具有从借到借的构造函数的特征无法推断使用的寿命

Trait with constructor that takes borrows to borrows cannot infer liftime on usage

我需要一个特征,它允许我构造一个借用对象的对象。在下面的示例中是 PaperBin。 https://play.rust-lang.org/?version=stable&mode=debug&edition=2018&gist=78fb3f88b71bc226614912001ceca65b

trait GarbageBin<'a,'b>{
    fn new(rubbish: &'b Paper<'a>) -> Self;
}
struct PaperBin<'a,'b> {
    rubbish: &'b Paper<'a>
}
struct Paper<'a> {
    matter: &'a [u8]
}
impl<'a,'b> GarbageBin<'a,'b> for PaperBin<'a,'b> {
    fn new(rubbish: &'b Paper<'a>) -> Self{
        Self {
            rubbish: rubbish
        }
    }
}
fn create_bin_with_rubbish<'a,'b, T>()
    where T: GarbageBin<'a,'b>
{
    let matter = &[1][..];
    let rubbish = Paper{matter};
    This gives an error:
    //let garbage_bin = T::new(&rubbish);
}

#[test]
fn run() {
    create_bin_with_rubbish::<PaperBin>();
}

我需要创建一般在函数调用时确定的 GarbageBins,如示例中所示。对于 Trait 的 new() 关联函数中的固定类型 Paper,此示例看起来可能是不必要的。这应该成为真实代码中的 Trait 对象。

如何实现garbage_bin泛型T的创建?

这是我收到的错误消息,如果我取消注释该行:

error[E0495]: cannot infer an appropriate lifetime for lifetime parameter `'a` due to conflicting requirements
  --> src\reference_to_reference\mod.rs:23:23
   |
23 |     let garbage_bin = T::new(&rubbish);
   |                       ^^^^^^
   |
note: first, the lifetime cannot outlive the lifetime `'a` as defined on the function body at 18:28...
  --> src\reference_to_reference\mod.rs:18:28
   |
18 | fn create_bin_with_rubbish<'a,'b, T>()
   |                            ^^
note: ...but the lifetime must also be valid for the lifetime `'b` as defined on the function body at 18:31...
  --> src\reference_to_reference\mod.rs:18:31
   |
18 | fn create_bin_with_rubbish<'a,'b, T>()
   |                               ^^
note: ...so that the types are compatible
  --> src\reference_to_reference\mod.rs:23:23
   |
23 |     let garbage_bin = T::new(&rubbish);
   |                       ^^^^^^
   = note: expected `GarbageBin<'_, '_>`
              found `GarbageBin<'a, 'b>`

您的直接问题是 T: GarbageBin<'a, 'b>,其中 'a'bcreate_bin_with_rubbish 的参数,因此必须长于对该函数的调用——但传递给的实际生命周期T::new 仅在函数内部,因此不满足这些界限。

不是使用生命周期 'a'b 来参数化 create_bin_with_rubbish,解决这个问题的一种方法是使用 HRTB(更高级别的特征边界):

fn create_bin_with_rubbish<T>()
where
    T: for<'a, 'b> GarbageBin<'a, 'b>,

但是,PaperBin 隐含地要求 'a: 'b,并且 HRTB 当前不支持此类界限。但这本身就表明第二个生命周期参数是不必要的:记住,一个生命周期参数只意味着“生命至少与……一样长”,同一参数的多个实例可以代表不同的具体生命周期(只要它们各自满足相同的界限)。所以让我们简化一下:

struct PaperBin<'a> {
    rubbish: &'a Paper<'a>,
}

最后,使用 HRTB 确实需要打破 GarbageBin 及其实现者 PaperBin 生命周期之间的直接关系 link——但这是有道理的,因为您正试图通过在不知道其生命周期参数的 PaperBin 类型中,然后在其上调用 GarbageBin 的构造函数方法以获得具有特定生命周期参数的 PaperBin。所以现在我们实现了 GarbageBin<'a> for PaperBin<'_>——但是我们仍然需要构造函数来 return 一个 PaperBin<'a> 的实例,为此我们添加了一个关联类型:

trait GarbageBin<'a> {
    type Constructed: GarbageBin<'a>;
    fn new(rubbish: &'a Paper<'a>) -> Self::Constructed;
}

struct Paper<'a> {
    matter: &'a [u8]
}

struct PaperBin<'a> {
    rubbish: &'a Paper<'a>
}

impl<'a> GarbageBin<'a> for PaperBin<'_> {
    type Constructed = PaperBin<'a>;
    fn new(rubbish: &'a Paper<'a>) -> PaperBin<'a> {
        PaperBin { rubbish }
    }
}

fn create_bin_with_rubbish<T>()
where
    T: for<'a> GarbageBin<'a>,
{
    let matter = &[1][..];
    let rubbish = Paper { matter };
    let garbage_bin = T::new(&rubbish);
}

fn main() {
    create_bin_with_rubbish::<PaperBin>();
}

See it on the Playground.

使用 GAT(通用关联类型),这是一个 currently 不稳定的功能,您可以将生命周期参数从特征推入关联类型,从而不需要 HRTB:

#![feature(generic_associated_types)]

trait GarbageBin {
    type Constructed<'a>: GarbageBin;
    fn new<'a>(rubbish: &'a Paper<'a>) -> Self::Constructed<'a>;
}

struct Paper<'a> {
    matter: &'a [u8]
}

struct PaperBin<'a> {
    rubbish: &'a Paper<'a>
}

impl GarbageBin for PaperBin<'_> {
    type Constructed<'a> = PaperBin<'a>;
    fn new<'a>(rubbish: &'a Paper<'a>) -> PaperBin<'a> {
        PaperBin { rubbish }
    }
}

fn create_bin_with_rubbish<T: GarbageBin>() {
    let matter = &[1][..];
    let rubbish = Paper { matter };
    let garbage_bin = T::new(&rubbish);
}

fn main() {
    create_bin_with_rubbish::<PaperBin>();
}

See it on the Playground