如何获得impl Trait使用适当的生存期来对其中包含另一个生存期的值进行可变引用?

问题描述

我有一个有生命的结构:

I have a struct with a lifetime:

struct HasLifetime<'a>( /* ... */ );

有一个特征 Foo 的实现:

impl<'a, 'b: 'a> Foo for &'a mut HasLifetime<'b> { }

我要实现以下功能:

fn bar_to_foo<'a, 'b: 'a>(bar: &'a mut Lifetime<'b>) -> impl Foo {
    bar
}

这不会编译，因为返回的 impl 仅对'a 有效.但是，指定 impl Foo +'a 会导致:

This won't compile because the returned impl is only valid for 'a. However, specifying impl Foo + 'a results in:

error[E0909]: hidden type for `impl Trait` captures lifetime that does not appear in bounds
 --> src/main.rs:7:60
  |
7 | fn bar_to_foo<'a, 'b: 'a>(bar: &'a mut HasLifetime<'b>) -> impl Trait + 'a {
  |                                                            ^^^^^^^^^^^^^^^
  |
note: hidden type `&'a mut HasLifetime<'b>` captures the lifetime 'b as defined on the function body at 7:1
 --> src/main.rs:7:1
  |
7 | fn bar_to_foo<'a, 'b: 'a>(bar: &'a mut HasLifetime<'b>) -> impl Trait + 'a {
  | ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

带有盒装特征对象的看似等效的函数将编译:

The seemingly equivalent function with a boxed trait object compiles:

fn bar_to_foo<'a, 'b: 'a>(bar: &'a mut Lifetime<'b>) -> Box<Foo + 'a> {
    Box::new(bar)
}

如何使用 impl Trait 定义 bar_to_foo ?

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推荐答案

您需要指出返回的值是建立在多个生存期上的.但是，您不能对 impl Trait 使用多个生存期限制，而尝试使用没有有用的错误消息.

You need to indicate that the returned value is built upon multiple lifetimes. However, you can't use multiple lifetime bounds with impl Trait, and attempting to do so doesn't have a useful error message.

有您可以使用的技巧，其中包括创建一个具有生存期参数的虚拟特征:

There's a trick you can use that involves creating a dummy trait that has a lifetime parameter:

trait Captures<'a> {}
impl<'a, T: ?Sized> Captures<'a> for T {}

fn bar_to_foo<'a, 'b: 'a>(bar: &'a mut HasLifetime<'b>) -> impl Trait + Captures<'b> + 'a {
    bar
}

非常感谢，此仅当隐藏"生存期不变时才会发生，这是因为引用是可变的.

Thankfully, this only occurs when the "hidden" lifetime is invariant, which occurs because the reference is mutable.

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