Is it possible to await a &dyn Future?

Question

I'm trying to asynchronously invoke a Future that is returned from a function. Is it possible?

use core::future::Future;

fn choose_your_adventure<'a>(i: usize) -> Box<&'a dyn Future<Output = ()>> {
    match i {
        0 => Box::new(&async {}),
        _ => Box::new(&async {})
    }
}

async fn does_not_work() -> () {
    let choice = choose_your_adventure(0);
    choice.await; // error[E0277]: `&dyn Future<Output = ()>` is not a future
}

If `&T` is a `Future`, just return `&T` as a `dyn Future`. Although normally you would be working with owned future types. — Ibraheem Ahmed, May 03 '21 at 18:21

score 5 · Accepted Answer · edited May 20 '21 at 21:12

No. Polling a future requires that it be mutable. An immutable reference cannot be mutated, by design.

In this case you don't need a Box<&dyn ...>. I'd write your code without the trait object:

async fn choose_your_adventure(i: usize) {
    match i {
        0 => (),
        _ => (),
    }
}

If you have your heart set on a trait object, then there's no need to have a boxed reference:

fn choose_your_adventure(i: usize) -> Box<dyn Future<Output = ()>> {
    match i {
        0 => Box::new(async {}),
        _ => Box::new(async {}),
    }
}

Although it may be better to use type aliases from the futures crate:

use futures::{future::BoxFuture, FutureExt}; // 0.3.14

fn choose_your_adventure(i: usize) -> BoxFuture<'static, ()> {
    match i {
        0 => async {}.boxed(),
        _ => async {}.boxed(),
    }
}

Notably, that resolves to Pin<Box<dyn Future<Output = T> + 'a + Send>>, which indicates that the future may be unpinned (if T: Unpin).

Is it possible to await a &dyn Future?

1 Answers1

Linked