rust - 如何借用对Option 内部内容的引用？

Question

如何从Option中提取引用，并将其与调用者的特定使用期限一起传递回去？

具体来说，我想从其中包含Box<Foo>的Bar借用对Option<Box<Foo>>的引用。我以为我可以做到:

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

...但是导致:

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`

嗯好也许不吧。看起来像我想做的事与 Option::as_ref 有关，好像我可以做的那样:

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，这也不起作用。

完整代码我遇到了麻烦:

#[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);
    }
}

我有合理的把握要尝试做的事在这里有效。

Answer 1

首先，您不需要&mut self。

匹配时，应匹配e作为引用。您正在尝试返回e的引用，但是其生存期仅适用于该match语句。

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)
        }
    }
}