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@ -0,0 +1,325 @@
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# 栈上的链表
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在之前的章节中，无一例外，我们创建的都是数据存储在堆上的链表，这种链表最常见也最实用：堆内存在动态分配的场景非常好用。
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但是，既然是高级技巧章节，那栈链表也应该拥有一席之地。但与堆内存的简单分配相比，栈内存就没那么友好了，你们猜大名鼎鼎的 C 语言的 `alloca` 是因为什么而出名的 :)
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限于章节篇幅，这里我们使用一个简单的栈分配方法：调用一个函数，获取一个新的、拥有更多空间的栈帧。说实话，该解决方法要多愚蠢有多愚蠢，但是它确实相当实用，甚至...有用。
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任何时候，当我们在做一些递归的任务时，都可以将当前步骤状态的指针传递给下一个步骤。如果指针本身就是状态的一部分，那恭喜你：你在创建一个栈上分配的链表！
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新的链表类型本身就是一个 Node，并且包含一个引用指向另一个 Node:
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```rust
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pub struct List<'a, T> {
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pub data: T,
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pub prev: Option<&'a List<'a, T>>,
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}
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```
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该链表只有一个操作 `push`，需要注意的是，跟其它链表不同，这里的 `push` 是通过回调的方式来完成新元素推入，并将回调返回的值直接返回给 `push` 的调用者:
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```rust
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impl<'a, T> List<'a, T> {
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pub fn push<U>(
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prev: Option<&'a List<'a, T>>,
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data: T,
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callback: impl FnOnce(&List<'a, T>) -> U,
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) -> U {
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let list = List { data, prev };
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callback(&list)
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}
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}
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```
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搞定，提前问一句：你见过回调地狱吗？
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```rust
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List::push(None, 3, |list| {
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println!("{}", list.data);
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List::push(Some(list), 5, |list| {
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println!("{}", list.data);
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List::push(Some(list), 13, |list| {
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println!("{}", list.data);
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})
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})
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})
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```
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不禁让人感叹，这段回调代码多么的美丽动人😿。
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用户还可以简单地使用 `while-let` 的方式来编译遍历链表，但是为了增加一些趣味，咱们还是继续使用迭代器:
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```rust
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impl<'a, T> List<'a, T> {
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pub fn iter(&'a self) -> Iter<'a, T> {
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Iter { next: Some(self) }
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}
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}
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impl<'a, T> Iterator for Iter<'a, T> {
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type Item = &'a T;
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fn next(&mut self) -> Option<Self::Item> {
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self.next.map(|node| {
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self.next = node.prev;
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&node.data
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})
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}
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}
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```
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测试下：
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```rust
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#[cfg(test)]
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mod test {
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use super::List;
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#[test]
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fn elegance() {
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List::push(None, 3, |list| {
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assert_eq!(list.iter().copied().sum::<i32>(), 3);
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List::push(Some(list), 5, |list| {
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assert_eq!(list.iter().copied().sum::<i32>(), 5 + 3);
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List::push(Some(list), 13, |list| {
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assert_eq!(list.iter().copied().sum::<i32>(), 13 + 5 + 3);
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})
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})
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})
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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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running 18 tests
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test fifth::test::into_iter ... ok
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test fifth::test::iter ... ok
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test fifth::test::iter_mut ... ok
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test fifth::test::basics ... ok
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test fifth::test::miri_food ... ok
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test first::test::basics ... ok
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test second::test::into_iter ... ok
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test fourth::test::peek ... ok
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test fourth::test::into_iter ... ok
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test second::test::iter_mut ... ok
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test fourth::test::basics ... ok
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test second::test::basics ... ok
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test second::test::iter ... ok
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test third::test::basics ... ok
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test silly1::test::walk_aboot ... ok
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test silly2::test::elegance ... ok
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test second::test::peek ... ok
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test third::test::iter ... ok
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test result: ok. 18 passed; 0 failed; 0 ignored; 0 measured; 0 filtered out;
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```
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部分读者此时可能会有一些大胆的想法：咦？我能否修改 Node 中的值？大胆但貌似可行，不妨来试试。
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```rust
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pub struct List<'a, T> {
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pub data: T,
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pub prev: Option<&'a mut List<'a, T>>,
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}
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pub struct Iter<'a, T> {
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next: Option<&'a List<'a, T>>,
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}
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impl<'a, T> List<'a, T> {
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pub fn push<U>(
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prev: Option<&'a mut List<'a, T>>,
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data: T,
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callback: impl FnOnce(&mut List<'a, T>) -> U,
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) -> U {
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let mut list = List { data, prev };
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callback(&mut list)
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}
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pub fn iter(&'a self) -> Iter<'a, T> {
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Iter { next: Some(self) }
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}
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}
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impl<'a, T> Iterator for Iter<'a, T> {
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type Item = &'a T;
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fn next(&mut self) -> Option<Self::Item> {
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self.next.map(|node| {
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self.next = node.prev.as_ref().map(|prev| &**prev);
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&node.data
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})
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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[E0521]: borrowed data escapes outside of closure
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--> src\silly2.rs:47:32
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46 | List::push(Some(list), 13, |list| {
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| ----
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| |
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| `list` declared here, outside of the closure body
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| `list` is a reference that is only valid in the closure body
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47 | assert_eq!(list.iter().copied().sum::<i32>(), 13 + 5 + 3);
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| ^^^^^^^^^^^ `list` escapes the closure body here
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error[E0521]: borrowed data escapes outside of closure
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--> src\silly2.rs:45:28
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44 | List::push(Some(list), 5, |list| {
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| ----
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| |
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| `list` declared here, outside of the closure body
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| `list` is a reference that is only valid in the closure body
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45 | assert_eq!(list.iter().copied().sum::<i32>(), 5 + 3);
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| ^^^^^^^^^^^ `list` escapes the closure body here
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<ad infinitum>
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```
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嗯，没想到是浓眉大眼的迭代器背叛了我们，为了验证到底是哪里出了问题，我们来修改下测试:
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```rust
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#[test]
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fn elegance() {
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List::push(None, 3, |list| {
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assert_eq!(list.data, 3);
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List::push(Some(list), 5, |list| {
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assert_eq!(list.data, 5);
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List::push(Some(list), 13, |list| {
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assert_eq!(list.data, 13);
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})
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})
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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[E0521]: borrowed data escapes outside of closure
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--> src\silly2.rs:46:17
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44 | List::push(Some(list), 5, |list| {
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| ----
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| |
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| `list` declared here, outside of the closure body
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| `list` is a reference that is only valid in the closure body
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45 | assert_eq!(list.data, 5);
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46 | / List::push(Some(list), 13, |list| {
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47 | | assert_eq!(list.data, 13);
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48 | | })
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| |______^ `list` escapes the closure body here
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error[E0521]: borrowed data escapes outside of closure
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--> src\silly2.rs:44:13
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42 | List::push(None, 3, |list| {
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| ----
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| |
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| `list` declared here, outside of the closure body
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| `list` is a reference that is only valid in the closure body
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43 | assert_eq!(list.data, 3);
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44 | / List::push(Some(list), 5, |list| {
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45 | | assert_eq!(list.data, 5);
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46 | | List::push(Some(list), 13, |list| {
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47 | | assert_eq!(list.data, 13);
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48 | | })
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49 | | })
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| |______________^ `list` escapes the closure body here
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```
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原因在于我们的链表不小心依赖了<ruby>型变<rt>variance</rt></ruby>。型变是一个[相当复杂的概念](https://doc.rust-lang.org/nomicon/subtyping.html)，下面来简单了解下。
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每一个节点( Node )都包含一个引用，该引用指向另一个节点， 且这两个节点是同一个类型。如果从最里面的节点角度来看，那所有外部的节点都在使用和它一样的生命周期，但这个显然是不对的：链表中的每一个节点都会比它指向的节点活得更久，因为它们的作用域是嵌套存在的。
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那之前的不可变引用版本为何可以正常工作呢？原因是在大多数时候，编译器都能自己判断：虽然某些东东活得太久了，但是这是安全的。当我们把一个 List 塞入另一个时，编译器会迅速将生命周期进行收缩以满足新的 List 的需求，**这种生命周期收缩就是一种型变**。
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如果大家还是觉得不太理解，我们来考虑下其它拥有继承特性的编程语言。在该语言中，当你将一个 `Cat` 传递给需要 `Animal` 的地方时( `Animal` 是 `Cat` 的父类型)，型变就发生了。从字面来说，将一只猫传给需要动物的地方，也是合适的，毕竟猫确实是动物的一种。
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总之，可以看出无论是从大的生命周期收缩为小的生命周期，还是从 `Cat` 到 `Animal`，型变的典型特征就是：范围在减小，毕竟子类型的功能肯定是比父类型多的。
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既然有型变，为何可变引用的版本会报错呢？其实在于型变不总是安全的，假如之前的代码可以编译，那我们可以写出<ruby>释放后再使用<rt>use-after-free</rt></ruby> 的代码:
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```rust
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List::push(None, 3, |list| {
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List::push(Some(list), 5, |list| {
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List::push(Some(list), 13, |list| {
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// 哈哈，好爽，由于所有的生命周期都是相同的，因此编译器允许我重写父节点，并让它持有一个可变指针指向我自己。
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// 我将创建所有的 use-after-free !
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*list.prev.as_mut().unwrap().prev = Some(list);
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})
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})
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})
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```
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一旦引入可变性，型变就会造成这样的隐患：意外修改了不该被修改的代码，但这些代码的调用者还在期待着和往常一样的结果！例如以下例子：
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|
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```rust
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let mut my_kitty = Cat; // Make a Cat (long lifetime)
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let animal: &mut Animal = &mut my_kitty; // Forget it's a Cat (shorten lifetime)
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*animal = Dog; // Write a Dog (short lifetime)
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my_kitty.meow(); // Meowing Dog! (Use After Free)
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```
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我们将长生命周期的猫转换成短生命周期的动物，可变的！然后通过短生命周期的动物将指针重新指向一只狗。此时我们想去撸软萌猫的时候，就听到：`旺旺...呜嗷嗷嗷`，对，你没听错，不仅没有了猫叫，甚至于狗还没叫完，就可能在某个地方又被修改成狼了。
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因此，**虽然你可以修改可变引用的生命周期，但是一旦开始嵌套，它们就将失去型变，变成`不变( invariant )`**。此时，就再也无法对生命周期进行收缩了。
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具体来说: `&mut &'big mut T` 无法被转换成 `&mut &'small mut T`，这里 `'big` 代表比 `'small` 更大的生命周期。或者用更正式的说法：`&'a mut T` 对于 `'a` 来说是协变( `covariant` )的，但是对于 `T` 是不变的( `invariant` )。
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---
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说了这么多高深的理论，那么该如何改变链表的数据呢？答案就是：使用老本行 - 内部可变性。
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下面让我们回滚到之前的不可变版本，然后使用 `Cell` 来替代 `&mut`。
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```rust
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#[test]
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fn cell() {
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use std::cell::Cell;
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List::push(None, Cell::new(3), |list| {
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List::push(Some(list), Cell::new(5), |list| {
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List::push(Some(list), Cell::new(13), |list| {
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// Multiply every value in the list by 10
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for val in list.iter() {
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val.set(val.get() * 10)
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}
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let mut vals = list.iter();
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assert_eq!(vals.next().unwrap().get(), 130);
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assert_eq!(vals.next().unwrap().get(), 50);
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assert_eq!(vals.next().unwrap().get(), 30);
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assert_eq!(vals.next(), None);
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assert_eq!(vals.next(), None);
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})
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})
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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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running 19 tests
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test fifth::test::into_iter ... ok
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test fifth::test::basics ... ok
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test fifth::test::iter_mut ... ok
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test fifth::test::iter ... ok
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test fourth::test::basics ... ok
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test fourth::test::into_iter ... ok
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test second::test::into_iter ... ok
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test first::test::basics ... ok
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test fourth::test::peek ... ok
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test second::test::basics ... ok
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test fifth::test::miri_food ... ok
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test silly2::test::cell ... ok
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test third::test::iter ... ok
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test second::test::iter_mut ... ok
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test second::test::peek ... ok
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test silly1::test::walk_aboot ... ok
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test silly2::test::elegance ... ok
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test third::test::basics ... ok
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test second::test::iter ... ok
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test result: ok. 19 passed; 0 failed; 0 ignored; 0 measured; 0 filtered out;
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```
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简简单单搞定，虽然之前我们嫌弃内部可变性，但是在这里：真香！
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sunface
3 years ago