Ready or pending future depending on a condition

Question

I need to save a ready or pending future in a variable depending on a condition.

Would be nice if I could do this:

let f = futures::future::ready(true);

But the API provides two different functions, which have different return types, so, this does not work either:

let f = if true { futures::future::ready(()) } else { futures::future::pending::<()>() }

I understand that I can implement my own future for this, but I wonder if there is a way to make the if expression work?

This future you want should be always either ready or pending? — Aleksander Krauze, Oct 13 '22 at 18:12
Sure, there's [`futures::future::Either`](https://docs.rs/futures/latest/futures/future/enum.Either.html). — Sven Marnach, Oct 13 '22 at 18:16
@SvenMarnach, Thank you! That is exactly what I was searching for. — nicolai, Oct 13 '22 at 18:17

score 4 · Accepted Answer · answered Oct 13 '22 at 18:21

You can use the futures::future::Either type:

use futures::future::Either;
use std::future::{pending, ready, Future};

pub fn pending_or_ready(cond: bool) -> impl Future<Output = ()> {
    if cond {
        Either::Left(ready(()))
    } else {
        Either::Right(pending())
    }
}

(Playground)

Ready or pending future depending on a condition

1 Answers1