使用捕获模式匹配改变 Vec 元素

Question

我可以更改 vec 的最后一个元素：

#[derive(Debug)]
enum Type {
    A,
    B,
    C,
}

fn main() {
    let mut v = vec![Type::A, Type::B, Type::B];
    
    match v.last_mut(){
        None => v.push(Type::A),
        Some(last) => *last = Type::C,
    }
    
    println!("{:?}", v)
}

==> 打印 [A, B, C].

但是如果我的枚举常量有数据，我似乎无法捕获它们... 例如：

#[derive(Debug)]
enum Type {
    A(i32),
    B(i32),
    C(i32),
}

fn main() {
    let mut v = vec![Type::A(0), Type::B(1), Type::B(2)];
    
    match v.last_mut(){
        None => v.push(Type::A(0)),
        Some(last @ Type::B(_)) => *last = Type::C(42),
        Some(Type::A(_)) | Some(Type::C(_)) => {}
    }
    
    println!("{:?}", v);
}

==> 打印 [A(0), B(1), C(42)]

上面的工作只是因为我在 Type::B(_) 里面放了一个 _。如果我尝试使用此行捕获它以在 Type::C() 中使用：

        Some(last @ Type::B(p)) => *last = Type::C(*p),

我收到三个奇怪的错误：

error: borrow of moved value
  --> src/main.rs:13:14
   |
13 |         Some(last @ Type::B(p)) => *last = Type::C(*p),
   |              ----^^^^^^^^^^^-^
   |              |              |
   |              |              value borrowed here after move
   |              value moved into `last` here
   |              move occurs because `last` has type `&mut Type` which does not implement the `Copy` trait

error[E0658]: pattern bindings after an `@` are unstable
  --> src/main.rs:13:29
   |
13 |         Some(last @ Type::B(p)) => *last = Type::C(*p),
   |                             ^
   |
   = note: see issue #65490 <https://github.com/rust-lang/rust/issues/65490> for more information

error[E0382]: borrow of moved value
  --> src/main.rs:13:29
   |
13 |         Some(last @ Type::B(p)) => *last = Type::C(*p),
   |              ---------------^-
   |              |              |
   |              |              value borrowed here after move
   |              value moved here
   |
   = note: move occurs because value has type `&mut Type`, which does not implement the `Copy` trait
help: borrow this field in the pattern to avoid moving the value
   |
13 |         Some(ref last @ Type::B(p)) => *last = Type::C(*p),
   |              ^^^

error: aborting due to 3 previous errors

我怎样才能完成这项工作，即捕获 B (a &mut i32) 的当前值，并将其传递给 C？

Answer 1

以可变方式捕获 last 并以不变方式捕获其中的值意味着您同时拥有对同一数据的 mut 引用和共享引用。 Rust 的所有权模型禁止这种别名，因此借用检查器拒绝它。

How could I make this work, i.e. capture the current value of B (a &mut i32), and pass it to C?

您可以捕获最后一个值并在单独的内部匹配中检查它。这避免了对别名引用的初始请求，这对 Rust 来说是一个问题。相反，它从外部 last 引用创建内部 p 引用，只要在内部引用的生命周期内不使用外部引用，就允许这样做。换句话说，这编译：

match v.last_mut() {
    None => v.push(Type::A(0)),
    Some(last) => {
        match last {
            Type::B(p) => *last = Type::C(*p),
            Type::A(_) | Type::C(_) => (),
        }
    }
}

Playground

使用捕获模式匹配改变 Vec 元素

Mutate a Vec element with a capturing pattern match

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