mirror of https://github.com/sunface/rust-course
sunface
3 years ago
parent
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# 数据布局
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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。
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栈的实现类似于下图：
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```shell
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input list:
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[Some(ptr)] -> (A, Some(ptr)) -> (B, None)
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stack push X:
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[Some(ptr)] -> (X, Some(ptr)) -> (A, Some(ptr)) -> (B, None)
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stack pop:
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[Some(ptr)] -> (A, Some(ptr)) -> (B, None)
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```
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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:
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```shell
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input list:
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[Some(ptr)] -> (A, Some(ptr)) -> (B, None)
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flipped push X:
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[Some(ptr)] -> (A, Some(ptr)) -> (B, Some(ptr)) -> (X, None)
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```
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而如果将 `pop` 放在末端:
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```shell
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input list:
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[Some(ptr)] -> (A, Some(ptr)) -> (B, Some(ptr)) -> (X, None)
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flipped pop:
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[Some(ptr)] -> (A, Some(ptr)) -> (B, None)
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```
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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！
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其中一个解决办法就是保存一个指针指向末端:
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```rust
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use std::mem;
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pub struct List<T> {
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head: Link<T>,
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tail: Link<T>, // NEW!
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}
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type Link<T> = Option<Box<Node<T>>>;
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struct Node<T> {
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elem: T,
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next: Link<T>,
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}
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impl<T> List<T> {
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pub fn new() -> Self {
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List { head: None, tail: None }
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}
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pub fn push(&mut self, elem: T) {
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let new_tail = Box::new(Node {
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elem: elem,
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// 在尾端推入一个新节点时，新节点的下一个节点永远是 None
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next: None,
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});
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// 让 tail 指向新的节点，并返回之前的 old tail
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let old_tail = mem::replace(&mut self.tail, Some(new_tail));
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match old_tail {
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Some(mut old_tail) => {
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// 若 old tail 存在，则让该节点指向新的节点
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old_tail.next = Some(new_tail);
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}
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None => {
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// 否则，将 head 指向新的节点
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self.head = Some(new_tail);
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}
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}
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}
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}
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```
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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。
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但是如果你担心不再能看到错误，那就纯属多余了:
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```shell
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> cargo build
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error[E0382]: use of moved value: `new_tail`
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--> src/fifth.rs:38:38
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26 | let new_tail = Box::new(Node {
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| -------- move occurs because `new_tail` has type `std::boxed::Box<fifth::Node<T>>`, which does not implement the `Copy` trait
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...
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33 | let old_tail = mem::replace(&mut self.tail, Some(new_tail));
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| -------- value moved here
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...
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38 | old_tail.next = Some(new_tail);
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| ^^^^^^^^ value used here after move
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```
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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:
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```rust
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pub struct List<T> {
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head: Link<T>,
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tail: Option<&mut Node<T>>, // NEW!
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}
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type Link<T> = Option<Box<Node<T>>>;
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struct Node<T> {
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elem: T,
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next: Link<T>,
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}
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impl<T> List<T> {
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pub fn new() -> Self {
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List { head: None, tail: None }
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}
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pub fn push(&mut self, elem: T) {
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let new_tail = Box::new(Node {
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elem: elem,
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next: None,
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});
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let new_tail = match self.tail.take() {
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Some(old_tail) => {
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old_tail.next = Some(new_tail);
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old_tail.next.as_deref_mut()
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}
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None => {
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self.head = Some(new_tail);
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self.head.as_deref_mut()
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}
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};
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self.tail = new_tail;
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}
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}
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```
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```shell
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> cargo build
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error[E0106]: missing lifetime specifier
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--> src/fifth.rs:3:18
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3 | tail: Option<&mut Node<T>>, // NEW!
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| ^ expected lifetime parameter
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```
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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:
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```rust
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pub struct List<'a, T> {
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head: Link<T>,
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tail: Option<&'a mut Node<T>>, // NEW!
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}
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type Link<T> = Option<Box<Node<T>>>;
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struct Node<T> {
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elem: T,
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next: Link<T>,
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}
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impl<'a, T> List<'a, T> {
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pub fn new() -> Self {
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List { head: None, tail: None }
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}
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pub fn push(&mut self, elem: T) {
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let new_tail = Box::new(Node {
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elem: elem,
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next: None,
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});
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let new_tail = match self.tail.take() {
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Some(old_tail) => {
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old_tail.next = Some(new_tail);
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old_tail.next.as_deref_mut()
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}
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None => {
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self.head = Some(new_tail);
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self.head.as_deref_mut()
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}
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};
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self.tail = new_tail;
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}
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}
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```
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```shell
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cargo build
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error[E0495]: cannot infer an appropriate lifetime for autoref due to conflicting requirements
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--> src/fifth.rs:35:27
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35 | self.head.as_deref_mut()
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| ^^^^^^^^^^^^
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note: first, the lifetime cannot outlive the anonymous lifetime #1 defined on the method body at 18:5...
