我如何借用对 Option<T> 中内容的引用? [英] How do I borrow a reference to what is inside an Option&lt;T&gt;?

问题描述

如何从 Option 中提取引用并将其与调用者的特定生命周期一起传回?

How do I pull a reference out of an Option and pass it back with the specific lifespan of the caller?

具体来说，我想从具有 Option 的 Bar 借用对 Box 的引用代码>在里面.我以为我可以做到:

Specifically, I want to borrow a reference to a Box<Foo> from a Bar that has an Option<Box<Foo>> in it. I thought I would be able to do:

impl Bar {
    fn borrow(&mut self) -> Result<&Box<Foo>, BarErr> {
        match self.data {
            Some(e) => Ok(&e),
            None => Err(BarErr::Nope),
        }
    }
}

...但结果是:

error: `e` does not live long enough
  --> src/main.rs:17:28
   |
17 |             Some(e) => Ok(&e),
   |                            ^ does not live long enough
18 |             None => Err(BarErr::Nope),
19 |         }
   |         - borrowed value only lives until here
   |
note: borrowed value must be valid for the anonymous lifetime #1 defined on the body at 15:54...
  --> src/main.rs:15:55
   |
15 |       fn borrow(&mut self) -> Result<&Box<Foo>, BarErr> {
   |  _______________________________________________________^ starting here...
16 | |         match self.data {
17 | |             Some(e) => Ok(&e),
18 | |             None => Err(BarErr::Nope),
19 | |         }
20 | |     }
   | |_____^ ...ending here

error[E0507]: cannot move out of borrowed content
  --> src/main.rs:16:15
   |
16 |         match self.data {
   |               ^^^^ cannot move out of borrowed content
17 |             Some(e) => Ok(&e),
   |                  - hint: to prevent move, use `ref e` or `ref mut e`

嗯，好的.也许不会.看起来我想做的事情与 <有关code>Option::as_ref，也许我可以这样做:

Hm, ok. Maybe not. It looks vaguely like what I want to do is related to Option::as_ref, like maybe I could do:

impl Bar {
    fn borrow(&mut self) -> Result<&Box<Foo>, BarErr> {
        match self.data {
            Some(e) => Ok(self.data.as_ref()),
            None => Err(BarErr::Nope),
        }
    }
}

...但是，这也行不通.

...but, that doesn't work either.

我遇到问题的完整代码:

Full code I'm having trouble with:

#[derive(Debug)]
struct Foo;

#[derive(Debug)]
struct Bar {
    data: Option<Box<Foo>>,
}

#[derive(Debug)]
enum BarErr {
    Nope,
}

impl Bar {
    fn borrow(&mut self) -> Result<&Box<Foo>, BarErr> {
        match self.data {
            Some(e) => Ok(&e),
            None => Err(BarErr::Nope),
        }
    }
}

#[test]
fn test_create_indirect() {
    let mut x = Bar { data: Some(Box::new(Foo)) };
    let mut x2 = Bar { data: None };
    {
        let y = x.borrow();
        println!("{:?}", y);
    }
    {
        let z = x2.borrow();
        println!("{:?}", z);
    }
}

我有理由确定我尝试做的事情在这里是有效的.

I'm reasonably sure what I've trying to do is valid here.

推荐答案

首先，你不需要&mut self.

匹配时，应该匹配e作为参考.您正在尝试返回 e 的引用，但它的生命周期仅适用于该匹配语句.

When matching, you should match e as a reference. You are trying to return a reference of e, but the lifetime of it is only for that match statement.

enum BarErr {
    Nope
}

struct Foo;

struct Bar {
    data: Option<Box<Foo>>
}

impl Bar {
    fn borrow(&self) -> Result<&Foo, BarErr> {
        match self.data {
            Some(ref x) => Ok(x),
            None => Err(BarErr::Nope)
        }
    }
}

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