)`, `,`, or `@` here

推荐答案

不是直接的答案，而是可能的解决方法

Not a direct answer, but possible workarounds

if let Some(ref mut s) = score {
    let mut s = s;
    &mut s;
}

#[derive(Debug)]
struct X;

enum Foo<T> {
    Bar(T),
    _Baz,
}

fn main() {
    let mut score = Foo::Bar(X);

    if let Foo::Bar(ref mut s) = score {
        //let x = s;
        //println!("{:?}", **x); ! not possible
        let x = &mut &mut *s; // &mut &mut X
        println!("{:?}", **x);
    }
}

对于 Option 专门

if let Some(ref mut s) = score.as_mut() {
    s; //:&mut &mut i32
}

if let Some(mut s) = score.as_mut() {
    &mut s;
}

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