Question

我正在尝试在Rust中编写二进制搜索树，但我不明白发生了什么：

enum BST<'a, T: Ord> {
    Leaf,
    BinTree { value: T, left: &'a mut BST<'a, T>, right: &'a mut BST<'a, T> }
}

impl<'a, T: Ord> BST<'a, T> {
    fn new() -> BST<'a, T> {
        BST::Leaf
    }

    fn add(self, val: T) {
        match self {
            BST::Leaf => self = BST::BinTree {
                value: val,
                left: &mut BST::<'a, T>::new(),
                right: &mut BST::<'a, T>::new()
            },
            BST::BinTree{value: v, left: l, right: r} => if val < v {
                l.add(val);
            } else {
                r.add(val);
            }
        }
    }
}

fn main() {
}

当我尝试编译时，我收到以下错误：

error[E0309]: the parameter type `T` may not live long enough
 --> heap.rs:3:25
  |
3 |     BinTree { value: T, left: &'a mut BST<'a, T>, right: &'a mut BST<'a, T> }
  |                         ^^^^^^^^^^^^^^^^^^^^^^^^
  |
  = help: consider adding an explicit lifetime bound `T: 'a`...
note: ...so that the reference type `&'a mut BST<'a, T>` does not outlive the data it points at
 --> heap.rs:3:25
  |
3 |     BinTree { value: T, left: &'a mut BST<'a, T>, right: &'a mut BST<'a, T> }
  |                         ^^^^^^^^^^^^^^^^^^^^^^^^

好吧，经过大量的研究并做了编译器建议的事情，我想出了这段代码：

enum BST<'a, T: Ord + 'a> {
    Leaf,
    BinTree { 
        value: T,
        left: &'a mut BST<'a, T>,
        right: &'a mut BST<'a, T>
    }
}

impl<'a, T: Ord + 'a > BST<'a, T> {
    fn new() -> BST<'a, T> {
        BST::Leaf
    }

    fn add(&mut self, val: T) {
        match *self {
            BST::Leaf => *self = BST::BinTree {
                value: val,
                left: &mut BST::<'a, T>::new() as &'a mut BST<'a, T>,
                right: &mut BST::<'a, T>::new() as &'a mut BST<'a, T>
            },
            BST::BinTree{value: ref v, left: ref mut l, right: ref mut r} => if val < *v {
                l.add(val);
            } else {
                r.add(val);
            }
        }
    }
}

fn main() {
}

但我仍然会遇到错误：

error: borrowed value does not live long enough
  --> heap.rs:19:16
   |
19 |                left: &mut BST::<'a, T>::new() as &'a mut BST<'a, T>,
   |                           ^^^^^^^^^^^^^^^^^^^ does not live long enough
20 |                right: &mut BST::<'a, T>::new() as &'a mut BST<'a, T>
21 |            },
   |            - temporary value only lives until here
   |
note: borrowed value must be valid for the lifetime 'a as defined on the body at 15:27...
  --> heap.rs:15:28
   |
15 |    fn add(&mut self, val: T) {
   |  ____________________________^
16 | |      match *self {
17 | |          BST::Leaf => *self = BST::BinTree {
18 | |              value: val,
...  |
27 | |      }
28 | |  }
   | |__^

error: borrowed value does not live long enough
  --> heap.rs:20:17
   |
20 |                right: &mut BST::<'a, T>::new() as &'a mut BST<'a, T>
   |                            ^^^^^^^^^^^^^^^^^^^ does not live long enough
21 |            },
   |            - temporary value only lives until here
   |
note: borrowed value must be valid for the lifetime 'a as defined on the body at 15:27...
  --> heap.rs:15:28
   |
15 |    fn add(&mut self, val: T) {
   |  ____________________________^
16 | |      match *self {
17 | |          BST::Leaf => *self = BST::BinTree {
18 | |              value: val,
...  |
27 | |      }
28 | |  }
   | |__^

error: aborting due to 2 previous errors

我知道这可以通过使用Box而不是引用来修复，但我想让它像练习一样。

Answer 1

如错误消息所示，可以通过添加生命周期绑定T: 'a来修复该特定错误。但是你会得到许多其他错误，因为你要做的事情是不健全的：你试图存储对其他地方没有所有者的对象的引用。

当您执行诸如在节点中存储&mut BST::<'a, T>::new()之类的操作时，BST::<'a, T>::new()会返回一个临时值，该值很快就会被销毁，因此您无法存储对它的引用并期望它继续存在。

您需要您的节点拥有其子节点，而不是引用。您可以通过在创建新子节点时将子类型更改为left: Box<BST<T>>并使用Box::new来执行此操作。完成此操作后，您可以摆脱所有'a无处不在的错误。

另一个问题是，您的add会使用self参数，因此调用者将无法再使用它。您应该改为使用&mut self，以便它可以修改调用者拥有的树。

在编写二叉搜索树时，参数类型“T”可能不够长

1 个答案: