可变借款后的不变参考

Question

每次使用Rust时,我都会遇到与所有权/借用相关的类似问题,所以这里有一段最简单的代码来说明我的常见问题:

use std::cell::RefCell;

struct Res {
    name: String,
}

impl Res {
    fn new(name: &str) -> Res {
        Res {
            name: name.to_string(),
        }
    }

    // I don't need all_res to be mutable
    fn normalize(&mut self, all_res: &Vec<Res>) {
        // [...] Iterate through all_res and update self.name
        self.name = "foo".to_string();
    }
}

fn main() {
    let res = RefCell::new(vec![Res::new("res1"), Res::new("res2")]);

    for r in res.borrow_mut().iter_mut() {
        // This panics at runtime saying it's
        // already borrowed (which makes sense, I guess).
        r.normalize(&*res.borrow());
    }
}

读完之后RefCell我觉得这样可行.它编译,但在运行时恐慌.

在迭代同一向量时如何引用向量？是否有更好的数据结构允许我这样做？

Answer 1

你的程序很恐慌,因为你试图Vec在同一时间可变地和不可变地借用:这是不允许的.

你需要做的只是包装Strings in RefCell.这允许您在迭代时改变字符串Vec.

use std::cell::RefCell;

struct Res {
    name: RefCell<String>,
}

impl Res {
    fn new(name: &str) -> Res {
        Res {
            name: RefCell::new(name.to_string()),
        }
    }

    // I don't need all_res to be mutable
    fn normalize(&self, all_res: &Vec<Res>) {
        // [...] Iterate through all_res and update self.name
        *self.name.borrow_mut() = "foo".to_string();
    }
}

fn main() {
    let res = vec![Res::new("res1"), Res::new("res2")];

    for r in res.iter() {
        r.normalize(&res);
    }

    println!("{}", *res[0].name.borrow());
}