我正在学习Rust，但我不太明白为什么它不起作用。
#[derive(Debug)]
struct Node {
value: String,
}
#[derive(Debug)]
pub struct Graph {
nodes: Vec<Box<Node>>,
}
fn mk_node(value: String) -> Node {
Node { value }
}
pub fn mk_graph() -> Graph {
Graph { nodes: vec![] }
}
impl Graph {
fn add_node(&mut self, value: String) {
if let None = self.nodes.iter().position(|node| node.value == value) {
let node = Box::new(mk_node(value));
self.nodes.push(node);
};
}
fn get_node_by_value(&self, value: &str) -> Option<&Node> {
match self.nodes.iter().position(|node| node.value == *value) {
None => None,
Some(idx) => self.nodes.get(idx).map(|n| &**n),
}
}
}
#[cfg(test)]
mod tests {
use super::*;
#[test]
fn some_test() {
let mut graph = mk_graph();
graph.add_node("source".to_string());
graph.add_node("destination".to_string());
let source = graph.get_node_by_value("source").unwrap();
let dest = graph.get_node_by_value("destination").unwrap();
graph.add_node("destination".to_string());
}
}
(playground)
这有错误
error[E0502]: cannot borrow `graph` as mutable because it is also borrowed as immutable
--> src/main.rs:50:9
|
47 | let source = graph.get_node_by_value("source").unwrap();
| ----- immutable borrow occurs here
...
50 | graph.add_node("destination".to_string());
| ^^^^^ mutable borrow occurs here
51 | }
| - immutable borrow ends here
来自Rust编程的示例与我的示例非常相似，但是可以正常工作:
pub struct Queue {
older: Vec<char>, // older elements, eldest last.
younger: Vec<char>, // younger elements, youngest last.
}
impl Queue {
/// Push a character onto the back of a queue.
pub fn push(&mut self, c: char) {
self.younger.push(c);
}
/// Pop a character off the front of a queue. Return `Some(c)` if there /// was a character to pop, or `None` if the queue was empty.
pub fn pop(&mut self) -> Option<char> {
if self.older.is_empty() {
if self.younger.is_empty() {
return None;
}
// Bring the elements in younger over to older, and put them in // the promised order.
use std::mem::swap;
swap(&mut self.older, &mut self.younger);
self.older.reverse();
}
// Now older is guaranteed to have something. Vec's pop method // already returns an Option, so we're set.
self.older.pop()
}
pub fn split(self) -> (Vec<char>, Vec<char>) {
(self.older, self.younger)
}
}
pub fn main() {
let mut q = Queue {
older: Vec::new(),
younger: Vec::new(),
};
q.push('P');
q.push('D');
assert_eq!(q.pop(), Some('P'));
q.push('X');
let (older, younger) = q.split(); // q is now uninitialized.
assert_eq!(older, vec!['D']);
assert_eq!(younger, vec!['X']);
}
最佳答案
您的问题的MCVE可以简化为:
fn main() {
let mut items = vec![1];
let item = items.last();
items.push(2);
}
error[E0502]: cannot borrow `items` as mutable because it is also borrowed as immutable
--> src/main.rs:4:5
|
3 | let item = items.last();
| ----- immutable borrow occurs here
4 | items.push(2);
| ^^^^^ mutable borrow occurs here
5 | }
| - immutable borrow ends here
您正在遇到Rust旨在防止的确切问题。您有一个指向向量的引用，并且正在尝试插入向量中。这样做可能需要重新分配向量的内存，从而使任何现有引用无效。如果发生这种情况，并且您使用了
item中的值，那么您将访问未初始化的内存，从而可能导致崩溃。在这种情况下，您实际上并没有使用
item(或原始版本中的source)，因此您可以...不致电该行。我假设您是出于某种原因这样做的，所以您可以将引用包装在一个块中，以便在尝试再次更改值之前将它们删除:fn main() {
let mut items = vec![1];
{
let item = items.last();
}
items.push(2);
}
由于已实现non-lexical lifetimes，但在Rust 2018中不再需要此技巧，但是仍然存在潜在的限制-当存在对同一事物的其他引用时，您不能具有可变的引用。这是Rust编程语言中介绍的rules of references之一。修改后的示例仍不适用于NLL:
let mut items = vec![1];
let item = items.last();
items.push(2);
println!("{:?}", item);
在其他情况下，您可以复制或克隆向量中的值。该项目将不再是引用，您可以根据需要修改向量:
fn main() {
let mut items = vec![1];
let item = items.last().cloned();
items.push(2);
}
如果您的类型不可克隆，则可以将其转换为引用计数值(例如
Rc 或 Arc )，然后可以将其克隆。您可能也可能不需要使用interior mutability:struct NonClone;
use std::rc::Rc;
fn main() {
let mut items = vec![Rc::new(NonClone)];
let item = items.last().cloned();
items.push(Rc::new(NonClone));
}
不，不是，因为它根本不使用引用。
也可以看看