不能借用可变性，因为它也被借用为不可变的

Question

我正在学习Rust，我不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

Programming Rust 中的这个示例与我所拥有的非常类似，但它有效：

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']);
}

Answer 1

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

您遇到了 Rust旨在阻止的确切问题。你有一个指向矢量的参考，并试图插入矢量。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，从而可能导致崩溃。

在此特定的情况下，您实际上并未使用item（或原始版本中的source），因此您可能只是...线。我假设你出于某种原因这样做了，所以你可以将引用包装在一个块中，这样它们就会在你试图再次变异之前消失：

fn main() {
    let mut items = vec![1];
    {
        let item = items.last();
    }
    items.push(2);
}

Rust 2018中不再需要这个技巧，因为已经实现了non-lexical lifetimes，但是基本限制仍然存在 - 当存在对同一事物的其他引用时，你不能拥有可变引用。这是 The Rust Programming Language 中涵盖的rules of references之一。一个仍然不适用于NLL的修改示例：

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。该项目将不再是参考，您可以根据需要修改矢量：

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，您可以将其转换为可以克隆的引用计数值（例如Rc或Arc）。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));
}

  

这个来自Programming Rust的例子很相似

不，不是，看看它根本不使用引用。

另见

• 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"?
• Why does Rust want to borrow a variable as mutable more than once at a time?

Answer 2

尝试将不可变借用放在块{...}中。 这结束了阻止后的借用。

#[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());
    }
}