为什么我需要为不是该结构成员的结构的通用参数提供生命周期? [英] Why do I need to provide lifetimes for a struct's generic parameters which are not members of the struct?

问题描述

我正在努力弄清泛型和生命周期相互作用的方式.考虑:

I'm trying to get my head around the way that generics and lifetimes interact. Consider:

use std::ops::Add;

struct Gloop<'a, T: Add> {
    wumpus: &'a Wumpus<T>,
}

trait Wumpus<T: Add> {
    fn fleeb(&self, x: &T) -> bool;
}

struct Mimsy {
    jubjub: f64,
}

impl<T: Add> Wumpus<T> for Mimsy {
    fn fleeb(&self, x: &T) -> bool {
        return (x + x) > 0;
    }
}

fn main() {
    let a = Mimsy { jubjub: 1. };
    let b = Gloop::<i32> { wumpus: &a };
    println!("{}", b.fleeb(1));
}

哪个产量:

error[E0309]: the parameter type `T` may not live long enough
 --> src/main.rs:4:5
  |
3 | struct Gloop<'a, T: Add> {
  |                  -- help: consider adding an explicit lifetime bound `T: 'a`...
4 |     wumpus: &'a Wumpus<T>,
  |     ^^^^^^^^^^^^^^^^^^^^^
  |
note: ...so that the reference type `&'a Wumpus<T> + 'a` does not outlive the data it points at
 --> src/main.rs:4:5
  |
4 |     wumpus: &'a Wumpus<T>,
  |     ^^^^^^^^^^^^^^^^^^^^^

程序中的哪个对象可能寿命不长? Gloop或Mimsy(或任何Wumpus<T>)中都没有存储过T.

Which object in my program is the one which may not live long enough? Nowhere in either Gloop or Mimsy (or any Wumpus<T>) is a T ever stored.

推荐答案

注意:我回答了这个问题，但认为问题实际上是

Note: I answered this, but think the question is actually a duplicate of

参数类型'C'可能寿命不足"

如果其他人同意，可以删除此答案.

This answer can be deleted if others agree.

在Gloop或Mimsy(或任何Wumpus<T>)中都没有存储过T.

Nowhere in either Gloop or Mimsy (or any Wumpus<T>) is a T ever stored.

Rust不在乎您做什么，但在乎您可以做什么.这两种结构之间没有签名差异:

Rust doesn't care about what you do, but what you could do. There are no signature differences between these two structs:

struct Alpha<T> {
    a: T,
}

struct Beta<T> {
    b: fn(&T),
}

即，您不能说出两者之间的区别，而Beta可能会变成Alpha，而无需对消费者进行明显的外部更改.一旦有了通用类型，就需要处理所有可能.

Namely, you can't tell the difference between the two and Beta could become Alpha without noticeable external changes to a consumer. Once you have a generic type, you need to handle any possibility.

您可以添加T: 'static或添加生存期并添加T: 'a.

You can either add T: 'static or add a lifetime and add T: 'a.

另请参阅:

• 为什么终生需要生命
  • 为什么要存储参考& a T"需要绑定的`T:'a`?
  • 参数类型的寿命可能不够长?
  • Why the lifetime is needed at all
    • Why is the bound `T: 'a` required in order to store a reference `&'a T`?
    • Parameter type may not live long enough?
    • 为什么Rust中需要显式的生存期?
    • 如何编写特征绑定以添加两个泛型类型的引用?

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