What's the type of an async fn?

Question

I want to have a function that takes a function pointer of async fn.
What should the type of f be in fn run?

async fn foo() {}

fn run(f: /* ??? */) {}

According to the async/await RFC:

An async fn foo(args..) -> T is a function of the type fn(args..) -> impl Future<Output = T>.

However, if I write

fn run(f: fn() -> impl Future<()>)

I get the error message:

`impl Trait` not allowed outside of function and inherent method return types

There is an [RFC](https://github.com/rust-lang/rust/issues/50547) going on for the async/await — Akiner Alkan, Jan 18 '19 at 07:19
@hellow I updated my question and I think mine is quite different from the one you suggested. — xxks-kkk, Jan 18 '19 at 07:30
Okay, I can accept that, thanks for editing your question :) — hellow, Jan 18 '19 at 07:34
Maybe this is a duplicate? https://stackoverflow.com/questions/26577070/how-to-use-fn-traits-closures-in-signatures-in-rust — hellow, Jan 18 '19 at 07:43

score 3 · Answer 1 · edited Jan 20 '19 at 03:06

3

You have to introduce two type parameters to your function signature, one for the Fn and one for the Future, e.g.

#![feature(futures_api, async_await)]

async fn foo() {}

fn run<G: std::future::Future, F: FnOnce() -> G>(f: F) {
    f();
}

fn main() {
    bar(foo)
}

You can replace FnOnce with Fn or FnMut depending on your needs.

edited Jan 20 '19 at 03:06

Shepmaster

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answered Jan 18 '19 at 07:37

hellow

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score 2 · Answer 2 · edited Jan 20 '19 at 03:07

2

Using generic type parameter can help since an async fn returns an implementation of the Future trait.

#![feature(futures_api, async_await)]

use std::future::Future;

fn main() {
    run(foo);
}

async fn foo() {}

fn run<T: Future>(f: fn() -> T) {
    f();
}

edited Jan 20 '19 at 03:07

Shepmaster

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answered Jan 18 '19 at 07:38

Ömer Erden

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What's the type of an async fn?

2 Answers2