以下代码段给了我一个错误:
The following code segment gives me an error:
use std::rc::Rc; // Definition of Cat, Dog, and Animal (see the last code block) // ... type RcAnimal = Rc<Box<Animal>>; fn new_rc_animal<T>(animal: T) -> RcAnimal where T: Animal /* + 'static */ // works fine if uncommented { Rc::new(Box::new(animal) as Box<Animal>) } fn main() { let dog: RcAnimal = new_rc_animal(Dog); let cat: RcAnimal = new_rc_animal(Cat); let mut v: Vec<RcAnimal> = Vec::new(); v.push(cat.clone()); v.push(dog.clone()); for animal in v.iter() { println!("{}", (**animal).make_sound()); } } error[E0310]: the parameter type `T` may not live long enough --> src/main.rs:8:13 | 4 | fn new_rc_animal<T>(animal: T) -> RcAnimal | - help: consider adding an explicit lifetime bound `T: 'static`... ... 8 | Rc::new(Box::new(animal) as Box<Animal>) | ^^^^^^^^^^^^^^^^ | note: ...so that the type `T` will meet its required lifetime bounds --> src/main.rs:8:13 | 8 | Rc::new(Box::new(animal) as Box<Animal>) | ^^^^^^^^^^^^^^^^但这编译得很好:
use std::rc::Rc; // Definition of Cat, Dog, and Animal (see the last code block) // ... fn new_rc_animal<T>(animal: T) -> Rc<Box<T>> where T: Animal, { Rc::new(Box::new(animal)) } fn main() { let dog = new_rc_animal(Dog); let cat = new_rc_animal(Cat); }错误的原因是什么?唯一真正的区别似乎是运算符 as 的使用.类型怎么可能活得不够久?(游乐场)
What is the cause of the error? The only real difference seems to be the use of operator as. How can a type not live long enough? (playground)
// Definition of Cat, Dog, and Animal trait Animal { fn make_sound(&self) -> String; } struct Cat; impl Animal for Cat { fn make_sound(&self) -> String { "meow".to_string() } } struct Dog; impl Animal for Dog { fn make_sound(&self) -> String { "woof".to_string() } }附录
为了澄清,我有两个问题:
Just to clarify, I had two questions:
实际上有很多类型可以活得不够久":所有类型都有生命周期参数.
There are actually plenty of types that can "not live long enough": all the ones that have a lifetime parameter.
如果我要介绍这种类型:
If I were to introduce this type:
struct ShortLivedBee<'a>; impl<'a> Animal for ShortLivedBee<'a> {}ShortLivedBee 对任何生命周期都无效,但仅对 'a 有效.
ShortLivedBee is not valid for any lifetime, but only the ones that are valid for 'a as well.
所以在你的情况下
where T: Animal + 'static我可以输入您的函数的唯一 ShortLivedBee 是 ShortLivedBee.
the only ShortLivedBee I could feed into your function is ShortLivedBee<'static>.
造成这种情况的原因是,在创建 Box 时,您正在创建一个 trait 对象,该对象需要具有关联的生命周期.如果不指定,则默认为'static.所以你定义的类型实际上是:
What causes this is that when creating a Box<Animal>, you are creating a trait object, which need to have an associated lifetime. If you do not specify it, it defaults to 'static. So the type you defined is actually:
type RcAnimal = Rc<Box<Animal + 'static>>;这就是为什么您的函数要求将 'static 绑定添加到 T:不可能存储 ShortLivedBee 在 Box 除非 'a = 'static.
That's why your function require that a 'static bound is added to T: It is not possible to store a ShortLivedBee<'a> in a Box<Animal + 'static> unless 'a = 'static.
另一种方法是为您的 RcAnimal 添加生命周期注释,如下所示:
An other approach would be to add a lifetime annotation to your RcAnimal, like this:
type RcAnimal<'a> = Rc<Box<Animal + 'a>>;并更改您的函数以明确生命周期关系:
And change your function to explicit the lifetime relations:
fn new_rc_animal<'a, T>(animal: T) -> RcAnimal<'a> where T: Animal + 'a { Rc::new(Box::new(animal) as Box<Animal>) }更多推荐
参数类型可能不够长?
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