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Rustln
3 years ago
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# 还可以的单向链表
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在之前我们写了一个最小可用的单向链表，下面一起来完善下，首先创建一个新的文件 `src/second.rs`，然后在 `lib.rs` 中引入：
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```rust
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// in lib.rs
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pub mod first;
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pub mod second;
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```
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并将 `first.rs` 中的所有内容拷贝到 `second.rs` 中。
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# IterMut以及完整代码
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上一章节中我们讲到了要为 `List` 实现三种类型的迭代器并实现了其中两种: `IntoIter` 和 `Iter`。下面再来看看最后一种 `IterMut`。
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再来回顾下 `Iter` 的实现：
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```rust
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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> { /* stuff */ }
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}
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```
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这段代码可以进行下脱糖( desugar ):
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```rust
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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<'b>(&'b mut self) -> Option<&'a T> { /* stuff */ }
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}
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```
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可以看出 `next` 方法的输入和输出之间的生命周期并没有关联，这样我们就可以无条件的一遍又一遍地调用 `next`:
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```rust
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let mut list = List::new();
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list.push(1); list.push(2); list.push(3);
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let mut iter = list.iter();
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let x = iter.next().unwrap();
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let y = iter.next().unwrap();
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let z = iter.next().unwrap();
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```
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对于不可变借用而言，这种方式没有任何问题，因为不可变借用可以同时存在多个，但是如果是可变引用呢？因此，大家可能会以为使用安全代码来写 `IterMut` 是一件相当困难的事。但是令人诧异的是，事实上，我们可以使用安全的代码来为很多数据结构实现 `IterMut`。
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先将之前的代码修改成可变的：
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```rust
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pub struct IterMut<'a, T> {
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next: Option<&'a mut Node<T>>,
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}
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impl<T> List<T> {
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pub fn iter_mut(&self) -> IterMut<'_, T> {
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IterMut { next: self.head.as_deref_mut() }
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}
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}
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impl<'a, T> Iterator for IterMut<'a, T> {
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type Item = &'a mut 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.next.as_deref_mut();
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&mut node.elem
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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 build
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error[E0596]: cannot borrow `self.head` as mutable, as it is behind a `&` reference
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--> src/second.rs:95:25
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94 | pub fn iter_mut(&self) -> IterMut<'_, T> {
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| ----- help: consider changing this to be a mutable reference: `&mut self`
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95 | IterMut { next: self.head.as_deref_mut() }
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| ^^^^^^^^^ `self` is a `&` reference, so the data it refers to cannot be borrowed as mutable
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error[E0507]: cannot move out of borrowed content
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--> src/second.rs:103:9
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103 | self.next.map(|node| {
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| ^^^^^^^^^ cannot move out of borrowed content
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```
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果不其然，两个错误发生了。第一错误看上去很清晰，甚至告诉了我们该如何解决:
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```rust
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pub fn iter_mut(&mut self) -> IterMut<'_, T> {
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IterMut { next: self.head.as_deref_mut() }
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}
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```
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但是另一个好像就没那么容易了。但是之前的代码就可以工作啊，为何这里就不行了？
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原因在于有些类型可以 [Copy](https://course.rs/basic/ownership/ownership.html#拷贝浅拷贝)，有些不行。而`Option` 和不可变引用 `&T` 恰恰是可以 Copy 的，但尴尬的是，可变引用 `&mut T` 不可以，因此这里报错了。
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因此我们需要使用 `take` 方法来处理这种情况：
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```rust
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fn next(&mut self) -> Option<Self::Item> {
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self.next.take().map(|node| {
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self.next = node.next.as_deref_mut();
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&mut node.elem
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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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```
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老规矩，来测试下:
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```rust
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#[test]
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fn iter_mut() {
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let mut list = List::new();
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list.push(1); list.push(2); list.push(3);
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let mut iter = list.iter_mut();
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assert_eq!(iter.next(), Some(&mut 3));
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assert_eq!(iter.next(), Some(&mut 2));
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assert_eq!(iter.next(), Some(&mut 1));
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}
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```
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```shell
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> cargo test
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Running target/debug/lists-5c71138492ad4b4a
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running 6 tests
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test first::test::basics ... ok
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test second::test::basics ... ok
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test second::test::iter_mut ... ok
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test second::test::into_iter ... ok
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test second::test::iter ... ok
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test second::test::peek ... ok
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test result: ok. 7 passed; 0 failed; 0 ignored; 0 measured
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```
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最终，我们完成了迭代器的功能，下面是完整的代码。
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## 完整代码
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```rust
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pub struct List<T> {
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head: Link<T>,
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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 }
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}
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pub fn push(&mut self, elem: T) {
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let new_node = Box::new(Node {
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elem: elem,
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next: self.head.take(),
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});
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self.head = Some(new_node);
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}
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pub fn pop(&mut self) -> Option<T> {
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self.head.take().map(|node| {
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self.head = node.next;
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node.elem
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})
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}
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pub fn peek(&self) -> Option<&T> {
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self.head.as_ref().map(|node| {
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&node.elem
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})
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}
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pub fn peek_mut(&mut self) -> Option<&mut T> {
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self.head.as_mut().map(|node| {
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&mut node.elem
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})
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}
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pub fn into_iter(self) -> IntoIter<T> {
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IntoIter(self)
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}
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pub fn iter(&self) -> Iter<'_, T> {
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Iter { next: self.head.as_deref() }
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}
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pub fn iter_mut(&mut self) -> IterMut<'_, T> {
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IterMut { next: self.head.as_deref_mut() }
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}
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}
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impl<T> Drop for List<T> {
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fn drop(&mut self) {
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let mut cur_link = self.head.take();
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while let Some(mut boxed_node) = cur_link {
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cur_link = boxed_node.next.take();
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}
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}
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}
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pub struct IntoIter<T>(List<T>);
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impl<T> Iterator for IntoIter<T> {
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type Item = T;
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fn next(&mut self) -> Option<Self::Item> {
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// access fields of a tuple struct numerically
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self.0.pop()
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}
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}
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pub struct Iter<'a, T> {
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next: Option<&'a Node<T>>,
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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.next.as_deref();
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&node.elem
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})
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}
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}
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pub struct IterMut<'a, T> {
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next: Option<&'a mut Node<T>>,
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}
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impl<'a, T> Iterator for IterMut<'a, T> {
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type Item = &'a mut T;
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fn next(&mut self) -> Option<Self::Item> {
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self.next.take().map(|node| {
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self.next = node.next.as_deref_mut();
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&mut node.elem
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})
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}
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}
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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 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(3));
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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(5));
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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(1));
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assert_eq!(list.pop(), None);
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}
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#[test]
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fn peek() {
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let mut list = List::new();
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assert_eq!(list.peek(), None);
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assert_eq!(list.peek_mut(), None);
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list.push(1); list.push(2); list.push(3);
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assert_eq!(list.peek(), Some(&3));
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assert_eq!(list.peek_mut(), Some(&mut 3));
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list.peek_mut().map(|value| {
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*value = 42
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});
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assert_eq!(list.peek(), Some(&42));
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assert_eq!(list.pop(), Some(42));
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}
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#[test]
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fn into_iter() {
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let mut list = List::new();
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list.push(1); list.push(2); list.push(3);
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let mut iter = list.into_iter();
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assert_eq!(iter.next(), Some(3));
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assert_eq!(iter.next(), Some(2));
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assert_eq!(iter.next(), Some(1));
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assert_eq!(iter.next(), None);
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}
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#[test]
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fn iter() {
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let mut list = List::new();
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list.push(1); list.push(2); list.push(3);
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let mut iter = list.iter();
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assert_eq!(iter.next(), Some(&3));
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assert_eq!(iter.next(), Some(&2));
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assert_eq!(iter.next(), Some(&1));
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}
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#[test]
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fn iter_mut() {
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let mut list = List::new();
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list.push(1); list.push(2); list.push(3);
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let mut iter = list.iter_mut();
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assert_eq!(iter.next(), Some(&mut 3));
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assert_eq!(iter.next(), Some(&mut 2));
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assert_eq!(iter.next(), Some(&mut 1));
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}
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}
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```
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