rust - 如何返回对 optional 结构字段内值的引用？

Question

这个问题在这里已经有了答案:





Cannot move out of value which is behind a shared reference when unwrapping

(2 个回答)



How to get an Option's value or set it if it's empty?

(1 个回答)


11 个月前关闭。




我正在努力改进 Cacher The Rust Programming Language 中描述的类型.建议的改进之一指出 Cacher应该可以用于多种类型。为此，我编写了以下代码:

struct Cacher<T, U>
where
    T: Fn(&U) -> U,
{
    calculation: T,
    value: Option<U>,
}

impl<T, U> Cacher<T, U>
where
    T: Fn(&U) -> U,
{
    fn new(calculation: T) -> Cacher<T, U> {
        Cacher {
            calculation,
            value: None,
        }
    }

    fn value(&mut self, arg: &U) -> &U {
        match self.value {
            Some(v) => &v,
            None => {
                let v = (self.calculation)(arg);
                self.value = Some(v);
                &v
            }
        }
    }
}

编译器提示:

error[E0507]: cannot move out of `self.value.0` which is behind a mutable reference
  --> src/lib.rs:21:15
   |
21 |         match self.value {
   |               ^^^^^^^^^^ help: consider borrowing here: `&self.value`
22 |             Some(v) => &v,
   |                  -
   |                  |
   |                  data moved here
   |                  move occurs because `v` has type `U`, which does not implement the `Copy` trait

error[E0515]: cannot return reference to local variable `v`
  --> src/lib.rs:22:24
   |
22 |             Some(v) => &v,
   |                        ^^ returns a reference to data owned by the current function

error[E0515]: cannot return reference to local variable `v`
  --> src/lib.rs:26:17
   |
26 |                 &v
   |                 ^^ returns a reference to data owned by the current function

error[E0382]: borrow of moved value: `v`
  --> src/lib.rs:26:17
   |
24 |                 let v = (self.calculation)(arg);
   |                     - move occurs because `v` has type `U`, which does not implement the `Copy` trait
25 |                 self.value = Some(v);
   |                                   - value moved here
26 |                 &v
   |                 ^^ value borrowed here after move

我想这是因为 v是函数的局部。但是，鉴于实际数据存在于结构中，该结构存在于 value 之外。方法，是否不可能以任何方式返回对此数据的引用？如果不是这样，那么我需要做什么才能获得 Cacher 的功能拥有计算数据并在需要时返回引用？

Answer 1

你几乎明白了。您只需要返回对对象的引用 内 Option :

struct Cacher<T, U>
where
    T: Fn(&U) -> U
{
    calculation: T,
    value: Option<U>,
}

impl<T, U> Cacher<T, U>
where
    T: Fn(&U) -> U
{
    fn new(calculation: T) -> Cacher<T, U> {
        Cacher {
            calculation,
            value: None,
        }
    }

    fn value(&mut self, arg: &U) -> &U {
        if self.value.is_none() {
            let v = (self.calculation)(arg);
            self.value = Some(v);
        }
        self.value.as_ref().unwrap()
    }
}

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