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那么单向链表的队列长什么样？对于栈来说，我们向一端推入( push )元素，然后再从同一端弹出( pop )。对于栈和队列而言，唯一的区别在于队列从末端弹出。
栈的实现类似于下图：
input list:
[Some(ptr)] -> (A, Some(ptr)) -> (B, None)
stack push X:
[Some(ptr)] -> (X, Some(ptr)) -> (A, Some(ptr)) -> (B, None)
stack pop:
[Some(ptr)] -> (A, Some(ptr)) -> (B, None)
由于队列是首端进，末端出，因此我们需要决定将 push 和 pop 中的哪个放到末端去操作，如果将 push 放在末端操作:
input list:
[Some(ptr)] -> (A, Some(ptr)) -> (B, None)
flipped push X:
[Some(ptr)] -> (A, Some(ptr)) -> (B, Some(ptr)) -> (X, None)
而如果将 pop 放在末端:
input list:
[Some(ptr)] -> (A, Some(ptr)) -> (B, Some(ptr)) -> (X, None)
flipped pop:
[Some(ptr)] -> (A, Some(ptr)) -> (B, None)
但是这样实现有一个很糟糕的地方：两个操作都需要遍历整个链表后才能完成。队列要求 push 和 pop 操作需要高效，但是遍历整个链表才能完成的操作怎么看都谈不上高效！
其中一个解决办法就是保存一个指针指向末端:
#![allow(unused)] fn main() { use std::mem; pub struct List<T> { head: Link<T>, tail: Link<T>, // NEW! } type Link<T> = Option<Box<Node<T>>>; struct Node<T> { elem: T, next: Link<T>, } impl<T> List<T> { pub fn new() -> Self { List { head: None, tail: None } } pub fn push(&mut self, elem: T) { let new_tail = Box::new(Node { elem: elem, // 在尾端推入一个新节点时，新节点的下一个节点永远是 None next: None, }); // 让 tail 指向新的节点，并返回之前的 old tail let old_tail = mem::replace(&mut self.tail, Some(new_tail)); match old_tail { Some(mut old_tail) => { // 若 old tail 存在，则让该节点指向新的节点 old_tail.next = Some(new_tail); } None => { // 否则，将 head 指向新的节点 self.head = Some(new_tail); } } } } }
在之前的各种链表锤炼下，我们对于相关代码应该相当熟悉了，因此可以适当提提速 - 在写的过程中，事实上我碰到了很多错误，这些错误就不再一一列举。
但是如果你担心不再能看到错误，那就纯属多余了:
$ cargo build
error[E0382]: use of moved value: `new_tail`
--> src/fifth.rs:38:38
|
26 | let new_tail = Box::new(Node {
| -------- move occurs because `new_tail` has type `std::boxed::Box<fifth::Node<T>>`, which does not implement the `Copy` trait
...
33 | let old_tail = mem::replace(&mut self.tail, Some(new_tail));
| -------- value moved here
...
38 | old_tail.next = Some(new_tail);
| ^^^^^^^^ value used here after move
新鲜出炉的错误，接好！Box 并没有实现 Copy 特征，因此我们不能在两个地方进行赋值。好在，可以使用没有所有权的引用类型:
#![allow(unused)] fn main() { pub struct List<T> { head: Link<T>, tail: Option<&mut Node<T>>, // NEW! } type Link<T> = Option<Box<Node<T>>>; struct Node<T> { elem: T, next: Link<T>, } impl<T> List<T> { pub fn new() -> Self { List { head: None, tail: None } } pub fn push(&mut self, elem: T) { let new_tail = Box::new(Node { elem: elem, next: None, }); let new_tail = match self.tail.take() { Some(old_tail) => { old_tail.next = Some(new_tail); old_tail.next.as_deref_mut() } None => { self.head = Some(new_tail); self.head.as_deref_mut() } }; self.tail = new_tail; } } }
$ cargo build
error[E0106]: missing lifetime specifier
--> src/fifth.rs:3:18
|
3 | tail: Option<&mut Node<T>>, // NEW!
| ^ expected lifetime parameter
好吧，结构体中的引用类型需要显式的标注生命周期，先加一个 'a 吧:
#![allow(unused)] fn main() { pub struct List<'a, T> { head: Link<T>, tail: Option<&'a mut Node<T>>, // NEW! } type Link<T> = Option<Box<Node<T>>>; struct Node<T> { elem: T, next: Link<T>, } impl<'a, T> List<'a, T> { pub fn new() -> Self { List { head: None, tail: None } } pub fn push(&mut self, elem: T) { let new_tail = Box::new(Node { elem: elem, next: None, }); let new_tail = match self.tail.take() { Some(old_tail) => { old_tail.next = Some(new_tail); old_tail.next.as_deref_mut() } None => { self.head = Some(new_tail); self.head.as_deref_mut() } }; self.tail = new_tail; } } }
$ cargo build
error[E0495]: cannot infer an appropriate lifetime for autoref due to conflicting requirements
--> src/fifth.rs:35:27
|
35 | self.head.as_deref_mut()
| ^^^^^^^^^^^^
|
note: first, the lifetime cannot outlive the anonymous lifetime #1 defined on the method body at 18:5...
--> src/fifth.rs:18:5
|
18 | / pub fn push(&mut self, elem: T) {
19 | | let new_tail = Box::new(Node {
20 | | elem: elem,
21 | | // When you push onto the tail, your next is always None
... |
39 | | self.tail = new_tail;
40 | | }
| |_____^
note: ...so that reference does not outlive borrowed content
--> src/fifth.rs:35:17
|
35 | self.head.as_deref_mut()
| ^^^^^^^^^
note: but, the lifetime must be valid for the lifetime 'a as defined on the impl at 13:6...
--> src/fifth.rs:13:6
|
13 | impl<'a, T> List<'a, T> {
| ^^
= note: ...so that the expression is assignable:
expected std::option::Option<&'a mut fifth::Node<T>>
found std::option::Option<&mut fifth::Node<T>>
好长... Rust 为啥这么难... 但是，这里有一句重点:
the lifetime must be valid for the lifetime 'a as defined on the impl
意思是说生命周期至少要和 'a 一样长，是不是因为编译器为 self 推导的生命周期不够长呢？我们试着来手动标注下:
#![allow(unused)] fn main() { pub fn push(&'a mut self, elem: T) { }
当当当当，成功通过编译:
$ cargo build
warning: field is never used: `elem`
--> src/fifth.rs:9:5
|
9 | elem: T,
| ^^^^^^^
|
= note: #[warn(dead_code)] on by default
这个错误可以称之为错误之王，但是我们依然成功的解决了它，太棒了！再来实现下 pop:
#![allow(unused)] fn main() { pub fn pop(&'a mut self) -> Option<T> { self.head.take().map(|head| { let head = *head; self.head = head.next; if self.head.is_none() { self.tail = None; } head.elem }) } }
看起来不错，写几个测试用例溜一溜:
#![allow(unused)] fn main() { mod test { use super::List; #[test] fn basics() { let mut list = List::new(); // Check empty list behaves right assert_eq!(list.pop(), None); // Populate list list.push(1); list.push(2); list.push(3); // Check normal removal assert_eq!(list.pop(), Some(1)); assert_eq!(list.pop(), Some(2)); // Push some more just to make sure nothing's corrupted list.push(4); list.push(5); // Check normal removal assert_eq!(list.pop(), Some(3)); assert_eq!(list.pop(), Some(4)); // Check exhaustion assert_eq!(list.pop(), Some(5)); assert_eq!(list.pop(), None); } } }
$ cargo test
error[E0499]: cannot borrow `list` as mutable more than once at a time
--> src/fifth.rs:68:9
|
65 | assert_eq!(list.pop(), None);
| ---- first mutable borrow occurs here
...
68 | list.push(1);
| ^^^^
| |
| second mutable borrow occurs here
| first borrow later used here
error[E0499]: cannot borrow `list` as mutable more than once at a time
--> src/fifth.rs:69:9
|
65 | assert_eq!(list.pop(), None);
| ---- first mutable borrow occurs here
...
69 | list.push(2);
| ^^^^
| |
| second mutable borrow occurs here
| first borrow later used here
error[E0499]: cannot borrow `list` as mutable more than once at a time
--> src/fifth.rs:70:9
|
65 | assert_eq!(list.pop(), None);
| ---- first mutable borrow occurs here
...
70 | list.push(3);
| ^^^^
| |
| second mutable borrow occurs here
| first borrow later used here
....
** WAY MORE LINES OF ERRORS **
....
error: aborting due to 11 previous errors
🙀🙀🙀，震惊！但编译器真的没错，因为都是我们刚才那个标记惹的祸。
我们为 self 标记了 'a，意味着在 'a 结束前，无法再去使用 self，大家可以自己推断下 'a 的生命周期是多长。
那么该怎么办？回到老路 RefCell 上？显然不可能，那只能祭出大杀器：裸指针。
事实上，上文的问题主要是自引用引起的，感兴趣的同学可以查看这里深入阅读。
#![allow(unused)] fn main() { pub struct List<T> { head: Link<T>, tail: *mut Node<T>, // DANGER DANGER } type Link<T> = Option<Box<Node<T>>>; struct Node<T> { elem: T, next: Link<T>, } }
如上所示，当使用裸指针后， head 和 tail 就不会形成自引用的问题，也不再违反 Rust 严苛的借用规则。
注意！当前的实现依然是有严重问题的，在后面我们会修复
果然，编程的最高境界就是回归本质：使用 C 语言的东东。