我正在学习生锈，我不太明白为什么这不起作用。
#[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）
这有个错误
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
这个来自Programming Rust的例子与我的非常相似，但它是有效的：
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
你的问题可以归结为：
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
你遇到的正是铁锈用来防止的问题。您有一个指向向量的引用，正试图插入到向量中。这样做可能需要重新分配向量的内存，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，可能会导致崩溃。在这种特殊情况下，您实际上没有使用
item（或source，在原始版本中），因此您可以。。。别打那个电话。我假设您这样做是出于某种原因，所以您可以将引用包装在一个块中，以便它们在您尝试再次更改该值之前消失：fn main() {
let mut items = vec![1];
{
let item = items.last();
}
items.push(2);
}
这个技巧在Rust 2018中不再需要，因为MCVE已经实现了，但是底层的限制仍然存在——当有其他引用指向同一事物时，不能有可变引用。这是Rust编程语言中包含的non-lexical lifetimes之一。一个修改后的示例仍然不适用于NLL：
let mut items = vec![1];
let item = items.last();
items.push(2);
println!("{:?}", item);
在其他情况下，可以复制或克隆向量中的值。该项将不再是引用，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);
}
如果类型不可克隆，可以将其转换为引用计数值（如rules of references或
Rc），然后可以克隆。您可能需要也可能不需要使用Arc：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));
}
这个来自Rust编程的例子非常类似
不，不是，因为它根本不使用引用。
另见
interior mutability
Cannot borrow `*x` as mutable because it is also borrowed as immutable
Pushing something into a vector depending on its last element
Why doesn't the lifetime of a mutable borrow end when the function call is complete?
How should I restructure my graph code to avoid an "Cannot borrow variable as mutable more than once at a time" error?
Why do I get the error "cannot borrow x as mutable more than once"?