Question

我有两个具有相同键的HashMap<&str, String>，并且我希望创建一个具有相同键的HashMap，将值组合在一起。我不想保留对前两个HashMap的引用，但想将String移到新的HashMap。

use std::collections::HashMap;

#[derive(Debug)]
struct Contact {
    phone: String,
    address: String,
}

fn main() {
    let mut phones: HashMap<&str, String> = HashMap::new();
    phones.insert("Daniel", "798-1364".into());
    phones.insert("Ashley", "645-7689".into());
    phones.insert("Katie", "435-8291".into());
    phones.insert("Robert", "956-1745".into());

    let mut addresses: HashMap<&str, String> = HashMap::new();
    addresses.insert("Daniel", "12 A Street".into());
    addresses.insert("Ashley", "12 B Street".into());
    addresses.insert("Katie", "12 C Street".into());
    addresses.insert("Robert", "12 D Street".into());

    let contacts: HashMap<&str, Contact> = phones.keys().fold(HashMap::new(), |mut acc, value| {
        acc.entry(value).or_insert(Contact {
            phone: *phones.get(value).unwrap(),
            address: *addresses.get(value).unwrap(),
        });
        acc
    });

    println!("{:?}", contacts);
}

但是我有一个错误

error[E0507]: cannot move out of a shared reference
  --> src/main.rs:24:20
   |
24 |             phone: *phones.get(value).unwrap(),
   |                    ^^^^^^^^^^^^^^^^^^^^^^^^^^^ move occurs because value has type `std::string::String`, which does not implement the `Copy` trait

error[E0507]: cannot move out of a shared reference
  --> src/main.rs:25:22
   |
25 |             address: *addresses.get(value).unwrap(),
   |                      ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ move occurs because value has type `std::string::String`, which does not implement the `Copy` trait

Playground

Answer 1

HashMap::get返回一个Option<&V>，即对映射内值的引用。除非*实现V，否则您不能移出Copy的引用。您需要使用另一种方法将值移出映射，即HashMap::remove（请注意，它返回Option<V>）。

如果尝试使用remove重写相同的算法，则会收到不同的错误：

    let contacts: HashMap<&str, Contact> = phones.keys().fold(HashMap::new(), |mut acc, value| {
        acc.entry(value).or_insert(Contact {
            phone: phones.remove(value).unwrap(),
            address: addresses.remove(value).unwrap(),
        });
        acc
    });

error[E0502]: cannot borrow `phones` as mutable because it is also borrowed as immutable
  --> src/main.rs:22:79
   |
22 |     let contacts: HashMap<&str, Contact> = phones.keys().fold(HashMap::new(), |mut acc, value| {
   |                                            ------        ----                 ^^^^^^^^^^^^^^^^ mutable borrow occurs here
   |                                            |             |
   |                                            |             immutable borrow later used by call
   |                                            immutable borrow occurs here
23 |         acc.entry(value).or_insert(Contact {
24 |             phone: phones.remove(value).unwrap(),
   |                    ------ second borrow occurs due to use of `phones` in closure

error: aborting due to previous error

For more information about this error, try `rustc --explain E0502`.

此错误告诉您在对数据结构进行迭代时不能对其进行变异，因为对数据结构进行变异可能会使迭代器无效。 Sometimes you can solve this with interior mutability，但在这种情况下，您无需执行任何操作。迭代时，只需致电phones.into_iter()，即可将电话号码移出地图。然后，很容易使用map创建(&str, Contact)元组，最后collect将它们全部重新组合成HashMap。

    let contacts: HashMap<_, _> = phones
        .into_iter()
        .map(|(key, phone)| {
            (
                key,
                Contact {
                    phone,
                    address: addresses.remove(key).unwrap(),
                },
            )
        })
        .collect();

Playground

Answer 2

zip是您的朋友。但是这里的“业务逻辑”规定它仅适用于排序的映射。因此，如果可以使用BTreeMap代替HashMap，则可以使用以下方法：

fn main() {
    let mut phones: BTreeMap<&str, String> = BTreeMap::new();
    ...

    let mut addresses: BTreeMap<&str, String> = BTreeMap::new();
    ...

    let contacts: BTreeMap<&str, Contact> = phones
        .into_iter()
        .zip(addresses.into_iter())
        .map(|((name, phone), (_, addr))| {
            (
                name,
                Contact {
                    phone: phone,
                    address: addr,
                },
            )
        })
        .collect();

    println!("{:#?}", contacts);
}

如何使用从其他两个HashMap中移动的值创建HashMap？

2 个答案: