rust - FnOnce 的泛型，返回一个终生的 future

Question

我有一个结构需要一个回调，该回调返回一个输出具有生命周期的 future :

struct Foo;
struct Bar;

struct Baz<F>
where F: for<'a> FnOnce(&'a Foo) -> impl std::future::Future<Output=&'a Bar> // ?
{
    cb: F
}

这不能编译，自 impl 以来出现语法错误不能出现在特征边界中:

   Compiling playground v0.0.1 (/playground)
error[E0562]: `impl Trait` not allowed outside of function and inherent method return types
 --> src/lib.rs:6:37
  |
6 | where F: for<'a> FnOnce(&'a Foo) -> impl std::future::Future<Output=&'a Bar>,
  |                                     ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

error: aborting due to previous error

这也不能编译:

struct Foo;
struct Bar;

struct Baz<O, F>
where
    O: for<'a> std::future::Future<Output=&'a Bar>,
    F: for<'b> FnOnce(&'b Foo) -> O,
{
    cb: F
}

投诉 'a不被识别(生命周期 'a 和 'b 将无关):

   Compiling playground v0.0.1 (/playground)
error[E0582]: binding for associated type `Output` references lifetime `'a`, which does not appear in the trait input types
 --> src/lib.rs:7:36
  |
7 |     O: for<'a> std::future::Future<Output=&'a Bar>,
  |                                    ^^^^^^^^^^^^^^

error: aborting due to previous error

有没有办法指定这种关系？

Answer 1

这有点难看，但这是我能找到的最好方法(使用 struct Foo(Bar) 进行演示):
Playground

use std::future::Future;

struct Bar;
struct Foo(Bar);

trait FooClosure<'a> {
    type Fut: Future<Output = &'a Bar>;

    fn call(self, foo: &'a Foo) -> Self::Fut;
}

impl<'a, Fut: Future<Output = &'a Bar>, C: FnOnce(&'a Foo) -> Fut> FooClosure<'a> for C {
    type Fut = Fut;

    fn call(self, foo: &'a Foo) -> Fut {
        self(foo)
    }
}

struct Baz<F: for<'a> FooClosure<'a>>(F);

fn main() {
    let closure: Box<dyn for<'a> FnOnce(&'a Foo) -> futures::future::Ready<&'a Bar>> =
        Box::new(|foo| futures::future::ready(&foo.0));
    let baz = Baz(closure);
}

我无法让编译器正确推断类型，所以我不得不将闭包包装为盒装 trait 对象。从理论上讲，这不应该是必要的，可能有办法避免这种情况，但我无法弄清楚。
注意:我认为有计划让您的原始代码工作而无需这种黑魔法。