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--> src/fifth.rs:18:5
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18 | / pub fn push(&mut self, elem: T) {
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19 | | let new_tail = Box::new(Node {
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20 | | elem: elem,
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21 | | // When you push onto the tail, your next is always None
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... |
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39 | | self.tail = new_tail;
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40 | | }
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| |_____^
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note: ...so that reference does not outlive borrowed content
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--> src/fifth.rs:35:17
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35 | self.head.as_deref_mut()
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| ^^^^^^^^^
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note: but, the lifetime must be valid for the lifetime 'a as defined on the impl at 13:6...
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--> src/fifth.rs:13:6
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13 | impl<'a, T> List<'a, T> {
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| ^^
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= note: ...so that the expression is assignable:
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expected std::option::Option<&'a mut fifth::Node<T>>
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found std::option::Option<&mut fifth::Node<T>>
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```
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好长... Rust 为啥这么难... 但是，这里有一句重点:
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> the lifetime must be valid for the lifetime 'a as defined on the impl
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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:
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```rust
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pub fn push(&'a mut self, elem: T) {
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```
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当当当当，成功通过编译:
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```shell
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cargo build
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warning: field is never used: `elem`
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--> src/fifth.rs:9:5
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9 | elem: T,
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| ^^^^^^^
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= note: #[warn(dead_code)] on by default
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```
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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`:
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```rust
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pub fn pop(&'a mut self) -> Option<T> {
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self.head.take().map(|head| {
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let head = *head;
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self.head = head.next;
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if self.head.is_none() {
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self.tail = None;
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}
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head.elem
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})
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}
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```
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看起来不错，写几个测试用例溜一溜:
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```rust
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mod test {
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use super::List;
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#[test]
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fn basics() {
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let mut list = List::new();
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// Check empty list behaves right
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assert_eq!(list.pop(), None);
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// Populate list
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list.push(1);
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list.push(2);
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list.push(3);
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// Check normal removal
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assert_eq!(list.pop(), Some(1));
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assert_eq!(list.pop(), Some(2));
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// Push some more just to make sure nothing's corrupted
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list.push(4);
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list.push(5);
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// Check normal removal
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assert_eq!(list.pop(), Some(3));
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assert_eq!(list.pop(), Some(4));
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// Check exhaustion
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assert_eq!(list.pop(), Some(5));
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assert_eq!(list.pop(), None);
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}
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}
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```
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```shell
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cargo test
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error[E0499]: cannot borrow `list` as mutable more than once at a time
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--> src/fifth.rs:68:9
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65 | assert_eq!(list.pop(), None);
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| ---- first mutable borrow occurs here
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...
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68 | list.push(1);
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| ^^^^
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| |
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| second mutable borrow occurs here
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| first borrow later used here
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error[E0499]: cannot borrow `list` as mutable more than once at a time
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--> src/fifth.rs:69:9
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65 | assert_eq!(list.pop(), None);
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| ---- first mutable borrow occurs here
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...
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69 | list.push(2);
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| ^^^^
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| |
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| second mutable borrow occurs here
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| first borrow later used here
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error[E0499]: cannot borrow `list` as mutable more than once at a time
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--> src/fifth.rs:70:9
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65 | assert_eq!(list.pop(), None);
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| ---- first mutable borrow occurs here
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...
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70 | list.push(3);
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| ^^^^
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| |
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| second mutable borrow occurs here
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| first borrow later used here
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....
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** WAY MORE LINES OF ERRORS **
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....
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error: aborting due to 11 previous errors
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```
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🙀🙀🙀，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。
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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。
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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。
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> 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](https://course.rs/advance/circle-self-ref/intro.html)深入阅读。
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```rust
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pub struct List<T> {
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head: Link<T>,
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tail: *mut Node<T>, // DANGER DANGER
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}
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type Link<T> = Option<Box<Node<T>>>;
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struct Node<T> {
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elem: T,
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next: Link<T>,
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}
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```
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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。
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> 注意！当前的实现依然是有严重问题的，在后面我们会修复
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果然，编程的最高境界就是回归本质：使用 C 语言的东东。
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Reference in new issue