Question

当我尝试编译此代码时：

pub struct Context<'a> {
    pub outer: Option<&'a mut Context<'a>>,
}

impl<'a> Context<'a> {
    pub fn inside(outer: &'a mut Context) -> Context<'a> {
        Context { outer: Some(outer) }
    }
}

我收到此错误：

error[E0495]: cannot infer an appropriate lifetime for lifetime parameter `'a` due to conflicting requirements
 --> src/main.rs:7:9
  |
7 |         Context { outer: Some(outer) }
  |         ^^^^^^^
  |
note: first, the lifetime cannot outlive the lifetime 'a as defined on the impl at 5:1...
 --> src/main.rs:5:1
  |
5 | / impl<'a> Context<'a> {
6 | |     pub fn inside(outer: &'a mut Context) -> Context<'a> {
7 | |         Context { outer: Some(outer) }
8 | |     }
9 | | }
  | |_^
note: ...so that expression is assignable (expected Context<'a>, found Context<'_>)
 --> src/main.rs:7:9
  |
7 |         Context { outer: Some(outer) }
  |         ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
note: but, the lifetime must be valid for the anonymous lifetime #1 defined on the method body at 6:5...
 --> src/main.rs:6:5
  |
6 | /     pub fn inside(outer: &'a mut Context) -> Context<'a> {
7 | |         Context { outer: Some(outer) }
8 | |     }
  | |_____^
note: ...so that expression is assignable (expected &mut Context<'_>, found &mut Context<'_>)
 --> src/main.rs:7:31
  |
7 |         Context { outer: Some(outer) }
  |                               ^^^^^

为什么会这样？

Answer 1

这是因为您还没有履行所需的义务。

由于终身省略，您的代码相当于：

pub fn inside<'b>(outer: &'a mut Context<'b>) -> Context<'a>

将您的代码更改为

pub fn inside(outer: &'a mut Context<'a>) -> Context<'a>

它会编译。

由于递归结构中的冲突要求，无法推断出适当的生命周期

1 个答案: