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< h1 id = "迭代器" > < a class = "header" href = "#迭代器" > 迭代器< / a > < / h1 >
< p > 坏男孩最令人头疼，而链表实现中，迭代器就是这样的坏男孩，所以我们放在最后来处理。< / p >
< h2 id = "intoiter" > < a class = "header" href = "#intoiter" > IntoIter< / a > < / h2 >
< p > 由于是转移所有权，因此 < code > IntoIter< / code > 一直都是最好实现的:< / p >
< pre class = "playground" > < code class = "language-rust edition2021" > < span class = "boring" > #![allow(unused)]
< / span > < span class = "boring" > fn main() {
< / span > pub struct IntoIter< T> (List< T> );
impl< T> List< T> {
pub fn into_iter(self) -> IntoIter< T> {
IntoIter(self)
}
}
impl< T> Iterator for IntoIter< T> {
type Item = T;
fn next(& mut self) -> Option< Self::Item> {
self.0.pop_front()
}
}
< span class = "boring" > }< / span > < / code > < / pre >
< p > 但是关于双向链表，有一个有趣的事实，它不仅可以从前向后迭代，还能反过来。前面实现的是传统的从前到后，那问题来了，反过来该如何实现呢？< / p >
< p > 答案是: < code > DoubleEndedIterator< / code > ，它继承自 < code > Iterator< / code > ( 通过 < a href = "https://course.rs/basic/trait/advance-trait.html?highlight=supertrait#特征定义中的特征约束" > < code > supertrait< / code > < / a > )，因此意味着要实现该特征，首先需要实现 < code > Iterator< / code > 。< / p >
< p > 这样只要为 < code > DoubleEndedIterator< / code > 实现 < code > next_back< / code > 方法，就可以支持双向迭代了: < code > Iterator< / code > 的 < code > next< / code > 方法从前往后，而 < code > next_back< / code > 从后向前。< / p >
< pre class = "playground" > < code class = "language-rust edition2021" > < span class = "boring" > #![allow(unused)]
< / span > < span class = "boring" > fn main() {
< / span > impl< T> DoubleEndedIterator for IntoIter< T> {
fn next_back(& mut self) -> Option< Self::Item> {
self.0.pop_back()
}
}
< span class = "boring" > }< / span > < / code > < / pre >
< p > 测试下:< / p >
< pre class = "playground" > < code class = "language-rust edition2021" > < span class = "boring" > #![allow(unused)]
< / span > < span class = "boring" > fn main() {
< / span > #[test]
fn into_iter() {
let mut list = List::new();
list.push_front(1); list.push_front(2); list.push_front(3);
let mut iter = list.into_iter();
assert_eq!(iter.next(), Some(3));
assert_eq!(iter.next_back(), Some(1));
assert_eq!(iter.next(), Some(2));
assert_eq!(iter.next_back(), None);
assert_eq!(iter.next(), None);
}
< span class = "boring" > }< / span > < / code > < / pre >
< pre > < code class = "language-shell" > $ cargo test
Running target/debug/lists-5c71138492ad4b4a
running 11 tests
test fourth::test::basics ... ok
test fourth::test::peek ... ok
test fourth::test::into_iter ... ok
test first::test::basics ... ok
test second::test::basics ... ok
test second::test::iter ... ok
test second::test::iter_mut ... ok
test third::test::iter ... ok
test third::test::basics ... ok
test second::test::into_iter ... ok
test second::test::peek ... ok
test result: ok. 11 passed; 0 failed; 0 ignored; 0 measured
< / code > < / pre >
< h2 id = "iter" > < a class = "header" href = "#iter" > Iter< / a > < / h2 >
< p > 这里又要用到糟糕的 < code > Ref< / code > :< / p >
< pre class = "playground" > < code class = "language-rust edition2021" > < span class = "boring" > #![allow(unused)]
< / span > < span class = "boring" > fn main() {
< / span > pub struct Iter< 'a, T> (Option< Ref< 'a, Node< T> > > );
impl< T> List< T> {
pub fn iter(& self) -> Iter< T> {
Iter(self.head.as_ref().map(|head| head.borrow()))
}
}
< span class = "boring" > }< / span > < / code > < / pre >
< pre > < code class = "language-shell" > $ cargo build
< / code > < / pre >
< p > 迄今为止一切运行正常，接下来的 < code > next< / code > 实现起来会有些麻烦:< / p >
< pre class = "playground" > < code class = "language-rust edition2021" > < span class = "boring" > #![allow(unused)]
< / span > < span class = "boring" > fn main() {
< / span > impl< 'a, T> Iterator for Iter< 'a, T> {
type Item = Ref< 'a, T> ;
fn next(& mut self) -> Option< Self::Item> {
self.0.take().map(|node_ref| {
self.0 = node_ref.next.as_ref().map(|head| head.borrow());
Ref::map(node_ref, |node| & node.elem)
})
}
}
< span class = "boring" > }< / span > < / code > < / pre >
< pre > < code class = "language-shell" > $ cargo build
error[E0521]: borrowed data escapes outside of closure
--> src/fourth.rs:155:13
|
153 | fn next(& mut self) -> Option< Self::Item> {
| --------- `self` is declared here, outside of the closure body
154 | self.0.take().map(|node_ref| {
155 | self.0 = node_ref.next.as_ref().map(|head| head.borrow());
| ^^^^^^ -------- borrow is only valid in the closure body
| |
| reference to `node_ref` escapes the closure body here
error[E0505]: cannot move out of `node_ref` because it is borrowed
--> src/fourth.rs:156:22
|
153 | fn next(& mut self) -> Option< Self::Item> {
| --------- lifetime `'1` appears in the type of `self`
154 | self.0.take().map(|node_ref| {
155 | self.0 = node_ref.next.as_ref().map(|head| head.borrow());
| ------ -------- borrow of `node_ref` occurs here
| |
| assignment requires that `node_ref` is borrowed for `'1`
156 | Ref::map(node_ref, |node| & node.elem)
| ^^^^^^^^ move out of `node_ref` occurs here
< / code > < / pre >
< p > 果然，膝盖又中了一箭。< / p >
< p > < code > node_ref< / code > 活得不够久， ， < code > Ref< / code > ，从 < code > head.borrow()< / code > 中取出的 < code > Ref< / code > 只允许跟 < code > node_ref< / code > 活得一样久。< / p >
< p > 而我们想要的函数是存在的:< / p >
< pre class = "playground" > < code class = "language-rust edition2021" > < span class = "boring" > #![allow(unused)]
< / span > < span class = "boring" > fn main() {
< / span > pub fn map_split< U, V, F> (orig: Ref< 'b, T> , f: F) -> (Ref< 'b, U> , Ref< 'b, V> ) where
F: FnOnce(& T) -> (& U, & V),
U: ?Sized,
V: ?Sized,
< span class = "boring" > }< / span > < / code > < / pre >
< p > 喔，这个函数定义的泛型直接晃瞎了我的眼睛。。< / p >
< pre class = "playground" > < code class = "language-rust edition2021" > < span class = "boring" > #![allow(unused)]
< / span > < span class = "boring" > fn main() {
< / span > fn next(& mut self) -> Option< Self::Item> {
self.0.take().map(|node_ref| {
let (next, elem) = Ref::map_split(node_ref, |node| {
(& node.next, & node.elem)
});
self.0 = next.as_ref().map(|head| head.borrow());
elem
})
}
< span class = "boring" > }< / span > < / code > < / pre >
< pre > < code class = "language-shell" > $ cargo build
Compiling lists v0.1.0 (/Users/ABeingessner/dev/temp/lists)
error[E0521]: borrowed data escapes outside of closure
--> src/fourth.rs:159:13
|
153 | fn next(& mut self) -> Option< Self::Item> {
| --------- `self` is declared here, outside of the closure body
...
159 | self.0 = next.as_ref().map(|head| head.borrow());
| ^^^^^^ ---- borrow is only valid in the closure body
| |
| reference to `next` escapes the closure body here
< / code > < / pre >
< p > 额，借用的内容只允许在闭包体中使用，看起来我们还是得用 < code > Ref::map< / code > 来解决问题:< / p >
< pre class = "playground" > < code class = "language-rust edition2021" > < span class = "boring" > #![allow(unused)]
< / span > < span class = "boring" > fn main() {
< / span > fn next(& mut self) -> Option< Self::Item> {
self.0.take().map(|node_ref| {
let (next, elem) = Ref::map_split(node_ref, |node| {
(& node.next, & node.elem)
});
self.0 = if next.is_some() {
Some(Ref::map(next, |next| & **next.as_ref().unwrap()))
} else {
None
};
elem
})
}
< span class = "boring" > }< / span > < / code > < / pre >
< pre > < code class = "language-shell" > error[E0308]: mismatched types
--> src/fourth.rs:162:22
|
162 | Some(Ref::map(next, |next| & **next.as_ref().unwrap()))
| ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ expected struct `fourth::Node`, found struct `std::cell::RefCell`
|
= note: expected type `std::cell::Ref< '_, fourth::Node< _> > `
found type `std::cell::Ref< '_, std::cell::RefCell< fourth::Node< _> > > `
< / code > < / pre >
< p > 晕, 多了一个 < code > RefCell< / code > ，随着我们的对链表的逐步深入，< code > RefCell< / code > 的代码嵌套变成了不可忽视的问题。< / p >
< p > 看起来我们已经无能为力了，只能试着去摆脱 < code > RefCell< / code > 了。< code > Rc< / code > 怎么样？我们完全可以对 < code > Rc< / code > 进行完整的克隆:< / p >
< pre class = "playground" > < code class = "language-rust edition2021" > < span class = "boring" > #![allow(unused)]
< / span > < span class = "boring" > fn main() {
< / span > pub struct Iter< T> (Option< Rc< Node< T> > > );
impl< T> List< T> {
pub fn iter(& self) -> Iter< T> {
Iter(self.head.as_ref().map(|head| head.clone()))
}
}
impl< T> Iterator for Iter< T> {
type Item =
< span class = "boring" > }< / span > < / code > < / pre >
< p > 等等，那现在返回的是什么？< code > & T< / code > 还是 < code > Ref< T> < / code > ?< / p >
< p > 两者都不是，现在我们的 < code > Iter< / code > 已经没有生命周期了：无论是 < code > & T< / code > 还是 < code > Ref< T> < / code > 都需要我们在 < code > next< / code > 之前声明好生命周期。但是我们试图从 < code > Rc< / code > 中取出来的值其实是迭代器的引用。< / p >
< p > 也可以通过对 < code > Rc< / code > 进行 map 获取到 < code > Rc< T> < / code > ？但是标准库并没有给我们提供相应的功能，第三方倒是有< a href = "https://crates.io/crates/owning_ref" > 一个< / a > 。< / p >
< p > 但是，即使这么做了，还有一个更大的坑在等着：一个会造成迭代器不合法的可怕幽灵。事实上，之前我们对于迭代器不合法是免疫的，但是一旦迭代器产生 < code > Rc< / code > ，那它们就不再会借用链表。这意味着人们可以在持有指向链表内部的指针时，还可以进行 < code > push< / code > 和 < code > pop< / code > 操作。< / p >
< p > 严格来说，< code > push< / code > 问题不大，因为链表两端的增长不会对我们正在关注的某个子链表造成影响。< / p >
< p > 但是 < code > pop< / code > 就是另一个故事了，如果在我们关注的子链表之外 < code > pop< / code > , 那问题不大。但是如果是 < code > pop< / code > 一个正在引用的子链表中的节点呢？那一切就完了，特别是，如果大家还试图去 unwrap < code > try_unwrap< / code > 返回的 < code > Result< / code > ，会直接造成整个程序的 < code > panic< / code > 。< / p >
< p > 仔细想一想，好像也不错，程序一切正常，除非去 < code > pop< / code > 我们正在引用的节点，最美的是，就算遇到这种情况，程序也会直接崩溃，提示我们错误的发生。< / p >
< p > 其实我们大部分的努力都是为了实现隐藏的细节和优雅的 API， ， ， ， < / p >
< p > 总之，可以看出，内部可变性非常适合写一个安全性的应用程序，但是如果是安全性高的库，那内部可变性就有些捉襟见肘了。< / p >
< p > 最终，我选择了放弃，不再实现 < code > Iter< / code > 和 < code > IterMut< / code > ，也许努力下，可以实现，但是。。。不愉快，算了。< / p >
< / main >
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