不能借用`x`作为不可变的，因为它也被借用为可变的

Question

我需要迭代和修改

for

循环中的元素，但我还必须将

Vec

传递到修改它的函数中。有没有更好的方法用 Rust 编写这个？

最小可重现代码：

fn main() {
    let mut x: Vec<Ex> = Vec::new();
    for _ in 0..10 {
        x.push(Ex {});
    }

    for e in &mut x {
        e.do_stuff(&x)
    }
}

struct Ex {}
impl Ex {
    pub fn do_stuff(&mut self, x: &Vec<Ex>) {
        for _ in x {
            println!("Yay!")
        }
    }
}

编译器消息：

error[E0502]: cannot borrow `x` as immutable because it is also borrowed as mutable
 --> src\main.rs:8:20
  |
7 |     for e in &mut x {
  |              ------
  |              |
  |              mutable borrow occurs here
  |              mutable borrow later used here
8 |         e.do_stuff(&x)
  |                    ^^ immutable borrow occurs here

Answer 1

可使用间接法求解，例如

RefCell

:

use std::cell::RefCell;

fn main() {
    let x: RefCell<Vec<Ex>> = RefCell::new(Vec::new());
    for _ in 0..10 {
        x.borrow_mut().push(Ex {});
    }

    for e in x.borrow_mut().iter_mut() {
        e.do_stuff(&(x.borrow()))
    }
}

struct Ex {}
impl Ex {
    pub fn do_stuff(&mut self, x: &Vec<Ex>) {
        for _ in x {
            println!("Yay!")
        }
    }
}