Background
考虑一个玩具问题，我有一个Node struct ，它表示链表的 node ，并且我想创建一个函数，该函数用1到9的值构建一个列表.以下代码按预期工作:
struct Node {
val: i32,
next: Option<Box<Node>>,
}
fn build_list() -> Option<Box<Node>> {
let mut head = None;
let mut tail = &mut head;
for n in 1..10 {
*tail = Some(Box::new(Node {val: n, next: None}));
if let Some(ref mut x) = tail {
tail = &mut x.next;
};
}
head
}
但如果我将build_list函数中的匹配表达式修改为以下内容，它将无法编译:
fn build_list() -> Option<Box<Node>> {
let mut head = None;
let mut tail = &mut head;
for n in 1..10 {
*tail = Some(Box::new(Node {val: n, next: None}));
if let Some(x) = tail.as_mut() {
tail = &mut x.next;
};
}
head
}
编译错误:
error[E0506]: cannot assign to `*tail` because it is borrowed
--> src/main.rs:72:9
|
72 | *tail = Some(Box::new(Node {val: n, next: None}));
| ^^^^^
| |
| assignment to borrowed `*tail` occurs here
| borrow later used here
73 | {
74 | if let Some(x) = tail.as_mut() {
| ---- borrow of `*tail` occurs here
error[E0499]: cannot borrow `*tail` as mutable more than once at a time
--> src/main.rs:74:30
|
74 | if let Some(x) = tail.as_mut() {
| ^^^^ mutable borrow starts here in previous iteration of loop
Question
在这个例子中，两者的区别是什么
if let Some(ref mut x) = tail
和
if let Some(x) = tail.as_mut()
?
(作为一名学习Rust 的初学者)我本以为这些匹配表达式是等效的，但显然我忽略了一些细微的差别.
使现代化
I cleaned up the code from my original example so that I don't need a placeholder element for the head of the list. The difference (和 compilation error) still remains, I just get an additional compilation error for assigning to borrowed *tail.
使现代化 2
(这只是一个奇怪的观察，无助于回答最初的问题)
fn build_list() -> Option<Box<Node>> {
let mut head = None;
let mut tail = &mut head;
for n in 1..10 {
*tail = Some(Box::new(Node {val: n, next: None}));
if let Some(ref mut x) = tail {
if n % 2 == 0 {
tail = &mut x.next;
}
};
}
head
}
新错误:
error[E0506]: cannot assign to `*tail` because it is borrowed
--> src/main.rs:60:9
|
60 | *tail = Some(Box::new(Node {val: n, next: None}));
| ^^^^^
| |
| assignment to borrowed `*tail` occurs here
| borrow later used here
61 | if let Some(ref mut x) = tail {
| --------- borrow of `*tail` occurs here
error[E0503]: cannot use `*tail` because it was mutably borrowed
--> src/main.rs:61:16
|
61 | if let Some(ref mut x) = tail {
| ^^^^^---------^
| | |
| | borrow of `tail.0` occurs here
| use of borrowed `tail.0`
| borrow later used here
error[E0499]: cannot borrow `tail.0` as mutable more than once at a time
--> src/main.rs:61:21
|
61 | if let Some(ref mut x) = tail {
| ^^^^^^^^^ mutable borrow starts here in previous iteration of loop