`在构建排序链表时无法移出`&mut`-pointer`的取消引用

Question

所以，我正在学习 Rust 并决定构建一个排序链表。在我到达 add 方法之前一切看起来都不错，这里是代码：

struct NodeItem<'a, V:'a + Ord> {
  value : V,
  next : Box<Option<NodeItem<'a,V>>> // '
}

impl <'a, V:'a + Ord> NodeItem<'a,V> { // '

  fn new(value : V) -> NodeItem<'a,V> { // '
    NodeItem { value : value, next : box None }
  }

  fn add(&mut self, value : V) {

    match self.value.cmp(&value) {
      Less => {
        self.next = box Some(NodeItem {value: self.value, next : self.next });
        self.value = value;
      },
      Equal | Greater => {
        match *self.next {
          Some(ref mut next) => next.add(value),
          None => self.next = box Some(NodeItem::new(value)),
        }
      },
    }

  }

}

编译器抱怨：

/home/mauricio/projects/rust/data_structures/src/lists/mod.rs:16:47: 16:51 error: cannot move out of dereference of `&mut`-pointer
/home/mauricio/projects/rust/data_structures/src/lists/mod.rs:16         self.next = box Some(NodeItem {value: self.value, next : self.next });
                                                                                                               ^~~~
/home/mauricio/projects/rust/data_structures/src/lists/mod.rs:16:66: 16:70 error: cannot move out of dereference of `&mut`-pointer
/home/mauricio/projects/rust/data_structures/src/lists/mod.rs:16         self.next = box Some(NodeItem {value: self.value, next : self.next });

这里到底有什么问题？我知道我正在将引用移动到其他地方，但是生命周期参数不应该显示这些项目有相关的 "life"?

这是从 2014 年 12 月 21 日开始的每晚使用。

Answer 1

这是一个similar example：

enum E { Hello }
struct A(E);

fn main() {
    let mut a = A(E::Hello);
    let b = &mut a;
    let c = b.0;
}

错误：

<anon>:7:13: 7:14 error: cannot move out of dereference of `&mut`-pointer
<anon>:7     let c = b.0;
                     ^
<anon>:7:9: 7:10 note: attempting to move value to here
<anon>:7     let c = b.0;
                 ^
<anon>:7:9: 7:10 help: to prevent the move, use `ref c` or `ref mut c` to capture value by reference
<anon>:7     let c = b.0;
                 ^

请注意，编译器会告诉您如何避免这种情况下的错误。

问题是您的 self.value 不是 Copyable。这意味着当您分配它时，您将 将其移出 NodeItem (self)，从而使其不再完全定义！这将是一件坏事，因此 Rust 会阻止您这样做。

您必须决定解决问题的正确方法是什么。最简单的方法是确保 T 是可复制的（或者 Cloneable, depending on your data). However, you probably don't want to be copying your data all around. I would investigate changing your code to prevent copying the node around and just updating entries. You may need to use something like swap.

我从探索一门新语言中获得所有乐趣

#[derive(Debug)] struct Node<T> { v: T, next: Option<Box<Node<T>>>, } impl<T> Node<T> { fn new(v: T) -> Node<T> { Node { v: v, next: None } } fn push_front(self, head: T) -> Node<T> { Node { v: head, next: Some(Box::new(self)), } } fn push_back(&mut self, tail: T) { match self.next { Some(ref mut next) => next.push_back(tail), None => self.next = Some(Box::new(Node::new(tail))), } } fn push_after(&mut self, v: T) { let old_next = self.next.take(); let new_next = Node { v: v, next: old_next, }; self.next = Some(Box::new(new_next)); } } fn main() { let mut n = Node::new(2u8); n.push_back(3u8); let mut n = n.push_front(0u8); n.push_after(1u8); println!("{:?}", n); }

`在构建排序链表时无法移出`&mut`-pointer`的取消引用

`cannot move out of dereference of `&mut`-pointer` while building a sorted linked list

rust

data-structures

我从探索一门新语言中获得所有乐趣