rust - 可变引用在直接调用函数时生命周期足够长，但在通过中间函数调用时生命周期不够长

Question

对于以下 Rust 代码:

fn facing_of_mut<'a>(domain: &'a mut Domain, entity: Entity) -> Option<&'a mut Direction> {
    component_of_mut(&mut domain, entity)
}

...编译器输出:

error: `domain` does not live long enough
 --> src/facing.rs:5:27
  |
5 |     component_of_mut(&mut domain, entity)
  |                           ^^^^^^ does not live long enough
6 | }
  | - borrowed value only lives until here
  |
note: borrowed value must be valid for the lifetime 'a as defined on the body at 4:90...
 --> src/facing.rs:4:91
  |
4 |   fn facing_of_mut<'a>(domain: &'a mut Domain, entity: Entity) -> Option<&'a mut Direction> {
  |  ___________________________________________________________________________________________^ starting here...
5 | |     component_of_mut(&mut domain, entity)
6 | | }
  | |_^ ...ending here

我不明白错误消息，因为我认为声明这些生命周期的目的是明确要求作为 domain 参数传递的任何对象只要 Direction 返回值中的引用取决于 domain 分配的内存。

component_of_mut 的签名是:

pub fn component_of_mut<'a, C: 'static>(domain: &'a mut Domain, entity: Entity) -> Option<&'a mut C>

...而且我可以在单元测试中直接调用它，而不会在编译期间出现生命周期错误。只有当 facing_of_mut 调用它时，我才会遇到错误。

Answer 1

您正在引用域。 domain 是参数的局部变量并且已经是一个引用。 domain 仅在函数调用期间持续，因此该值的引用可以存在多长时间。

要修复它，请不要尝试引用引用:

type Domain = String;
type Entity = String;
type Direction = String;

pub fn component_of_mut<'a, C>(domain: &'a mut Domain, entity: Entity) -> Option<&'a mut C>
    where C: 'static
{
    unimplemented!()
}

fn facing_of_mut<'a>(domain: &'a mut Domain, entity: Entity) -> Option<&'a mut Direction> {
    component_of_mut(domain, entity)
}

fn main() {}