Rust“类型的可变性不同”

Question

我正在尝试从向量中删除一个元素（如果它存在于其中）：

use std::collections::HashMap;

fn test(map: HashMap<String, Vec<String>>, department: String, employee: String) {
    let &mut list = map.get(&department).unwrap();
    let index = list.iter().position(|x| x == &employee);
    match index {
        Some(i) => {
            list.remove(i);
        },
        None => {
            println!("No records of {} in {}!", &employee, &department);
        },
    }
}

我收到此错误：

error[E0308]: mismatched types
 --> src/lib.rs:4:9
  |
4 |     let &mut list = map.get(&department).unwrap();
  |         ^^^^^^^^^   ----------------------------- this expression has type `&std::vec::Vec<std::string::String>`
  |         |
  |         types differ in mutability
  |
  = note:      expected reference `&std::vec::Vec<std::string::String>`
          found mutable reference `&mut _`

^Playground

我以为我理解错误的含义（第 170 行的 RHS 返回对向量的不可变引用），但我不太确定如何解决它。如果尝试这样的事情：

let mut list = map.get(&department).unwrap();
let index = list.iter().position(|x| x == &employee);
match index {
    Some(i) => {
        list.remove(i);
    },
    ...
}

然后我得到

error[E0596]: cannot borrow `*list` as mutable, as it is behind a `&` reference
 --> src/lib.rs:8:13
  |
4 |     let mut list = map.get(&department).unwrap();
  |         -------- help: consider changing this to be a mutable reference: `&mut std::vec::Vec<std::string::String>`
...
8 |             list.remove(i);
  |             ^^^^ `list` is a `&` reference, so the data it refers to cannot be borrowed as mutable

^Playground

这些错误对我来说似乎是一种循环，这让我觉得我需要重新考虑我的设计。我该如何解决这个问题？

Answer 1

get 返回一个不可变的Option<&V>，它不允许你调用.remove(i)。使用get_mut 获得可变的Option<&mut V>。调用它要求地图本身是可变的，这表明我们正在做正确的事情。

use std::collections::HashMap;

fn test(map: &mut HashMap<String, Vec<String>>, department: String, employee: String) {
    //       ^^^^ 
    let list = map.get_mut(&department).unwrap();
    //             ^^^^^^^
    let index = list.iter().position(|x| x == &employee);
    match index {
        Some(i) => {
            list.remove(i);
        },
        None => {
            println!("No records of {} in {}!", &employee, &department);
        },
    }
}

^Playground