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10 KiB
10 KiB
数据布局
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:
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:
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:
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:
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:
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:
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。
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。