简体   繁体   English

锈蚀寿命误差预期具体寿命但发现约束寿命

[英]Rust lifetime error expected concrete lifetime but found bound lifetime

I am having an issue working with lifetime parameters for structs. 我在使用结构的生命周期参数时遇到问题。 I am not 100% sure how to describe the problem, but I created a trivial case that shows my compile time error. 我不是100%肯定如何描述问题,但我创建了一个显示我的编译时错误的简单案例。

struct Ref;

struct Container<'a> {
  r : &'a Ref
}

struct ContainerB<'a> {
  c : Container<'a>
}

trait ToC {
  fn to_c<'a>(&self, r : &'a Ref) -> Container<'a>;
}

impl<'a> ToC for ContainerB<'a> {
  fn to_c(&self, r : &'a Ref) -> Container<'a> {
    self.c
  }
}

The error I am getting with this is 我得到的错误是

test.rs:16:3: 18:4 error: method `to_c` has an incompatible type for trait: expected concrete lifetime, but found bound lifetime parameter 'a
test.rs:16   fn to_c(&self, r : &'a Ref) -> Container<'a> {
test.rs:17     self.c
test.rs:18   }
test.rs:16:48: 18:4 note: expected concrete lifetime is the lifetime 'a as defined on the block at 16:47
test.rs:16   fn to_c(&self, r : &'a Ref) -> Container<'a> {
test.rs:17     self.c
test.rs:18   }
error: aborting due to previous error

I have tried many variations and just can't get this thing to compile. 我已经尝试了很多变化,只是无法编译这个东西。 I found another post here ( How to fix: expected concrete lifetime, but found bound lifetime parameter ) but is appears to get around the problem instead of solving it. 我在这里找到了另一篇文章( 如何修复:预期的具体生命周期,但找到了绑定生命周期参数 )但似乎解决了问题而不是解决问题。 I can't really see why the problem even originates. 我真的不明白为什么问题甚至起源。 The &Ref is being passed along via moves so it should just work right? &Ref正在通过移动传递,所以它应该正常工作?

Any ideas? 有任何想法吗? Thanks for all the help. 谢谢你的帮助。

Let's compare the two definitions. 让我们比较两个定义。 First, the trait method: 一,特质方法:

fn to_c<'a>(&self, r: &'a Ref) -> Container<'a>;

And the implementation: 并实施:

fn to_c(&self, r: &'a Ref) -> Container<'a>;

See the difference? 看到不同? The latter doesn't have <'a> . 后者没有<'a> <'a> has been specified elsewhere; 其他地方已指明<'a> ; the fact that it has the same name does not matter: it is a different thing entirely. 它具有相同名称的事实并不重要:它完全是另一回事。

Functionally, your trait definition says that the returned container will have a reference inside it to something from r , but nothing from self . 从功能上讲,你的特征定义表明返回的容器里面会有一个来自r的引用,但是没有来自self It may use self inside the method, but it may not store any references to it in the returned value. 它可以在方法中使用self ,但它可能不会在返回的值中存储对它的任何引用。

Your method definition, however, is using a 'a that ties the lifetimes of r and the returned Container to self (that is, to the object itself, not the reference—the ρ₂ in &'ρ₁ T<'ρ₂> —it's a subtle but sometimes significant difference), whereas the trait definition had no such connection. 您的方法定义,然而,使用'a是关系的寿命r和返回的Containerself (即,对象本身,而不是基准的ρ₂在&'ρ₁ T<'ρ₂> -它是一个细微但有时显着的差异),而特质定义没有这种联系。

The two can be made to match by inserting the <'a> in the method definition in the implementation. 通过在实现中的方法定义中插入<'a> ,可以使两者匹配。 But bear in mind that that is shadowing the 'a from ContainerB<'a> ; 但请记住,这是影响'aContainerB<'a> ; it is not the same 'a ! 一样'a We're better to give it another name; 我们最好给它另一个名字; for convenience, I'll make the change the other way round, changing it on the impl instead of the method (either would do): 为方便起见,我将进行相反的更改,在impl而不是方法上更改它(要么会这样做):

impl<'b> ToC for ContainerB<'b> {
    fn to_c<'a>(&self, r: &'a Ref) -> Container<'a> {
        self.c
    }
}

But now of course you have a problem: the return value is of type Container<'b> (because that's what the field c in a ContainerB<'b> is), but your signature demands Container<'a> (something using a reference from r , not from self ). 但是现在你有一个问题:返回值的类型为Container<'b> (因为这就是ContainerB<'b>的字段c是什么),但是你的签名需要Container<'a> (使用的是r引用,而不是来自self )。

One way which would fix it is specifying the lifetime of &self as 'a in both the trait definition and the implementation; 解决这个问题的一种方法是将&self的生命周期指定为特征定义和实现中的'a ; in the implementation, this would then demand that 'b was greater than or equal to 'a (by virtue of the fact that you have successfully taken a reference with lifetime 'a to an object with lifetime 'b , and the object must outlive the reference) and so due to the subtyping ( 'a is a subtype of 'b ) Container<'b> would be safely coerced to Container<'a> . 在实现中,这将要求'b大于或等于'a (由于您已经成功地将具有生命周期'a的引用引用到具有生命周期'b的对象,并且该对象必须比由于子类型( 'a'b的子类型), Container<'b>将被安全地强制转换为Container<'a>

These sorts of lifetime matters are difficult to think about when you're not familiar with them; 当你不熟悉它们时,很难想到这些生活中的问题; but in time they become quite natural. 但随着时间的推移,它们变得很自然

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM