IterMut以及完整代码
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。
再来回顾下 Iter 的实现：
#![allow(unused)] fn main() { impl<'a, T> Iterator for Iter<'a, T> { type Item = &'a T; fn next(&mut self) -> Option<Self::Item> { /* stuff */ } } }
这段代码可以进行下脱糖( desugar ):
#![allow(unused)] fn main() { impl<'a, T> Iterator for Iter<'a, T> { type Item = &'a T; fn next<'b>(&'b mut self) -> Option<&'a T> { /* stuff */ } } }
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:
#![allow(unused)] fn main() { let mut list = List::new(); list.push(1); list.push(2); list.push(3); let mut iter = list.iter(); let x = iter.next().unwrap(); let y = iter.next().unwrap(); let z = iter.next().unwrap(); }
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。
先将之前的代码修改成可变的：
#![allow(unused)] fn main() { pub struct IterMut<'a, T> { next: Option<&'a mut Node<T>>, } impl<T> List<T> { pub fn iter_mut(&self) -> IterMut<'_, T> { IterMut { next: self.head.as_deref_mut() } } } impl<'a, T> Iterator for IterMut<'a, T> { type Item = &'a mut T; fn next(&mut self) -> Option<Self::Item> { self.next.map(|node| { self.next = node.next.as_deref_mut(); &mut node.elem }) } } }
$ cargo build
error[E0596]: cannot borrow `self.head` as mutable, as it is behind a `&` reference
--> src/second.rs:95:25
|
94 | pub fn iter_mut(&self) -> IterMut<'_, T> {
| ----- help: consider changing this to be a mutable reference: `&mut self`
95 | IterMut { next: self.head.as_deref_mut() }
| ^^^^^^^^^ `self` is a `&` reference, so the data it refers to cannot be borrowed as mutable
error[E0507]: cannot move out of borrowed content
--> src/second.rs:103:9
|
103 | self.next.map(|node| {
| ^^^^^^^^^ cannot move out of borrowed content
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:
#![allow(unused)] fn main() { pub fn iter_mut(&mut self) -> IterMut<'_, T> { IterMut { next: self.head.as_deref_mut() } } }
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？
原因在于有些类型可以 Copy，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。
因此我们需要使用 take 方法来处理这种情况：
#![allow(unused)] fn main() { fn next(&mut self) -> Option<Self::Item> { self.next.take().map(|node| { self.next = node.next.as_deref_mut(); &mut node.elem }) } }
$ cargo build
老规矩，来测试下:
#![allow(unused)] fn main() { #[test] fn iter_mut() { let mut list = List::new(); list.push(1); list.push(2); list.push(3); let mut iter = list.iter_mut(); assert_eq!(iter.next(), Some(&mut 3)); assert_eq!(iter.next(), Some(&mut 2)); assert_eq!(iter.next(), Some(&mut 1)); } }
$ cargo test
Running target/debug/lists-5c71138492ad4b4a
running 6 tests
test first::test::basics ... ok
test second::test::basics ... ok
test second::test::iter_mut ... ok
test second::test::into_iter ... ok
test second::test::iter ... ok
test second::test::peek ... ok
test result: ok. 7 passed; 0 failed; 0 ignored; 0 measured
最终，我们完成了迭代器的功能，下面是完整的代码。
完整代码
#![allow(unused)] fn main() { pub struct List<T> { head: Link<T>, } 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 } } pub fn push(&mut self, elem: T) { let new_node = Box::new(Node { elem: elem, next: self.head.take(), }); self.head = Some(new_node); } pub fn pop(&mut self) -> Option<T> { self.head.take().map(|node| { self.head = node.next; node.elem }) } pub fn peek(&self) -> Option<&T> { self.head.as_ref().map(|node| { &node.elem }) } pub fn peek_mut(&mut self) -> Option<&mut T> { self.head.as_mut().map(|node| { &mut node.elem }) } pub fn into_iter(self) -> IntoIter<T> { IntoIter(self) } pub fn iter(&self) -> Iter<'_, T> { Iter { next: self.head.as_deref() } } pub fn iter_mut(&mut self) -> IterMut<'_, T> { IterMut { next: self.head.as_deref_mut() } } } impl<T> Drop for List<T> { fn drop(&mut self) { let mut cur_link = self.head.take(); while let Some(mut boxed_node) = cur_link { cur_link = boxed_node.next.take(); } } } pub struct IntoIter<T>(List<T>); impl<T> Iterator for IntoIter<T> { type Item = T; fn next(&mut self) -> Option<Self::Item> { // access fields of a tuple struct numerically self.0.pop() } } pub struct Iter<'a, T> { next: Option<&'a Node<T>>, } impl<'a, T> Iterator for Iter<'a, T> { type Item = &'a T; fn next(&mut self) -> Option<Self::Item> { self.next.map(|node| { self.next = node.next.as_deref(); &node.elem }) } } pub struct IterMut<'a, T> { next: Option<&'a mut Node<T>>, } impl<'a, T> Iterator for IterMut<'a, T> { type Item = &'a mut T; fn next(&mut self) -> Option<Self::Item> { self.next.take().map(|node| { self.next = node.next.as_deref_mut(); &mut node.elem }) } } #[cfg(test)] 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(3)); 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(5)); assert_eq!(list.pop(), Some(4)); // Check exhaustion assert_eq!(list.pop(), Some(1)); assert_eq!(list.pop(), None); } #[test] fn peek() { let mut list = List::new(); assert_eq!(list.peek(), None); assert_eq!(list.peek_mut(), None); list.push(1); list.push(2); list.push(3); assert_eq!(list.peek(), Some(&3)); assert_eq!(list.peek_mut(), Some(&mut 3)); list.peek_mut().map(|value| { *value = 42 }); assert_eq!(list.peek(), Some(&42)); assert_eq!(list.pop(), Some(42)); } #[test] fn into_iter() { let mut list = List::new(); list.push(1); list.push(2); list.push(3); let mut iter = list.into_iter(); assert_eq!(iter.next(), Some(3)); assert_eq!(iter.next(), Some(2)); assert_eq!(iter.next(), Some(1)); assert_eq!(iter.next(), None); } #[test] fn iter() { let mut list = List::new(); list.push(1); list.push(2); list.push(3); let mut iter = list.iter(); assert_eq!(iter.next(), Some(&3)); assert_eq!(iter.next(), Some(&2)); assert_eq!(iter.next(), Some(&1)); } #[test] fn iter_mut() { let mut list = List::new(); list.push(1); list.push(2); list.push(3); let mut iter = list.iter_mut(); assert_eq!(iter.next(), Some(&mut 3)); assert_eq!(iter.next(), Some(&mut 2)); assert_eq!(iter.next(), Some(&mut 1)); } } }