如何为Option< closure>指定有效期? [英] How to specify a lifetime for an Option<closure>?

问题描述

我试图将一个字段放在应该包含Option<closure>的结构上.

I'm trying to put a field on a struct that should hold an Option<closure>.

但是，Rust对我大吼大叫，我必须指定生命周期(不是我真的会抱怨).我正在尽力做到这一点，但是Rust对我的想法永远不满意.看看我的内联注释，了解我得到的编译错误.

However, Rust is yelling at me that I have to specify the lifetime (not that I would have really grokked that yet). I'm trying my best to do so but Rust is never happy with what I come up with. Take a look at my inline comments for the compile errors I got.

struct Floor{
    handler: Option<|| ->&str> //this gives: missing lifetime specifier 
    //handler: Option<||: 'a> // this gives: use of undeclared lifetime name `'a`
}

impl Floor {
    // I guess I need to specify life time here as well 
    // but I can't figure out for the life of me what's the correct syntax
    fn get(&mut self, handler: || -> &str){
        self.handler = Some(handler);
    }
}

推荐答案

这有点棘手.

作为一般经验法则，每当在数据结构中存储借来的引用(即&类型)时，都需要命名其生存期.在这种情况下，使用'a使您处在正确的轨道上，但是必须在当前作用域中引入'a.与引入类型变量的方式相同.因此，定义您的Floor结构:

As a general rule of thumb, whenever you're storing a borrowed reference (i.e., an & type) in a data structure, then you need to name its lifetime. In this case, you were on the right track by using a 'a, but that 'a has to be introduced in the current scope. It's done the same way you introduce type variables. So to define your Floor struct:

struct Floor<'a> {
    handler: Option<|| -> &'a str>
}

但是这里还有另一个问题.闭包本身也是带有生存期的引用，也必须命名.因此，这里有两个不同的生命！试试这个:

But there's another problem here. The closure itself is also a reference with a lifetime, which also must be named. So there are two different lifetimes at play here! Try this:

struct Floor<'cl, 'a> {
    handler: Option<||:'cl -> &'a str>
}

对于您的impl Floor，您 还需要将这些生存期引入范围:

For your impl Floor, you also need to introduce these lifetimes into scope:

impl<'cl, 'a> Floor<'cl, 'a> {
    fn get(&mut self, handler: ||:'cl -> &'a str){
        self.handler = Some(handler);
    }
}

从技术上讲，您可以将其缩减为一个生存期，并使用||:'a -> &'a str，但这意味着返回的&str始终具有与闭包本身相同的生存期，我认为这是一个错误的假设.

You could technically reduce this down to one lifetime and use ||:'a -> &'a str, but this implies that the &str returned always has the same lifetime as the closure itself, which I think is a bad assumption to make.

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