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

我不理解错误消息，因为我认为声明这些生命周期的目的特别要求，只要domain参考Direction参数存在，就会传递任何对象。在返回值中，该值取决于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>

...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。只有在facing_of_mut调用它时我才会收到错误。

Answer 1

您正在参考domain。 domain是参数的局部变量，并且已经是引用。 domain仅持续函数调用的持续时间，因此可以存在对该值的引用的长度。

要修复它，请不要尝试引用参考：

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() {}

当直接调用函数时，可变引用的寿命足够长，但在通过中间函数调用时，它的寿命不够长

1 个答案: